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Practice Problem in Physics for the JEE Main and Advanced

Abhay Kumar

Chapter 5

Work, Energy, Power and Circular Motion - all with Video Answers

Educators

VS

Section 1

Section A

01:13

Problem 1

$\Lambda$ circular race track is banked at $45^{\circ}$ and has a radius of $40 \mathrm{~m} . \Lambda \mathrm{t}$ what speed does a car have no tendency to slip? If the coefficient of friction between the wheels and the track is $\frac{1}{2}$, find the maximum speed at which the car can travel round the track without skidding.
Solution
(a) Banking angle is given by
$$
\tan \theta-\frac{v^{2}}{r g} \quad \therefore \quad v^{2}-\sqrt{g r \operatorname{lan} \theta}-\sqrt{400}-20 \mathrm{~m} / \mathrm{s}
$$
(b) Normal to plane, $\mathrm{N}-m g \cos 45^{\circ}+\frac{m v^{2}}{r} \cos 45^{\circ}-\frac{m}{\sqrt{2}}\left(g+\frac{v^{2}}{r}\right)$
$$
f_{x a t}-\mu N-\frac{1}{2} \frac{m}{\sqrt{2}}\left(g+\frac{v^{2}}{r}\right)
$$
Nlong the planc, friction $+m g \sin 45^{\circ}-\frac{m v^{2}}{r} \cos 45^{\circ}$
$$
\begin{aligned}
&\frac{m}{2 \sqrt{2}}\left(g+\frac{v^{2}}{r}\right)+\frac{m g}{\sqrt{2}}-\frac{m v^{2}}{\sqrt{2} r} \Rightarrow \frac{g}{2}+\frac{v^{2}}{2 r}+g-\frac{v^{2}}{r} \Rightarrow \frac{v^{2}}{2 r}-\frac{3 g}{2} \\
&\Rightarrow v^{2}-3 g r-3 \times 10 \times 40-1200 \quad v_{\mathrm{m} n}-\sqrt{1200} \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:24

Problem 2

A small ball is hung as shown on a string of length $L$.
(a) If $v_{e}>\sqrt{2 g L}$, find the angle $\theta\left(<90^{\circ}\right)$ with the upward vertical at which the string becomes slack.
(b) \Gammaind the value of $v_{0}$ if the particle passes through point of suspension.
Solution
(a) At the angle $\theta$, when the string becomes slack For circular motion, $\frac{m v^{2}}{L}-m g \cos \theta \quad \ldots(1)$
\Gammarom T.M.E. conservation law $\frac{1}{2} m v_{0}^{2}-\frac{1}{2} m v^{2} \mid m g L(1+\cos \theta) \quad \ldots(2)$
Solving Eqs. (1) and (2) gives $v_{n}-\sqrt{g L(2 \mid 3 \cos \theta)} \quad \Rightarrow \cos \theta-\frac{v_{0}^{3} 2 g L}{3 g L}$
(b) After the string become slack, the ball follows the path of projectile. For it to pass through point of suspension $L \sin \theta=(\cos \theta) t \quad(x-$ direction $)$
and, $-L \cos \theta=(v \sin \theta) t-\frac{1}{2} g t^{2} \quad(y$-direction $)$
Solving Eqs. (3) and (4) gives, $\tan \theta-\sqrt{2}$ $v_{0}-\sqrt{g L(2+3 \cos \theta)}$
$\Rightarrow v_{c}-\sqrt{g L(2+\sqrt{3)}}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:31

Problem 3

A hemispherical bowl of radius $R$ is rotating about its axis of symmetry which is kept vertical, A small ball kept in the bowl rotates with the bow! without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is $\alpha$. Find the angular speed at which the bowl is rotating.
Solution Lel $\omega$ be the angular speed of rotation of the bowl two forces are acting on the ball.
(a) normal reaction $N$
(b) weight $m g$ The ball is rolating in a circle of radius $r(=R \sin \alpha)$ with centre at $A$ at an angular specd $\omega$. Thus, $N \sin \alpha=m r \omega^{2}=m R \omega^{2} \sin \alpha \quad \ldots$ (i) $\quad$ and $\quad N \cos \alpha=m g \quad \ldots$ (ii)
Dividing Eqs. (i) by (ii), we gel $\frac{1}{\cos \alpha}-\frac{\omega^{2} R}{g}$
$\omega-\sqrt{\frac{g}{R \cos \alpha}}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:34

Problem 4

$\Lambda$ particle of mass $m$ is attached to onc end of a weightless and inextensible string of lengih $L$. The particle is on a smooth horizontal Lable. The string passes through a hole in the table and to its other end is atlached a small paricle of equal mass $m .$ Both the particles describe horizontal circular motion with angular velocitics $\omega_{1}$ and $\omega_{2}$. What is the ratio of surings on cither side of the hole? $\Lambda$ lso prove that $\left(\frac{1}{\omega_{1}^{2}}\right)$ ? $\left(\frac{1}{\omega_{2}^{2}}\right)<\frac{L}{g}$
Solution We will apply Newton's second law to both the masses. Equation for mass on table: $T-m \omega_{1}^{2} r_{1}$ $\ldots(1)$
Equations for hanging mass: $T \sin \theta-m \omega_{2}^{2} r_{2} \quad \ldots(2)$
$T \cos \theta=m g \quad \ldots(3)$
where $r_{2}=\left(L-r_{1}\right) \sin \theta$
Trom Eq. (2), $T \sin \theta=m \omega_{2}^{2}\left(L-r_{1}\right) \sin \theta \quad$ or $\quad T=m \omega_{2}^{2}\left(L-r_{1}\right)=m \omega_{1}^{2} r_{1}$
From Eq. (l) Therefore, $\frac{r_{1}}{L-r_{1}}=\frac{\omega_{2}^{2}}{\omega_{1}^{2}}$
$r_{1}=\frac{T}{m \omega_{1}^{2}} \quad$ and $\quad\left(L-r_{1}\right)=\frac{T}{m \omega_{2}^{2}}$
or $L=\frac{T}{m \omega_{1}^{2}}+\frac{T}{m o_{2}^{2}}=\frac{T}{m}\left(\frac{1}{\omega_{1}^{2}}+\frac{1}{\omega_{2}^{2}}\right)$
From Eqs. (3). $T=\frac{m g}{\cos \theta} \quad$ or $\quad L=\frac{g}{\cos \theta}\left(\frac{1}{\omega_{1}^{2}}+\frac{1}{\omega_{2}^{2}}\right)$ or $\left(\frac{1}{\omega_{1}^{2}}+\frac{1}{\omega_{2}^{2}}\right)=\frac{L \cos \theta}{g}$
Since $\cos \theta<1 \quad$ Therefore, $\left(\frac{1}{\omega_{1}^{2}}+\frac{1}{\omega_{2}^{2}}\right)<\frac{L}{g}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
07:18

Problem 5

A very small cube of mass $m$ is placed on the inside of a funnel rotating about a vertical axis at constant rate $a$ radian $/ \mathrm{sec}$. The wall of the funnel makes an angle $\theta$ with the horizontal, If the centre of the cube is at a distance $r$ from the axis of rotation of the funnel, what are the largest and the smallest values of $\omega$ for which the block neither slip down nor skid up?
Solution
If $v$ as well as $\omega$ is large, the block moves up and if $v$ is small, the block moves down.
If $N$ is the normal reaction,
$$
N-m g \cos \theta+\frac{m v^{2}}{r} \cos (90-\theta)-m g \cos \theta+\frac{m v^{2}}{r} \sin \theta
$$
$\therefore$
The frictional force $=\mu \mathrm{N}=\mu m\left(g \cos \theta+v^{2} \sin \theta\right)$
This acts down because the cube has a tendency to move up when $\omega$ is large. The force must be just sufficient to balance the force up the funnel surface. 'lhis is $\frac{m v^{2}}{r} \cos \theta-m g \sin \theta$
$\therefore \mu m\left(g \cos \theta \mid \frac{v^{2}}{r} \sin \theta\right)=m\left(\frac{v^{2}}{r} \cos \theta \quad g \sin \theta\right)$
$\mu g \cos \theta+g \sin \theta-\frac{v^{2}}{r}(\cos \theta-\mu \sin \theta)$
$\therefore \quad v^{2}-\frac{g r(\mu \cos \theta \mid \sin \theta)}{\cos \theta-\mu \sin \theta} \quad \Rightarrow \quad v-\sqrt{\frac{g r(\mu \cos \theta \mid \sin \theta}{\cos \theta-\mu \sin \theta}}$
$\therefore \quad \omega^{\max }-\frac{v}{r}-\sqrt{\frac{g(\mu \cos \theta \mid \sin \theta)}{r(\cos \theta-\mu \sin \theta)}}$
When $\omega$ is small, $\mu\left(m g \cos \theta \times \frac{m v^{2}}{r} \sin \theta\right)$ acts upwards.
$\therefore \mu\left(m g \cos \theta+\frac{m v^{2}}{r} \sin \theta\right)-m g \sin \theta+\frac{m v^{2}}{r} \cos \theta-0$or $g(\sin \theta \quad \mu \cos \theta)-\frac{v^{2}}{r}(\mu \sin \theta \mid \cos \theta)$
$\therefore \quad v^{2}-\frac{r g(\sin \theta-\mu \cos \theta)}{\mu \sin \theta+\cos \theta} \Rightarrow v-\sqrt{\frac{r g(\sin \theta-\mu \cos \theta)}{\mu \sin \theta+\cos \theta}}$
$\therefore \quad \omega^{\min }-\frac{v}{r}-\sqrt{\frac{g(\sin \theta-\mu \cos \theta)}{r(\mu \sin \theta+\cos \theta)}}$

VS
Vivek Singh
Numerade Educator
01:02

Problem 6

A wet open unbrella is held upright with its rim of radius $a$ at a height $h$ from the ground. It is rotated about the handle with uniform velocity $\omega$. Show that the drops of water which fly ofl from the rim will on reaching the ground be circle of radius $a\left(1+\frac{2 \omega^{2} h}{g}\right)^{1 / 2}$.

Find the diameter of the circle if $a=0.6125 \mathrm{~m}$ and $h=1.6 \mathrm{~m}$. The handle is rotated 49 times in 154 seconds.
Solution Velocity $=\omega a$ is along the tangent to the rim The drop falls to the ground in time $-\sqrt{\frac{2 h}{g}} \quad \therefore$ I Iorizontal distance $-\omega a \sqrt{\frac{2 h}{g}}$
$\therefore$ Radius of the circle $-\sqrt{a^{2} ? \omega^{2} a^{2} \frac{2 h}{g}}-a \sqrt{\left(1+\frac{2 \omega^{2} h}{g}\right)}$
IIcre, $\omega=\frac{2 \pi \times 49}{154}-2 ; a-0.6125 \mathrm{~m}$ and $h-1.6$
$\therefore$ Diameter of circle $-2 a \sqrt{1+\frac{2 \omega^{2} h}{g}}-1.225 \sqrt{1+\frac{2 \times 4 \times 1.6}{9.8}}-\frac{1.225}{7} \sqrt{49+64}-1.86 \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:08

Problem 7

In the figure, the block $B$ begins to rise when the frame rotates at $38.2 \mathrm{rpm}$. Find the coefficient of friction under block $\Lambda$ if that under block $B$ is $0.27 .$
Solution
If $T$ is the tension in the string, for $B$ to be in equilibrium, $200 g+\mu m \omega^{2} r=T$
$$
\therefore T-200 g+0.2 \times 200\left(\frac{2 \pi \times 38.2}{60}\right)^{2} \times 4-200 g+\frac{8}{45} \pi^{2}(38.2)^{2}
$$
Similarly, for $\Lambda$ to be in equilibrium,

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:20

Problem 8

A car goes on a horizontal circular road of radius $R$, the speed increasing at a constant rate $\frac{d v}{d t}-\alpha$. The friction coefficient betwecn the road and the tyre is $\mu$. Find the speed at which the car
will skid.
Solution
$$
f_{s} \cos \theta-\frac{m v^{2}}{R} \quad \text {...(1) } \quad f_{x} \sin \theta-m \frac{d v}{d t}-m \alpha
$$
for just to slide,
$$
f_{s}^{\mathrm{max}}-\mu m g \cos \theta-\frac{m v^{\max 2}}{R} \quad \ldots(3) \quad \mu m g \sin \theta-m \alpha
$$
$$
\begin{aligned}
&\therefore \tan \theta-\left(\frac{\alpha R}{v^{\max 2}}\right) \Rightarrow v^{\mathrm{mus}}-\sqrt{\frac{\alpha R}{\tan \theta}} \\
&\Lambda \operatorname{gain},\left(\frac{v^{\max }}{R}\right)^{2} \mid(m \alpha)^{2}-\mu^{2} m^{2} g^{2} \quad \therefore v^{\max }-\left[\left(\mu^{2} g^{2} \alpha^{2}\right) R^{2}\right]^{1 / 4}
\end{aligned}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:10

Problem 9

The given rod is rotating uniformly about one end. Find the value of tension at fixed cnd and at frec end Also find the variation of tension along its length.
Solution Every point of the given rotating rod is in circular motion and the tension in the rod provides the centripetal force. The value of tension is minimum at its free end (and it is zero) and is maximum at the fixed cnd.
$\therefore$ centripetal force on $\mathrm{d} m-d T$
$\Rightarrow-d T-(d m) \omega^{2} x \Rightarrow-d T-\left(\frac{m \omega^{2}}{L}\right) x d x \quad \Rightarrow \quad \int-d T-\left(\frac{m \omega^{2}}{L}\right) \int x d x$
$\Rightarrow \quad T-\frac{m \omega^{2}}{L}\left(\frac{x^{2}}{2}\right) ? c$
$\Lambda \mathrm{t} \quad x-L, T-0, \quad \therefore 0-\frac{m \omega^{2}}{L}\left(\frac{L^{2}}{2}\right) \mid c \quad \therefore \quad c-\frac{m \omega^{2}}{L}\left(\frac{L^{2}}{2}\right)$
$\therefore \quad T-\frac{m \omega^{2}}{2 L}\left(L^{2} \quad x^{2}\right)$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:53

Problem 10

$\Lambda$ metal ring of mass $m$ and radius $R$ is placed on a smooth horizontal table and is set rotating about its own axis in such a way that cach part of the ring moves with a specd $v$. Find the tension in the ring.Solntion Resolving the forces along the radius $\mathrm{CO}$, $T \cos \left(90^{\circ} \quad \frac{\Delta \theta}{2}\right) \wedge T \cos \left(90^{\circ}-\frac{\Delta \theta}{2}\right)-(\Delta m)\left(\frac{v^{2}}{R}\right)$
or $2 T \sin \frac{\Delta \theta}{2}-(\Delta m)\left(\frac{v^{2}}{R}\right)$
$\ldots(1)$
The length of the part $\Lambda C B$ is $R \Delta \theta \cdot \Lambda s$ the total mass of the ring is $m$, the mass of the part $A C B$ will be
$\Delta m-\frac{m}{2 \pi R} R \Delta \theta-\frac{m \Delta \theta}{2 \pi}$
Putting $\mathrm{Am}$ in Eq. (1), $2 T \sin \frac{\Delta \theta}{2}-\frac{m}{2 \pi} \wedge \theta\left(\frac{v^{2}}{R}\right)$ or $\quad T-\frac{m^{2}}{2 \pi R} \frac{\Delta \theta}{\sin \left(\frac{\Delta \theta}{2}\right)}$
As $\Lambda \theta$ is very small, $\frac{2}{\sin \left(\frac{\Delta \theta}{2}\right)}-1$ and $T-\frac{m v^{2}}{2 \pi R}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
07:26

Problem 11

Figure shows two blocks of mass $m_{1}=2 \mathrm{~kg}$ and $m_{2}=4 \mathrm{~kg}$ connected by an ideal spring of force constant $k=1000 \mathrm{~N} / \mathrm{m}$. The system is kept on a smooth inclined plane inclined at $30^{\circ}$ with the horizontal. A third block $m_{y}$, attached to $m_{2}$, hangs vertically. A force $F=15 \mathrm{~N}$ is applied on $m_{1}$ and the system is released from rest. Assume that the spring is initially unstretched and the system is released from rest. Find the
(a) maximum extension of the spring.
(b) acceleration of the system at this instant.
Solution Let the displacement of blocks $m_{1}$ and $m_{2}$ be $x_{1}$ and $x_{2}$ downward respectively and $x_{2}>x_{1}$. The extension in spring, $x=x_{2}-x_{1}$. Now we consider the force parallel to inclinc.
$m_{?} g \sin \theta+k x-F=m_{1} a_{1}$
and $m_{\xi} g-T=m_{3} a_{2}$
$\ldots(3)$From eqns. (2) and (3), we have $a_{2}-\frac{m_{2} g \sin \theta-k x+m_{3} g}{m_{2} \mid m_{3}}$
$\ldots(4)$
Acceleration of $m_{2}$ relative to $m_{1}$,
$$
\begin{aligned}
&a-a_{2} \quad a_{1}-\frac{m_{2} g \sin \theta \quad k x \mid m_{3} g}{m_{2}+m_{3}} \quad \frac{m_{1} g \sin \theta \perp k x \quad F}{m_{1}} \\
&-\frac{\left(4 \times 10 \times \frac{1}{2}-1000 x+1 \times 10\right)}{5}-\frac{\left(2 \times 10 \times \frac{1}{2}+1000 x-15\right)}{5}
\end{aligned}
$$
or $a-8.5-700 x$ or $\quad v \frac{d y}{d x}-8.5-700 x$
At the maximum values of $x, v=0$
So we have $\int_{0}^{0} v d v-\int_{0}^{x_{24}}(8.5700 x) d x \quad$ or $\quad x_{\text {nux }}-2.4 \mathrm{~cm}$
Subsitituting the value of $x_{\operatorname{tm} x}$ in eqn. (5), we get
$$
a-8.5 \mathrm{~m} / \mathrm{s}^{2}
$$
Hence the acecleration ol the system is up the incline.

VS
Vivek Singh
Numerade Educator
01:40

Problem 12

$\Lambda$ string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of $2 \mathrm{~m}$ from the wall, has a point mass $M=2 \mathrm{~kg}$ attached to it at a distance of $1 \mathrm{~m}$ from the wall. $\Lambda$ mass $m=0.5 \mathrm{~kg}$ attached at the frec end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pullcy. What will be the speed with which the mass $M$ will hit the wall when the mass $m$ is relcased?
Solution When mass $m$ is released from rest, the heavier mass $M$ loses potential energy and it strikes the wall with velocity $v$ and the string remains stretched. It increases the kinetic energy of mass $M$ and increases the kinetic and potential energies of mass $m$, Decrease in potential energy of mass $M$ $=M_{g} \times 1=2.0 \times g \times 1=2 g \quad$ where $g=9.8 \mathrm{~ms}^{-2}$
Increase in kinetic energy of mass $M-\frac{1}{2} M v^{2}-\frac{1}{2} \times 2.0 \times v^{2}-v^{2}$
Increase in potential cnergy of mass $m-m g\left(\sqrt{5} \quad\right.$ 1) $-\frac{g}{2}(\sqrt{5} \quad 1)$
$\{\because$ Distance through which mass $m$ is raised $-A B+B C-A C-1+\sqrt{5}-2-(\sqrt{5}-1) \mathrm{m}\}$
Increase in kinctic energy of mass $m-\frac{1}{2} m v^{2}-\frac{1}{2} \times 0.5 \times v^{2}-\frac{v^{2}}{4}$
\Lambdas the total energy must be conserved, the decrease in the potential energy of the mass $M$ cquals the increase in the potential and kinetic energies of both the masses.$\therefore \quad M g \times 1-\frac{1}{2} M v^{2} \mid m g\left(\sqrt{5} \quad\right.$ 1) $1 \frac{1}{2} m v^{2}$
$\therefore 2 g-v^{2}+\frac{1}{2} g(\sqrt{5}-1)+\frac{1}{4} v^{2} \quad \Rightarrow \quad g\left[2-\frac{\sqrt{5}}{2}+\frac{1}{2}\right]-\frac{5}{4} v^{2}-g(3.6)$
or $v^{2}-3.6 \times \frac{9.8 \times 4}{5} \quad \therefore \quad v-5.31 \mathrm{~m} / \mathrm{s}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:05

Problem 13

In the arrangement shown, the masses $m$ of the bar and $M$ of the wedge, as well as the wedge angle $\alpha$, arc known. The masses of the pullcy and the thread are negligible. Find the acceleration of the wedge $M$. The friction is absent cverywhere.
Solution Let $M$ move to the right when $m$ moves down. If $M$ moves to the right by $x, m$ moves down also by $x$. So, its P.E. decreases by $m g x \sin \alpha$. The increase of $\mathrm{K} . \mathrm{E}$. of the system is
$$
\left.\frac{1}{2} M v^{2}+\frac{1}{2} m \mid v^{2}+v^{2}+2 v, v \cos (180-\alpha)\right]
$$
since $m$ has a velocity $v$ down the plane and $v$ to the right along with wedge, both inclined at $(180-\alpha)$. $\therefore m g x \sin \alpha-\frac{1}{2} v^{2}\{M$ ? $2 m(1 \quad \cos \alpha)\}$
$\therefore \quad v^{2}=\frac{2 m g x \sin \alpha}{M+2 m(1 \quad \cos \alpha)}$
But $v^{2}-2 a x$ (where $a$ is acceleration of wedge)
$\therefore \quad a-\frac{m g \sin \alpha}{M+2 m(1-\cos \alpha)}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:30

Problem 14

Find the acccleration of the block of mass $M$ assuming friction to be absent.
Solution When $m^{\prime}$ moves downward, the block $M$ moves to the right. If $M$ moves to the right by $x, m$ moves $2 x$ and $m^{\prime}$ moves $3 x$ down the plane.
Loss of P.E. $-\left(m^{\prime} g\right) 3 x \sin \alpha$
gain in K.E. $-\frac{1}{2} M v^{2}+\frac{1}{2} m(2 v)^{2}+\frac{1}{2} m^{\prime}\left[(3 v)^{2}+v^{2}+2 \times 3 v \cdot v \cos (180-\alpha) \mid\right.$
$-\frac{1}{2} M v^{2} \mid 2 m v^{2}, \frac{m^{\prime}}{2} v^{2}[10 \quad 6 \cos \alpha]-\frac{y^{2}}{2}\left[M|4 m| m^{\prime}(10 \quad 6 \cos \alpha)\right]$
$\therefore \quad m^{\prime} g \cdot 3 x \sin \alpha-\frac{v^{2}}{2}\left[M|4 m| m^{\prime}(10 \quad 6 \cos \alpha)\right]$
$\therefore \quad y^{2}-\frac{2 m^{\prime} g \cdot 3 x \sin \alpha}{M+4 m+m^{\prime}(10-6 \cos \alpha)} \quad$ But $v^{2}-2 a x$, where $a$ is acceleration of $M$.
$\therefore \quad a-\frac{3 m^{\prime} g \sin \alpha}{M|4 m| m^{\prime}(10 \quad 6 \cos (x)}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:02

Problem 15

Find the acceleration of $M$, assuming all surfaces of contact are smooth.
Solution When $M$ moves to the right by a distance $x, m$ moves down by $2 x$. Loss of P.E. $=m g .2 x$ Gain in K.E. $-\frac{1}{2} M v^{2}+\frac{1}{2} m^{\prime} v^{2}+\frac{1}{2} m\left[v^{2}+(2 v)^{2}-\frac{v^{2}}{2}\left|M+m^{\prime}+5 m\right|\right.$
$\therefore \quad m g 2 x-\frac{v^{2}}{2}\left[M+m^{\prime} \mid 5 m\right]$
$\therefore \quad v^{2}-\frac{4 m g x}{M+m^{\prime}+5 m}=2 a x$, where $a$ is accclcration of $M$. $\therefore \quad a-\frac{2 m g}{M+m^{\prime}+5 m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:51

Problem 16

Find the accelerations of $A, B, C$, all having the same mass.
Solution When $C$ moves down $x$, the length of the string attached to $C$ has its part increased by $x$. This is shared by all the other parts of the string. So, cach of the masses $\Lambda$ and $B$ moves up by $x / 4$.
$\therefore$ Loss in P.E. $-\quad m g x+m g \cdot \frac{x}{4}, \frac{m g x}{4}-\frac{m g x}{2}$. $A \mid m$
Gain in K.E. $-\frac{1}{2} m v^{2}$ ? $\frac{1}{2} m\left(\frac{v}{4}\right)^{2}$ ? $\frac{1}{2} m\left(\frac{y}{4}\right)^{2}-\frac{1}{2} m v^{2}\left[11 \frac{1}{16}, \frac{1}{16}\right]-\frac{9}{16} m v^{2}$
$\therefore \frac{9}{16} m v^{2}-m g \frac{x}{2} \quad$ or $\quad v^{2}-\frac{8}{9} g x-2 a x \quad$ where $a$ is acceleration.
$\therefore$ accelcration of $C, a-\frac{4 g}{9} ;$ acceleration of $\Lambda$ or $B, a^{\prime}-\frac{g}{9}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
05:39

Problem 17

Two sliders $\Lambda$ and $B$ connected by a light rigid rod $10 \mathrm{~m}$ long, move in two frictionless shafts as shown in the figure. If $B$ starts from rest, determine the velocity of $B$ when $x=6 \mathrm{~m}$. \Lambdassume $m_{A}=m_{s}=200 \mathrm{~kg}$ and $m_{c}=100 \mathrm{~kg}$
Solution
$x^{2}+y^{2}-L^{2} \quad 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}-0$
$\frac{x d x}{d t}-\frac{y d y}{d t} ; \quad \frac{d x}{d t}-\frac{y}{x} \frac{d y}{d t}$
$\therefore$ Velocity of $B=-\left(\frac{y}{x}\right) \times$ velocity of $\Lambda . \Lambda$ s when $x=6, y=8$,
$\therefore$ Velocity of $B=\frac{8}{6} v_{4}=\frac{4}{3} v_{A}$$\Lambda$ pplying the law of conservation of energy, $\quad$ P.E. lost $=m_{A} \cdot g \cdot 2+m_{c} \cdot g \cdot 6$ Since when $B$ moves $6 \mathrm{~m}$ from $O, A$ moves down $2 \mathrm{~m}$ from its original position and $C$ moves down $6 \mathrm{~m}$. $\begin{aligned} \therefore & \text { P.E. lost }=200 \times 2 g+100 \times 6 g=1000 \mathrm{~g} \\ & \text { K.E. gaincd }=\frac{1}{2} m_{A} v_{A}^{2} ? \frac{1}{2} m_{n} v_{n}^{2} \text { ? } \frac{1}{2} m_{C} v_{C}^{2} \\ &-\frac{1}{2} \times 200 \times \frac{9}{16} v_{R}^{2}+\frac{1}{2} \times 200 \times v_{8}^{2}+\frac{1}{2} \times 100 \times v_{8}^{2} \\ &-\frac{1}{2} \times 200 \times v_{B}^{2}\left\{\frac{9}{16}+1+\frac{1}{2}\right\}-100\left\{\frac{33}{16}\right\} v_{B}^{2} \\ \therefore & 100 \times \frac{33}{16} v_{n}^{2}-1000 \times 9.8 \quad \therefore \quad v_{n}^{2}-\frac{98 \times 16}{33} \\ \therefore & v_{n}-7 \times 4 \sqrt{\frac{2}{33}}-6.9 \mathrm{~m} / \mathrm{sec} \end{aligned}$

VS
Vivek Singh
Numerade Educator
01:44

Problem 18

Two bodies $A, B$ weighing $400,300 \mathrm{~kg}$, respectively are connected by a rigid bar of negligible weight and move along the smooth surfaces shown 1
in the figure. I' they start from rest at the given position, determine the acceleration of $B$ at this instant.
$$
4 \textrm{ } L=
$$
Solution
$$
\frac{x d x}{d t}--\frac{y d y}{d t} ; \frac{x d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}--\left[\frac{y d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}\right]
$$
$\frac{4}{0}+\ldots$
when they start, $\frac{d x}{d t}-\frac{d y}{d t}-0$
$\therefore \quad \frac{x d^{2} x}{d t^{2}}-\frac{y d^{2} y}{d t^{2}}$, numcrically. $\frac{d^{2} x}{d t^{2}}-\frac{y}{x} \frac{d^{2} y}{d t^{2}}$
If $T$ is the tension along the bar, $300 g-T \cos \theta-300 a_{1} ; \quad T \sin \theta=400 a_{2}-400 \frac{y}{x} a_{1}-\frac{1600}{3} a_{1}\left(\right.$ but $\left.\frac{y}{x}-\frac{4}{3}\right)$
$300 g-\frac{4}{5} T=300 a_{1} ; \quad \frac{3 T}{5}-\frac{1600}{3} a_{1}$ or dividing, $\frac{300 g \quad 4 T / 5}{3 T / 5}-\frac{900}{1600}$
or $300 g-\frac{4}{5} T-\frac{3 T}{5} \times \frac{9}{16} 300 g-T\left\{\frac{4}{5}+\frac{27}{80}\right\}-T\left\{\frac{91}{80}\right\}$
$\therefore \quad \mathrm{T}-300 \times \frac{80}{91} \quad \therefore \quad 300 g \quad \frac{4}{5} T-300 a_{1}$
gives $300 g \quad \frac{4}{5} \times 300 g \times \frac{80}{91}-300 g\left\{1 \quad \frac{64}{91}\right\}-300 g \times \frac{27}{91}-300 a_{1}$
$\therefore \quad a_{1}-\frac{27 g}{91}-2.91 \mathrm{~m} / \mathrm{sec}^{2}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:21

Problem 19

A small disc of mass $m$ slides down a smooth hill of height $h$ without any initial velocity and lands upon a plank of mass $M$ lying along the horizontal plane at the base of the hill. Due to friction between the disc and the plank the disc slows down till the two move together as a single piece with a certain common velocity. Find the work done by the force of friction in the process.
Solution Since mass $m$ falls through hcight $h$ just before landing on the plank of mass $M$, so velocity of at this moment is $v-\sqrt{2 g h}$. Now, the frictional force betwcen $M$ and $m$ is the internal force the system so momentum of system remains conserved. I lence, $m v-(M \mid m) v^{\prime}$ or $v^{\prime}-\frac{m v}{M+m}$ or $v^{\prime}-\frac{m \sqrt{2 g h}}{M+m}$
So, initial kinetic cnergy of the system $K_{i}-\frac{1}{2} m v^{2} \quad$ or $\quad K_{i}-m g h$
$\ldots(2)$
and linal kinctic cnergy of the system $K_{f}-\frac{1}{2}(M+m) v^{\prime 2}-\frac{1}{2}(M+m) \frac{m^{2} \times 2 g h}{(M+m)^{2}} \quad \Rightarrow K_{f}-\frac{m^{2} g h}{M+m} \quad \ldots(3)$
So, change of cnergy $\Delta K-K_{f} K_{t} \Rightarrow \Delta K-\frac{m M g h}{M \mid m}$
Hence, work-done by friction $W-\Delta K \quad \therefore W--\frac{m M g h}{M \mathrm{I} m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:56

Problem 20

Two bars of masses $m_{1}$ and $m_{2}$ connccted by a non-deformed light spring rest on a horizontal plane. 'Ihe coefficient of friction between the surface and the bars is equal to $\mu$. What minimum constant force has to be applied in the horizontal direction to the bar of mass $m_{1}$ in order to shift the other bar? $m_{2}=00000000^{2} m_{1} \rightarrow F$
Solution Let the motion of $m_{2}$ start when the spring is slowly (without producing velocity) stretched by $x$. By the work-cnergy theorem Work done by the various forces $=$ change in kinctic energy
$\therefore F \cdot x-\left(\mu m_{1} g\right) x-\frac{1}{2} k x^{2}-0$
where $k$ is the force constant of the spring. But $k x=\mu m_{2} g$
(for just shifting $m_{2}$ )
$\therefore \quad F \cdot x \quad \mu m_{1} g x \quad \frac{1}{2}\left(\frac{\mu m_{2} g}{x}\right) x^{2}-0$
or $\quad F-\mu m_{1} g+\frac{1}{2} \mu m_{2} g-\mu\left(m_{1}+\frac{m_{2}}{2}\right) g$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:34

Problem 21

A $2 \mathrm{~kg}$ mass $M$ slides from rest at $A$ along the frictionless rod bent into elliptical shape. The spring with spring constant $k=17.5$ $\mathrm{N} / \mathrm{m}$ has an unstretched length of $450 \mathrm{~mm}$. Determine the speed of $M$ at $B$.
Solution
At $A$, the energy $-\frac{1}{2} k x^{2}+m g h$ $=\frac{1}{2} \times 17.5 \times\left[\frac{60 \quad 45}{100}\right]^{2}$ ? $2 \times 9.8 \times \frac{60}{100}-\frac{1}{2} \times \frac{35}{2} \times\left(\frac{15}{100}\right)^{2}$ ? $19.6 \times \frac{3}{5}-\frac{63}{320}$ ? $11.76$
At $B$, energy $-\frac{1}{2} k x^{2}+\frac{1}{2} m v^{2}-\frac{1}{2} 17.5 \frac{(90-45)^{2}}{100 \times 100}+\frac{1}{2} m v^{2}$
$=\frac{17.5}{2} \times \frac{45}{100} \times \frac{45}{100}+\frac{1}{2} \times 2 \times v^{2}=\frac{567}{2 \times 160}+v^{2}-\frac{63}{320}+11.76-\frac{567}{320}+v^{2}$
$0.1969111 .76 \frac{354375}{2}-v^{2}$
$\therefore \quad v^{2}-10.18 \quad \therefore \quad v-3.19 \mathrm{~m} / \mathrm{s}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:00

Problem 22

A particle of mass $1 \mathrm{~g}$ executes an oscillatory motion on a concave surface of spherical dish of radius $2 \mathrm{~m}$ placed on
a horizontal plane. If the motion of the particle begins from a point on the dish at the height of $1 \mathrm{~cm}$ from the horizontal planc and coclficient of friction is $0.01$, find the total distance covered by the particle before it comes to rest, Solution Since the particle starts from rest from a height $h$,
P.E. lost $-m g h-\frac{1}{1000} \times 9.8 \times \frac{1}{100}-$ work done against friction
$$
\begin{aligned}
&-\mu m g S-0.01 \times m \times 9.8 S-0.01 \times \frac{1}{1000} \times 9.8 S \\
S=& 1 \mathrm{~m}
\end{aligned}
$$
llere $r=2 \mathrm{~m}$, So $\theta$ will be very small.
$\therefore \quad$ Frictional forec $\mu m g \cos \theta=\mu m g$
and gravitational force down the surface $m g \sin \theta$ will be very small.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:31

Problem 23

Two plates of masses $m_{1}$ and $m_{2}$ are connected by a spring of force constant $k$. What force should be applied to the upper plate for it to raise the lower one after the force is removed? Disregard the mass of the spring. Solution $F$ be the force required. Let $x_{1}$ be the compression produced in the spring. Then for cquilibrium
$$
\Gamma+m_{1} g=k x_{1}
$$
When released, let the spring extend by $x_{2}$. The lower mass will be lifted if $k x_{2} \geq m_{2} g$
Considering conservation of energy between the initial and final positions of the spring
$$
\frac{1}{2} k x_{1}^{2}+m_{1} g\left(l_{n}-x_{1}\right)-\frac{1}{2} k x_{2}^{2}+m_{1} g\left(l_{0}+x_{2}\right)
$$or $\frac{1}{2} k x_{1}^{2}-\frac{1}{2} k x_{2}^{2} \mid m_{1} g\left(x_{1} \mid x_{2}\right) \quad$ or $\quad k^{2} x_{1}^{2}=k^{2} x_{2}^{2}+\left(2 m_{1} g x_{1}+2 m_{1} g x_{2}\right) k$
or $\left(k x_{1}-m_{1} g\right)^{2}=\left(k x_{2}+m_{1} g\right)^{2}$
or $k x_{1}-m_{1} g=k x_{2}+m_{1} g$
or $\quad F+m_{1} g-m_{1} g=k x_{2}+m_{1} g$
or $\quad F=k x_{2}+m_{1} g$
or $k x_{2}=F-m_{1} g$
For lifting the lower mass $k x_{2} \geq m_{2} g \quad$ or
$\Gamma-m_{1} g \geq m_{2} g$ or $F \geq m_{1} g+m_{2} g$
\Gammaor just lifting $\quad F=m_{1} g+m_{2} g$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:48

Problem 24

Two masses $M$ and $m$ connected by a light spring of force constant $k$ fall in a $m$ vertical plane with the spring unstretched until $M$ strikes a completely inelastic horizontal table. Find the height $h$ through which $M$ should fall if it is to rise from the table some time after hitting it. Solution Let $l_{v}$, be the length of the spring and $h$ be the height from which the mass $M$ falls. When $M$ strikes the ground, its velocity becomes zero as its impact with the ground is completely inelastic. The other mass $m$ continues to move downward compressing the spring. Let it come to rest after compressing spring through $x$, From the principle of conservation of energy $m g l_{0}\left|\frac{1}{2} m v^{2}-m g\left(l_{a} \quad x\right)\right| \frac{1}{2} k x^{2}$
But $\frac{1}{2} m v^{2}=m g h$
$\therefore m g l_{\infty}+m g h=m g\left(l_{0}-x\right)+\frac{1}{2} k x^{2}$
or $\frac{1}{2} k x^{2}-m g x-m g h=0 \quad$ or $\quad k x^{2}-2 m g x-2 m g h=6$
or $x=\frac{2 m g \pm \sqrt{4 m^{2} g^{2}+8 m g h k}}{2 k}$
or $k x=m g \pm \sqrt{m^{2} g^{2}+2 m g h k}$
Since $k x$ is positive, $k x=m g+\sqrt{m^{2} g^{2}+2 m g h k}$ Let the mass $m$ go up $x$ ' above the dotted line Then $m g\left(l_{n} \mid x^{\prime}\right)\left|\frac{1}{2} k x^{2}-m g\left(l_{\rho} \quad x^{\prime}\right)\right| \frac{1}{2} k x^{2}$
$x \quad x^{\prime}-\frac{2 m g}{k}$ or $k x \quad k x^{\prime}-2 m g$
$\begin{array}{llll}\text { or } & k x^{\prime}-m g & \sqrt{m^{2} g^{2} \mid 2 m g h k} & 2 m g-\sqrt{m^{2} g^{2} \mid 2 m g h k} & m g\end{array}$
The mass $M$ will be lifted if $k x^{\prime}>M g$
or $\sqrt{m^{2} g^{2}+2 m g h k}-m g \geq M g \quad$ or $\quad m^{2} g^{2}+2 m g h k \geq(M+m)^{2} g^{2}$
or $2 m g h k \geq M(M+2 m) g^{2} \quad$ or $\quad h \geq \frac{M g}{k}\left(1+\frac{M}{2 m}\right)$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:08

Problem 25

$\Lambda$ block weighing $1 \mathrm{~kg}$ is released from rest at point $\Lambda$ on a track which is one quadrant of a circle of radius $1.2 \mathrm{~m}$. It slides down the track and reaches point $B$ with a velocity of $3.6 \mathrm{~ms}^{-1}$. From the point $B$ it slides on a level surface a distance of $2.7 \mathrm{~m}$ to point $C$, where it comes to rest. (a) What was the coefficient of friction on the horizontal surface? (b) I low much work was done against friction as the body slide down the circular track from $A$ to $B$ ?
Solution
(b) Work done by gravitational pull $=m g h=1 \times g \times 1.2$, work done by frictional force $=W$ (say) Net work done $=1.2 g+W \quad$ By the work-cnergy theorem $1.2 g \ W-\frac{1}{2} \times 1 \times 3.6^{2} \quad 0 \quad$ or $\quad W-\frac{1}{2} \times 3.6^{2}-6.48 \quad 11.76-5.28 \mathrm{~J}$
(a) Work done by frictional foree $=\mu \times(1 \times g) \times 2.7$ Work done by the gravitational pull $=0$ By the work-cnergy theorem $-\mu \times g+2.7-0-\frac{1}{2} \times 1 \times 3.62 \quad \Rightarrow \quad \mu-0.24$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:38

Problem 26

A small body of mass $m=0.1 \mathrm{Kg}$ moves in a reference frame, rotating about a stationary axis with a constant angular velocity $\omega=5.0 \mathrm{rad} / \mathrm{s}$. What work does the centrifugal force perform during the transfer of the body along an arbitrary path from point 1 to 2 which are located at distances $r_{1}=30 \mathrm{~cm}$ and $r_{2}=50 \mathrm{~cm}$ from the rotation axes?
Solution Centrifugal force $=m \omega^{2} r$ Work donc to displace it by a small distance $d r=m \omega^{2} r d r$
$\therefore$ The tolal work done $-\int_{r}^{3} m \omega^{2} r d r-\left[\frac{m \omega^{2} r^{2}}{2}\right]_{i}^{5}-\frac{m \omega^{2}}{2}\left[r_{2}^{2} \quad r_{1}^{2}\right]$
$$
\begin{aligned}
&-0.1 \times \frac{25}{2}\left[\left(\frac{50}{100}\right)^{2}-\left(\frac{30}{100}\right)^{2}\right]-\frac{2.5}{2}[0.25-0.09]-\frac{2.5}{2} \times 0.16 \\
&-2.5 \times 0.08-0.2 \mathrm{~J}
\end{aligned}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:59

Problem 27

A cyclist ridos along a circular path in a horizontal plane whore the coellicicnt of friction varics from the centre $O$ of the circular path as $\mu-\mu_{\circ}\left(1-\frac{r}{R}\right)$ where $R$ is the maximum distance up to which the road is rough. Find the radius of the circular path along which the cyclist can ride with maximum velocity. Find that maximum velocity.Solution When cyclist moves on a circular path, the frictional force supplies the centripetal force.
So, $\frac{m v^{2}}{r}-\mu m g-\mu_{\circ}\left(1-\frac{r}{R}\right) m g ;$
$v^{2}-\mu_{0}\left(r \quad \frac{r^{2}}{R}\right) g$
For maximum $\frac{d}{d r}-0$
So, $2 v \frac{d v}{d r}-\mu_{0} g\left(1 \quad \frac{2 r}{R}\right)-0$
or $R-2 r$
$\therefore \quad r-\left(\frac{R}{2}\right) \quad \therefore \quad v_{\max }-\mu_{0}\left(\frac{R}{2}-\frac{R^{2}}{4 R}\right) g-\frac{\mu_{0} R}{4} g-\frac{1}{2} \sqrt{\mu_{0} R g}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:43

Problem 28

$75 \%$ of the power of a car is used in accelerating it and the rest in overcoming friction. What is the coellicient ol friction i $\left[\right.$ the acceleration of the car is $2.45 \mathrm{~m} / \mathrm{sec}^{2} ?$
Solution Power $=$ Force $\times$ Velocity $=m a v$ (where $a$ is the acceleration) $=-75$ and $\mu m g v=-25$
$\therefore \quad \frac{\mu g}{a}-\frac{25}{75}-\frac{1}{3}=\frac{\mu \times 9.8}{2.45}-4 \mu$
$\therefore \mu-\frac{1}{12}-0.83$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:16

Problem 29

In figures (a) and (b), $\Lambda C, D G$ and $G F$ are fixed inclined planes $B C=E F=x \& \Lambda B=D E=y$. $\Lambda$ small block is released from rest from the point $\Lambda$. It slides down $A C$ and reaches $C$ with a velocity $v_{c}$. The same block is relcased from rest from point $D$. It slides down $D G F$ and reaches point $F$ with a speed $v_{k}$. The coefficient of kinetic friction between the block and both the surfaces $\Lambda C$ and $D G F$ is $\mu$. Calculate $v_{c}$ and $v_{c-}$
(a)
(b)
Solution For motion on $\Lambda C$, T.M.E. $_{c}-$ (T.M.E.) $_{A}=$ Work done against friction $(B C$ as R.L.)
$\therefore m g y-\mu m g \cos \alpha . \Lambda C$ ? $\frac{1}{2} m v_{C}^{2}$
or $g y-\mu g \Lambda C \cos \alpha+\frac{1}{2} v_{c}^{2}-\mu g x+\frac{1}{2} v_{c}^{2} \quad\left(\sin \alpha \frac{x}{\Lambda C}-\cos \alpha\right)$
$\therefore \quad g y-\mu g x-\frac{1}{2} v_{c}^{2}-g(y-\mu x)$
$\therefore \quad v_{c}-\sqrt{2 g(y-\mu x)}$
For motion on DGF, (T.M.L.) $_{F}-\left(\right.$ T.M.E. $_{D}=$ work donc against friction
- work done against friction on $D G$ and $G F$ ? $\frac{1}{2} m v_{F}^{2}$ $-\mu m g \cos \alpha_{1} \cdot D G\left|\mu m g \cos \alpha_{1} \cdot G F\right| \frac{1}{2} m_{F}^{2}$$\begin{aligned} &-\mu m g\left(D G \cos \alpha_{1}+G F \cos \alpha,\right)+\frac{1}{2} m v_{F}^{2}-\mu m g\left(x_{1}+x\right)+\frac{1}{2} m v_{r}^{2} \\ &-\mu m g(x)+\frac{1}{2} m v_{F}^{2} \quad\left(\text { Since } D G \cos \alpha_{1}-x \text { and } F G \cos \alpha_{i}-x_{7}\right) \\ \therefore m g y-\mu m g x-\frac{1}{2} m v_{F}^{2} \quad \therefore & v_{F}-\sqrt{2 g(y-\mu x)} \end{aligned}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:26

Problem 30

A body with a mass of $1 \mathrm{~kg}$ slides down an inclined plane $1 \mathrm{~m}$ high and $10 \mathrm{~m}$ long. Find:
(i) the K.E. of the body at the base of the plane.
(ii) the velocity of the body at the base of the plane, and
(iii) the distance travelled by the body over the horizontal part of the track till it stops. (Triction coc|ficient is cqual to $0.05$ over the track.)
Solution Foree down the plane $=(m g \sin \alpha-\mu m g \cos \alpha)$
$\therefore \quad$ Acceleration down the plane $-g(\sin \alpha \quad \mu \cos \alpha)$
$\sin \alpha-\frac{1}{10}$ and $\cos \alpha-\sqrt{1-\frac{1}{100}}-\sqrt{\frac{99}{100}}-0.995$
$\therefore$ Velocity $(v)$ on reaching the bottom is given by $v^{2}-2 a l-2 g(\sin \alpha \quad \mu \cos \alpha) \times 10-2 \times 98\left(\frac{1}{10} \quad 0.05 \times 0.995\right)$
$-196(0.1 \quad 0.04975)-196 \times 0.5025$
$\therefore \quad v-3.1 \mathrm{~m} / \mathrm{scc} \quad \therefore \quad$ K.E. at the basc $-\frac{1}{2} \times 1 \times(3.1)^{2}-4.9 \mathrm{~J}$
K.E. of the body is used for doing work against friction on the plane ground.
So, $f s-\mu m g s-\frac{1}{2} m v^{2} \quad \therefore s-\frac{v^{2}}{2 \mu g}-\frac{(3.1)^{2}}{2 \times 0.05 \times 9.8}-\frac{(3.1)^{2}}{0.98}-10 \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:16

Problem 31

$\Lambda$ body projected up has potential cnergy of $1080 \mathrm{~J}$ and a kinetic energy $880 \mathrm{~J}$ at a point $P$ in its path vertically upwards. If its mass is $10 \mathrm{~kg}$. what is the velocity of projection? What is the maximum height reached by it? What is the height of $P$ expressed as a fraction of the maximum height reached?
Solution Total cncrgy at any point $=1080+880=1960=$ intial K.E. $=\frac{1}{2} m v^{2}-\frac{1}{2} \times 10 \times v^{2}$
$\therefore$ Velocity of projection $-v-\sqrt{\frac{1960}{5}}-\sqrt{392}-14 \sqrt{2} \mathrm{~m} / \mathrm{sec}$
Total energy is also equal to the P.E. at the highest point. $\therefore \quad 1960=m \mathrm{gh}=10 \times 9.8 \times h$
$\therefore$ Maximum height reached $=20 \mathrm{~m}$
$$
\frac{\text { height of } \mathrm{P}}{\text { maximum height }}-\frac{m g h}{m g H}-\frac{h}{H}-\frac{1080}{1960}-\frac{27}{49} \quad \therefore \quad h-\frac{27}{49} \mathrm{H}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
06:47

Problem 32

$\Lambda$ small bar $\Lambda$ resting on a smooth, horizontal surface is attached by threads to the wall and by means of a massless pullcy, to a block $B$ of the same mass as the bar itself. Further, the bar is attached to a point $O$ by means of a light unstretched spring of lenght $l_{0}$ and spring constant $k=5 m g /_{0}$, where $m$ is mass of the bar. When the thread is burnt, the bar starts moving. Find the velocity when $\Lambda$ breaks off the surface.
Solution Let us consider the instantaneous diplacement of $A$ to be $x$ and the spring be inclined to the vertical by $\theta$. spring force $F-k \sqrt{l_{*}^{2} \mid x^{2}} \quad l_{o}$
Trom figurc (a), $T-k\left(\sqrt{l_{0}^{2}+x^{2}}-l_{a}\right) \sin \theta-m \frac{d v}{d t}$
and $N+k\left(\sqrt{l_{0}^{2} \mid x^{2}} \quad l_{0}\right) \cos \theta-m g$
From figure (b), $m g \quad T-m \frac{d v}{d t}$
Combining (i) and (iii) $m g \quad k\left(\sqrt{l_{0}^{2} \mid x^{2}}-l_{0}\right) \sin \theta-2 m \frac{d v}{d t}$
When $A$ breaks of the planc, $N=0$
$\therefore k\left(\sqrt{l_{n}^{2}+x^{2}}-l_{0}\right) \cos \theta=m g$
or $\frac{5 m g}{l_{n}}\left(\sqrt{l_{n}^{2}+x^{2}}-l\right) \frac{l_{n}}{\sqrt{l_{0}^{2} \mid x^{2}}}-m g$
or $x-\frac{3}{4} l \quad\left(\because k-\frac{5 m g}{l_{o}}\right.$ (given) and $\cos \theta-\frac{l_{o}}{\sqrt{l_{n}^{2} \mid x^{2}}}$ from the figure)
\mathrm{\{} T r o m ~ t h e ~ p r i n c i p l e ~ o f ~ c o n s e r v a t i o n ~ o f ~ c n e r g y ~
$-m g h-2\left(\frac{1}{2} m v^{2}\right)+\frac{1}{2} k \Delta l^{2}-m g(h+x)$
where $h$ is the initial distance of $B$ below the table and $\Delta l$ is the increase in length.
When $A$ breaks of $x-\frac{3}{4} l$$\therefore \quad A-\sqrt{l_{0}^{2}+\frac{9}{16} l_{0}}-l_{o}-\frac{l_{o}}{4} \quad \therefore \quad-m g h-m v^{2}+\frac{1}{2} k \frac{l_{o}^{2}}{16}-m g\left(h+\frac{3 l_{v}}{4}\right)$
or $m g \frac{3 l_{o}}{4}-m v^{2}+\frac{1}{2} \frac{5 m g}{l_{o}} \frac{l_{o}^{2}}{16} \quad$ or $\quad \frac{3 g l_{o}}{4}-m v^{2}+\frac{5 g l_{o}}{16 \times 2}$
or $v^{2}-\frac{3}{4} g l, \frac{5 g l_{o}}{32} \times \frac{19 g l_{n}}{32}$
$\therefore \quad v-\sqrt{\frac{19 g l_{n}}{32}}$

VS
Vivek Singh
Numerade Educator
01:07

Problem 33

$\Lambda$ locomotive of mass $m$ staris moving so that its velocily varics according to the law $v=a \sqrt{s}$, where $a$ is a constant, and $s$ is the distance covered. Find the total work performed by all the forees which are acting on the locomotive during the first $t$ scconds.
Solution By the work-energy theorem $W$ (work done) $=$ change in the kinetic energy
$$
-\frac{1}{2} m v^{2}-0-\frac{1}{2} m v^{2}-\frac{1}{2} m(a \sqrt{s})^{2}-\frac{1}{2} a^{2} s
$$
$\because v-a \sqrt{s}, \frac{d s}{d t}-a \sqrt{s}\left(\because v-\frac{d s}{d t}\right) \Rightarrow d t-\frac{d s}{a \sqrt{s}} \Rightarrow \int d t-\int \frac{d s}{a \sqrt{s}}$
$\therefore t-\frac{2}{a} \sqrt{s} \mathrm{l} c$
when $f-0, s-0 \quad \therefore \quad c-0$
$\therefore t-\frac{2}{a} \sqrt{s}$
$$
\text { or } t^{2}-\frac{4 s}{a^{2}}
$$
$\therefore W-\frac{1}{2} m a^{2} \frac{a^{2} t^{2}}{4}-\frac{1}{8} m a^{4} t^{2}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:12

Problem 34

A particle moves along a circle of radius $R$ with radial acceleration changing with time as $a_{r}=k t^{n}$ where $k$ is constant and $n>1$. Find the power developed by the net force on the particle as a function of time.
Solution In circular motion, work done by centripetal force is always zero, hence power is developed by tangential force only. Given radial accelcration
$\therefore$ Power developed $-\vec{\Gamma} \cdot \vec{v}-m \frac{n}{2} \sqrt{R k} t^{(n-1)} \cdot \sqrt{R k} t^{n / 2}-\frac{m n R k}{2} t^{n-1}$A particle suspended from the ceiling by inextensible light string is moving along a horizontal circle of radius $1.5 \mathrm{~m}$ as shown. The string traces a cone of height $2 \mathrm{~m}$. The string breaks and the particle finally hits the floor (which is $x y$ plane $5.76 \mathrm{~m}$ below the circle) at point $P$. Find the distance $O P$
Solution Let the string breaks when the particle is $1.5 \mathrm{~m}$ right of point $O$ and dircction of its velocity $v$ is along $y$-axis.
$$
\begin{aligned}
& T \sin \theta=\frac{m v^{2}}{r} \quad \text { and } \quad T \cos \theta=m g \\
\therefore \quad & v=\sqrt{r g \tan \theta}
\end{aligned}
$$Now, time to reach the floor $: t-\sqrt{\frac{2 h_{2}}{g}}$ $\therefore$ Before it hits the floor: $\Delta y-v t-\sqrt{2 h_{2} r \tan \theta}$
where $\tan \theta-\frac{r}{h_{1}} \quad \wedge y-\sqrt{2 h_{2} \frac{r^{2}}{h_{1}}}-\sqrt{2 \times \frac{144}{25} \times \frac{(1.5)^{2}}{2}}-\frac{18}{5} \mathrm{~m}-3.6 \mathrm{~m}$
Its position from $O$, when it hits the floor $-1.5 \hat{i}+3.6 \hat{j}$ $O P-\sqrt{(1.5)^{2} 1(3.6)^{2}}-3.9 \mathrm{~m}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:25

Problem 36

As a conical pendulum, a thin uniform rod of length $l$ and mass $m$, rotates uniformly about a vertical axis with angular velocity $\omega$ (the upper end of the rod is hinged). Find the angle $\theta$ between rod and the vertical.
Solution Let us consider an clement of the rod of width $d x$ at a distance $x$ from hinge. Mass of the element, $d m-\frac{m}{l} d x$. The centrifugal force on this element
$$
d F=(d m) \omega^{2}(x \sin \theta)
$$
Its torque of force about the hinge $d \tau=d F \cdot x \cos \theta=(d m) \omega^{2}(x \sin \theta)(x \cos \theta)$
For the torque of force of whole length of rod, integrating
$$
\tau-\frac{m \omega^{2}}{2 l} \sin \theta \int_{v}^{1} x^{2} d x-\frac{m \omega^{2} l^{2}}{6} \sin 2 \theta
$$
In the rotating frame, apart from other forces the centrifugal force also act. For rotational equilibrium of rod we have $\sum \tau=0$
Taking torque of all forces about hinge and pul their algebric sum zuro, we gel
$$
m g \frac{1}{2} \sin \theta-\frac{m \omega l^{2}}{6} \sin 2 \theta
$$
or $\cos \theta-\frac{3 g}{2 \omega^{2} l}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:17

Problem 37

A smooth light horizontal $\operatorname{rod} A B$ can rotate about a vertical axis passing through its end $A$. The rod is fited with a small slecve of mass $m$ attached to the end $A$ by weightless spring of length $l$ and stiffness $k$. What work must be performed to slowly gets this system going and reaching the angular velocity $\omega ?$
Solution Let $x$ be the suretehing in the spring. Then spring force will be $k x$ which counterbalanced by the centrifugal foree $m \omega^{2}\left(l_{0}+x\right)$ acting on the slecve.Therefore we have $k x=m \omega^{2}\left(l_{0}+x\right) \quad \therefore \quad x=m a^{2} l_{0} l\left(k-m \omega^{2}\right) \quad \ldots(1)$
Now from the work energy theorem Work done $=K . E$. gained by the sleeve $+$ energy stored in the spring in stretching it by $x$.
or $W-\frac{1}{2} m \omega^{2}(l, x) \mid \frac{1}{2} k x^{2}$
Solving equations (1) and (2), we gel
where $\eta-\frac{m \omega^{2}}{k}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:42

Problem 38

A small body of mass $m$ tied to a non-stretchable thread moves over a smooth horizontal plane. The other end of the thread is being drawn into a hole $O$ shown in the fig. with a constant velocity. Find the thread tension as a function of the distance $r$ between the body and the hole if at $r=r_{0}$ the angular velocity of the thread is equal to $\omega_{i}$.
Solution The thrcad uension $T$ is cqual to the centrifugal force $m \omega^{2} r$. Since the net torque due to all the forces acting on the body of mass $m$ is zero, and therelore the angular momcntum must be constant.
$\therefore m \omega_{a} r_{0}^{2}-m \omega r^{2} \quad$ or $\quad \omega-\frac{m \omega_{n} r^{2}}{m r^{2}}-\frac{\omega r_{0}^{2}}{r^{2}}$
Tension in the string will be cqual to the centrifugal force

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:18

Problem 39

Find the acceleration of a small body $A$ which slides without initial velocity down a helical groove with a pitch $h$ and radius $R$ at the end of the $n^{\text {th }}$ turn Neglect friction.
Solution In $n$ turns the particle will fall through a vertical height $n h$. So, by the conservation of energy
$$
m g(n h)-\frac{1}{2} m v^{2} \Rightarrow v-\sqrt{2 n g h}
$$
The particle will have two accelerations at any instant:
the tangential acceleration $a_{t}=g \sin \alpha$, where $\alpha=$ inclination of the groove with the horizontal and the normal acceleration
$$
a_{n}-\frac{(v \cos \alpha)^{2}}{R}
$$To find $\alpha$, we mentally flatten the surlace of the cylinder with the helical groove into a plane surface. The groove will become the hypotenuse of a right-angled triangle of height $n h$ and base $2 \pi n R$ and so
$$
\begin{aligned}
& \tan \alpha-\frac{n h}{2 \pi n R}-\frac{h}{2 \pi R} \\
\therefore & a-\sqrt{a_{t}^{2}+a_{n}^{2}}-\sqrt{g^{2} \sin ^{2} \alpha+\frac{v^{4} \cos ^{4} \alpha}{R^{2}}} \\
\Rightarrow & a-\sqrt{g^{2} \frac{h^{2}}{h^{2}+4 \pi^{2} R^{2}}+\frac{4 n^{2} g^{2} h^{2}}{R^{2}} \frac{16 \pi^{4} R^{4}}{\left(h^{2}+4 \pi^{2} R^{2}\right)^{2}}} \\
\Rightarrow & a-\frac{g h \sqrt{h^{2} \mid 4 \pi^{2} R^{2}+64 \pi^{4} n^{2} R^{2}}}{h^{2}+4 \pi^{2} R^{2}}
\end{aligned}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:21

Problem 40

A man and mass $m$ are initially situated on a diametrically opposite ends as shown in figure. Al somc instant, they start moving with constant speeds $v_{1}$ and $v_{2}$. The man moves in $\hat{j}$ direction and mass moves in a circle. Find the linear momentum of the mass with respect to the man as a function of time.
Solution
$\vec{v}_{2}-\hat{i} v_{2} \sin \theta \quad \hat{j} v_{2} \cos \theta-\hat{i} v_{2} \sin \left(\frac{2 v_{2}}{r} t\right) \mid \hat{j} v_{2} \cos \left(\frac{2 v_{2}}{r} t\right)\left[\because \theta-\omega t-\left(\frac{v_{2}}{(r / 2)}\right) t\right]$
The relative velocity of the mass with respect to man is given by $\vec{v}_{2}-\vec{v}_{1}$ $\vec{v}_{2}-\vec{v}_{1}-\left[-\hat{i} v_{2} \sin \left\{\frac{2 v_{2}}{r} t\right\}+\hat{j}_{2} \cos \left\{\frac{2 v_{2}}{r} t\right\}\right]-\hat{j} v_{1}--\hat{i} v_{2} \sin \left(\frac{2 v_{2}}{r} t\right)+\hat{j}\left[v_{2} \cos \left\{\frac{2 v_{2}}{r} t\right\}-v_{1}\right]$
$\therefore$ The relative momentum of mass with respeet to man $=$ mass $\times$ relative velocily $-m\left[\hat{\lambda}_{2} \sin \left(\frac{2 v_{2}}{r} t\right) \mid \hat{j} v_{2} \cos \left(\frac{2 v_{2}}{r} t\right) \quad v_{1}\right]$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:41

Problem 41

A uniform chain of length $l$ is released from rest by pulling the right end slightly down when the chain was in equilibrium. Find the speed of the chain when it loses contact with the pulley.
Solution
$\frac{d v}{d t}-\frac{v d y}{d x}-\frac{g x}{l} \Rightarrow \int_{0}^{v} v d v-\frac{g}{l} \int_{0}^{x} x d x \Rightarrow v-\sqrt{\frac{g}{l}} x$
Then the speed of the chain is $v^{\prime}-\frac{v}{2}-\frac{1}{2} \sqrt{\frac{g}{l}} x$.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:11

Problem 42

Sixteen beads in a suring are placed on a smooth inclined planc of inclination $\sin ^{-1}(1 / 3)$ such that some of them lic along the incline whereas the rest hang over the top of the plane. If acceleration at verlex bead is $g / 2$, the arrangemem of bcads is that
(a) 12 hang vertically
(b) 10 lie along inclined planc.
(c) 8 lic along inclined planc.
(d) 10 hang vertically.
Solution If $n$ beads each of mass $m$ are hanging vertically, then $n m g-T-(n m) a \Rightarrow n m g-T(n m) \frac{g}{2} \quad \therefore T-n\left(\frac{m g}{2}\right) \quad \ldots(1)$
Also, $T-(16-n) m g \sin \theta-(16-n) m\left(\frac{g}{2}\right)$
$\Rightarrow n\left(\frac{m g}{2}\right)-(16-n) m g \sin \theta=(16-n)\left(\frac{m g}{2}\right)$
$\therefore \quad n=10$
$\Rightarrow\left(\frac{n}{2}\right)-(16-n)\left(\frac{1}{3}\right)-\left(\frac{16 n}{2}\right) \Rightarrow \frac{n}{2}-(16-n) \frac{5}{6}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
00:53

Problem 43

A closed chain of mass $m$ is attached to a vertical rotating shaft by means of a thread and rotates with a constant angular velocity $\omega$. The thread forms an angle $\theta$ with the vertical. Find the distance between the chain's centre of gravity and the axis of rotation, and the tension of the thread.
Solution We can take the chain to be stationary with respect to chain-frame by considering pscudo force at the centre of gravily, of the chain. The c.g. of the chain moves in a circle ofradius $r$ about the axis of the shaft. The real forces acting on the chin are: (a) $m g$, at the c.g. of the chain vertically downwards (b) $T$, tension of the string.
(c) centrifugal force $m \omega^{2} r$ at the $c . g$. of the chain. Resolving forces along the vertical and horizontal.
$$
T \cos \theta=m g \text { and } \quad T \sin \theta=m \omega^{2} r
$$
$\therefore T-\frac{m g}{\cos \theta}$
and $r-\frac{T \sin \theta}{m \omega^{2}}-\frac{g \tan \theta}{\omega^{2}}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
05:05

Problem 44

$\Lambda$ chain of mass $m$ forming a circle of radius $R$ is slipped on a smooth cone with half-angle $\theta$. Tind the tension of the chain if it rotates with constant angular velocity $\omega$ about a vertical axis coinciding with the axis of the cone.
Solution Consider an clement of the chain. Lct its length be $\Delta l$ and let it subtend an angle $2 \alpha$ al the cenure.The mass of the element $-\frac{m}{2 \pi} \times 2 \alpha-\frac{m \alpha}{\pi}$
The forees acting on the clement are:
(a) $\mathrm{W} \operatorname{cigh}\left(\frac{m \alpha}{\pi}\right) g$ downwards,
(b) reaction of the cone $A N$ on the element at inclination $\theta$,
(c) tensions $T$ and $T$ tangential to the element in the plane of the ring.
Taking the resolved part of all the forces towards the cente $O$, the centripetal forec $-T \sin \alpha \mid T \sin \alpha \quad \Delta N \cos \theta-2 T \alpha \quad \Delta N \cos \theta-\left(\frac{m g}{\pi}\right) \omega^{2} R$
As there is no vertical acceleration, $(\because \alpha$ is small, $\sin \alpha=\alpha$ ) $\Delta N \sin \theta-\left(\frac{m \alpha}{\pi}\right) g$
$\therefore \quad 2 T \alpha-\frac{m \alpha g}{\pi \sin \theta} \cos \theta-\frac{m \alpha}{\pi} \omega^{2} R$
or $2 T-\frac{m g}{\pi} \cot \theta \mid \frac{m g}{\pi} \omega^{2} R \quad \Rightarrow \quad T-\frac{m g}{2 \pi}\left(\cot \theta \mid \frac{\omega^{2} R}{g}\right)$

VS
Vivek Singh
Numerade Educator
00:44

Problem 45

Consider the length of a chain to be $l$ and coefficient of static friction $\mu$, find the maximum lenghth a chain which can be held outside a table without sliding. Solution
Since $W=f \ldots$ (i) (for equilibrium of the chain) Basically, weight of hanging part is balanced by force of friction on the postion on the table.
But mass per unit lenghl $-\frac{M}{l}$ $(M=$ mass of chain, $l=$ total length
$\therefore W-\frac{M}{l} \cdot y \cdot g$
$\ldots(i i)$
and $R-W^{\prime}-\frac{M}{l}(l \quad y) g$
$\therefore \quad f-\mu R-\mu \frac{M}{l}(l-y) g \quad \ldots$
(iii)
Putting Eqs. (ii) and (iii) in Eq. (i), we get:
$\frac{M}{l} \cdot y \cdot g-\mu \frac{M}{l}(l \quad y) g$
On solving, we get $y-\frac{\mu l}{1+\mu}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:06

Problem 46

A semicrcular chain of mass $m$ slides with a specd $v$ over a smooth hemisphere. Find the lincar momentum of the chain at the given position.Solution The momentum of the system is given as

Hast Aggarwal
Hast Aggarwal
Numerade Educator
05:39

Problem 47

A thin flexible unifom chain of mass $m$ and lenght $L$ is suspended so that its lower end just touches a smooth inelastic plane inclined at $45^{\circ}$. When it is released, find:
(a) the force exerted by the chain on the plane at any subsequent instant and
(b) the total impulse on the plane while the chain falls.
Solution
(a) When the chain is released all its links are in free fall, so the chain remains vertical but without tension, After time $t$, the tip link has fallen a distance
$\frac{1}{2} g t^{2}$
Weight of chain on inclined plane $-\frac{m g}{L} \frac{1}{2} g t^{2}-\frac{m g^{2} l^{2}}{2 L}$
Normal force on the plane exerted due to weight, $N-\frac{m g^{2} f^{2}}{2 L} \cos 45^{\circ}-\frac{m g^{2} t^{2}}{2 \sqrt{2} L}$
In addition to weight the impulsive foree duc to change in momentum also acts. During a smail interval of time $\Lambda t$, a length $g t N$ of mass $\frac{m g t \Delta t}{L}$ has the component of its velocity perpendicular to the plane reduced from $\frac{g t}{\sqrt{2}}$ to zero. We have
$F \cdot \Delta i-\frac{m g t \Delta t}{L}\left(0-\frac{g t}{\sqrt{2}}\right)$
or $F--\frac{m g^{2} t^{2}}{L \sqrt{2}}$
Minus sign indicates that the force on the chain is opposite to its velocity. The total normal force at any instant, while the chain is falling, is $F_{N}-\frac{m g^{2} t^{2}}{2 L \sqrt{2}}+\frac{m g^{2} t^{2}}{L \sqrt{2}}-\frac{3 m g^{2} t^{2}}{2 \sqrt{2} L}$

The time of fall, $t-\sqrt{\frac{2 L}{g}}$
(b) The total impulse $-\int_{0}^{\sqrt{2 / \lambda_{g}}} F_{N} d t-\int_{a}^{\sqrt{2 I . / g}} \frac{3 m g^{2} t^{2}}{2 \sqrt{2} L} d t-\frac{3 m g^{2}}{2 \sqrt{2} L}\left[\frac{t^{3}}{3}\right]_{0}^{\sqrt{2 L /_{*}}}-m(\sqrt{g L})$

VS
Vivek Singh
Numerade Educator
04:44

Problem 48

A uniform chain of length $l$ and mass $m$ is hanging vertically from its ends $A$ and $B$ which are close together, At a given instant the end $B$ is released. What is the tension at $A$ when $B$ has fallen a distance $x \mid x<l]$ ?
Solution Suppose $C$ be the mid-point of chain. So, $\Lambda C=B C=l / 2$. Let $C^{\prime}$ be the bend when $B$ has gone down by $x$. Then $B C^{\prime}-\frac{l}{2} \frac{x}{2}$ The velocity of the bend $-\sqrt{2 g \frac{x}{2}}-\sqrt{g x}$
as the bend has fallen through $\frac{x}{2} . \Lambda$ n clement of length $d x$ at $C$ has the same velocity before it is transferred to the left side. Immediately after it is transferred $A$
to left, its velocity becomes zero. Ilence, change of momentum of element.
$$
-0-\left(\frac{m}{l} d x\right) v--\frac{m}{l} d x v \text { (downward) }-\frac{m}{l} d x v \text { (upward) }
$$
The change takes place in time $d t$.
$\therefore$ Rate of change of momentum $-\frac{m}{l} v \cdot \frac{d x}{d t}-\frac{m}{l} v^{2}-\frac{m}{l} g x$
$\therefore$ Force on left part $-\frac{m g x}{l}$ (downward)
The tension at $A$ is the sum of this force and the weight of the left hanging part.
$$
T_{A}-\left(\frac{l}{2}, \frac{x}{2}\right) \frac{m}{l} g+\frac{m g x}{l}-\frac{m g}{2}, \frac{3 m g x}{2 l}-\frac{m g}{2}\left[1+\frac{3 x}{l}\right]
$$

VS
Vivek Singh
Numerade Educator
01:34

Problem 49

$\Lambda$ chain of mass $m$ and length $l$ is held vertical, such that its lower end just touches the floor. I released from rest. Find the force exeried by the chain on the table when upper end is about to hit the foor.
Solution
Force $F$ exerted by chain consists of two components
(a) $F_{1}$ weight of the fallen portion of the chain,
(b) $F_{2}$ thrust of the falling part of chain.
Now consider an clement of chain of length $d y$ at a height $y$ from the floor. It will strike the floor with
a velocity $v-\sqrt{2 g y}$. Thus we have,
$\Gamma_{1}=\lambda y g$
Here $\lambda$ is the mass per unit length of chain
and $\Gamma_{2}-v_{\mathrm{rel}} \frac{d m}{d t}$
We have $v_{\mathrm{rel}}=v \quad$ and $\quad d m=\lambda d x \quad \therefore F=-v \frac{\lambda d x}{d t}-\lambda v^{2}$
'Ihe force exerted by chain on the floor,$$
F=F_{1}+F_{2}=\lambda y g+\lambda v^{2}-\lambda y g+\lambda(\sqrt{2 g y})^{2}=\lambda y g+2 \lambda g=3 \lambda y g
$$
When upper end is about to hit the floor, $y=l$ $\therefore \quad F=3 \lambda / g=3 m g$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
06:46

Problem 50

$\Lambda$ uniform chain of mass $m$ and length $l$ overhangs a table with its $2 / 3^{\text {rd }}$ part on the table.
(a) Find the work to be done by a person to pul the hanging part back on the uable.
(b) Calculate the P.E of the hanging pari assuming the table surface as reference level.
(c) Calculate the K.E. of the chain as it completely slips ofr the table.
(d) If the table surface is rough and friction co-efficient is $\mu$ then find the work done by the friction during the period when the chain slips of the table.
(e) \Lambdalso, find the kinetic energy at that instant.
Solution
(a) liner mass density $(\lambda)-\left(\frac{m}{l}\right)$ $d m-\lambda d x-\left(\frac{m}{l}\right) d x ; d W-(d m) g x$
$\therefore \quad d W-\left(\frac{m}{l}\right) g x d x$
$W-\int d W-\left(\frac{m g}{l}\right) \int_{x 0}^{x-1 / 3} x d x-\left(\frac{m g}{l}\right)\left(\frac{x^{2}}{2}\right)_{x=0}^{n / 3}-\left(\frac{m g}{l}\right)\left(\frac{1 / 3}{2}\right)^{2}-\left(\frac{m g}{l}\right) \frac{l^{2}}{18}$
$\therefore W-\frac{m g l}{18}$(b) I lere, the gravitational P.E. of the chain is the work done by the external agent in bringing the hanging chain to the table surface with (-) sign,
$d U-d W-(d m) g x \quad \Rightarrow \quad U-\int d U \quad U-\frac{-m g l}{18}$
(c) (loss in P.E.) = (gain in K.E.)
$\left(U, U_{i}\right)-\left(\frac{1}{2} m v^{2} \quad 0\right) \Rightarrow U_{1}-\int d U-\left(\frac{m g}{l}\right) \int_{x=0}^{x} x d x-\frac{m g l}{18}$
$U_{z}-\int d U--\left(\frac{m g}{l}\right) \int_{x}^{x=l} x d x--\frac{m g}{l} \times \frac{l^{2}}{2}--\frac{m g l}{2}$
$\frac{1}{2} m v^{2}-\frac{m g l}{18}, \frac{m g l}{2}-\frac{8 m g l}{18}-\frac{4}{9} m g l \Rightarrow v-\sqrt{\frac{4 m g l}{9} \times \frac{2}{m}}-\sqrt{\frac{8}{9} g l}$1- $\cdots-2 l / 3 \cdots$
(d) $\lambda-\left(\frac{m}{l}\right)$
$d m-\lambda d x-\left(\frac{m}{l}\right) d x \quad \mu(d m) g=f \underbrace{\downarrow}$
work-done by friction on $(d m)$ is $\begin{aligned} & d W=\int d \cos 180^{\circ}=-\mu(d m) g x ; \\ d W &--\left(\frac{\mu m g}{l}\right) x d x \\ W-& \int d W--\left(\frac{\mu m g}{l}\right) \int_{x}^{x=2 l / 3} x d x--\frac{\mu m g}{l}\left(\frac{x^{2}}{2}\right)_{x}^{x} 2 / / 3 \\-& \frac{\mu m g}{l}\left\{\frac{(2 l / 3)^{2}}{2}\right\}--\frac{\mu m g}{l}\left(\frac{4 l^{2}}{18}\right) \quad W--\frac{2}{9}(\mu m g l) \end{aligned}$
(c) W - E Theorem:
$W_{N}\left|W_{\text {mg }}\right| W_{\text {frimtion }}-K_{f} \quad K_{i} ; \quad 0 \quad \mid \frac{m g l}{2},\left(\frac{2}{9} \mu m g l\right)-\frac{1}{2} m v^{2} \quad 0$
$\frac{9 m g l-4 \mu m g l}{18}-\frac{1}{2} m v^{2} \quad \Rightarrow m\left(\frac{9 g l-4 \mu g l}{18}\right)-\frac{1}{2} m^{2}$
$\therefore v-\sqrt{\frac{9 g l 4 \mu g l}{9}}$\Lambdalternative
(f) T.M.E. Conscrvation Law:
$\begin{array}{ll}(T . M . E .)_{j}-(T . M . E .)_{i}=W_{N c} ; & \text { N.c. }=\text { Non-conservative force }\end{array}$
$\Rightarrow\left(\frac{1}{2} m v_{j}^{2} \mid U_{2}\right)\left(\frac{1}{2} m v_{i}^{2} \mid U_{1}\right)-W_{\text {friclien }}$
$\Rightarrow\left\{\frac{1}{2} m^{2}+\left(\frac{m g l}{2}\right)\right\}-\left(\frac{m g l}{18}\right)=\frac{2}{9} \mu m g l$
$\Rightarrow \frac{1}{2} m v^{2}-\frac{9 m g l^{4} \mu m g l}{18} \Rightarrow v-\sqrt{\frac{9 g l^{4} \mu g l}{9}}$

VS
Vivek Singh
Numerade Educator
06:57

Problem 51

A chain of length $l$ and mass $m$ lies on the surface of a smooth sphere of radius $R>l$ with one end tied to the top of the sphere.
(a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere.
(b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle $\alpha$.
(c) \Gammaind the tangential acceleration $\frac{d v}{d t}$ of the chain when the chain starts sliding down.Solution
(a) The mass of the element $d m-\left(\frac{m}{l} R d \theta\right)$ The gravilational pountial cnergy of the clement $d U=(d m) g y$ Thus the gravitational potential energy of whole
$-\frac{m R^{2} g}{l} \int_{n}^{(/ / k)} \cos \theta d \theta-\frac{m g R^{2}}{l}[\sin \theta]_{0}^{t / R}$
$-\frac{m g R^{2}}{2} \sin \left(\frac{l}{R}\right)$
(b)
$\begin{aligned} &\text { (T.M.E.) }=\text { T.M.E. })_{f} \\ \Rightarrow & K_{f}-K_{i}=U_{i}-U_{f} & \Rightarrow \quad K_{f}=U_{i}-U_{f} \\ \therefore & U_{i}-\int d U-\frac{m R^{2} g}{l} \int_{0}^{11 / \alpha^{\prime}} \cos \theta d \theta \\ & \quad-\frac{m g R^{2}}{l}[\sin \theta]_{0}^{I / R}-\frac{m g R^{2}}{2} \sin \left(\frac{l}{R}\right) \end{aligned}$
and $U_{f}-\int d U-\frac{m R^{2} g}{l} \int_{\theta}^{\alpha} \cos \theta d \theta$
$-\frac{m g R^{2}}{l}[\sin \theta]_{\theta=a}^{\alpha \perp \theta}$
$\Rightarrow v_{f}-R \sqrt{\frac{2 g}{l}\left[\sin \left(\frac{l}{R}\right) \sin \left(\theta \mid \frac{l}{R}\right) 1 \sin \theta\right]}$(c) Tangential force on $d m-(d m) g \sin \theta-\left(\frac{m R_{g}}{l}\right) \sin \theta d \theta$
$\therefore \quad$ Tangential force on the chain $-\left(\frac{m R g}{l}\right) \int_{\theta}^{1 / R} \sin \theta d \theta$
$-\left(\frac{m R g}{l}\right)[-\cos \theta]_{0}^{t / K}-\left(\frac{m R g}{l}\right)\left[1-\cos \frac{l}{R}\right]$
$\therefore \quad$ Tangential acceleration force on the chain $=\left(\frac{R g}{l}\right)\left[1-\cos \frac{l}{R}\right]$

VS
Vivek Singh
Numerade Educator
04:32

Problem 52

A smooth chain $A B$ of mass $m$ rests against a surface in the form of a quarter of a circle of radius $R$, If it released from rest, what is the velocity of the chain after it comes over the horizontal part of the surface?
Solution
(T.M.E.) $=$ (T.M.E.) $\quad \Rightarrow \quad \mathrm{K} \cdot \mathrm{E}_{i}+\mathrm{P.E}_{i}=\mathrm{K} \cdot \mathrm{E}_{y}+\mathrm{P} \mathrm{L}_{f}$
$\Rightarrow U_{1}-U_{f}-K_{f}-K_{i}-\frac{1}{2} m v^{2}-0$
$d U=(d m) g h=(\lambda R d \theta) g(R-R \sin \theta)-\frac{m R^{2} g}{(\pi R / 2)}(1 \quad \sin \theta) d \theta$
$\Rightarrow d U-\left(\frac{2 m R g}{\pi}\right)(1 \sin \theta) d \theta$
$\therefore U_{i}-\int d U-\left(\frac{2 m R g}{\pi}\right) \int_{o=0}^{\pi / 2}(1 \quad \sin \theta) d \theta-\left(\frac{2 m g R}{\pi}\right)\left[\int_{0-0}^{\pi / 2} d \theta \int_{0=f}^{\pi / 2} \sin \theta d \theta\right]$
$-\left(\frac{2 m g R}{\pi}\right)\left[\frac{\pi}{2}-(\cos \theta)_{0}^{\pi / 2}\right]-\left(\frac{2 m g R}{\pi}\right)\left[\frac{\pi}{2}-\left(-\cos \frac{\pi}{2}+\cos 0\right)\right]-\left(\frac{2 m R g}{\pi}\right)\left[\frac{\pi}{2}-1\right]$
and $U_{f}-0 ; K_{i}-0$
$\therefore$ From (1) $K_{f}-U_{i}-U_{f}-\left(\frac{2 m R g}{\pi}\right)\left(\frac{\pi}{2}-1\right) \Rightarrow \frac{1}{2} m v^{2}-m R g\left(1-\frac{2}{\pi}\right)$
$\therefore U-\sqrt{2 R g\left(1 \frac{2}{\pi}\right)}$

VS
Vivek Singh
Numerade Educator
01:14

Problem 53

'he flexible chain of length $\frac{\pi r}{2}$ and mass per unit length $\lambda$ is
released from rest with $\theta=0^{\circ}$ in the sinooth circular channel and falls through the hole in the supporting surface. Determine the velocity $v$ ' of the chain as the last link leaves the slot.

Hast Aggarwal
Hast Aggarwal
Numerade Educator
02:06

Problem 54

A chain of length $L$ and mass $M$ is arranged as shown in following four cases. The correct decreasing order of potential energy (assumed zero at horizontal surface) is:
(a)
(b)
(c)
(d)
(c)
Solution
(a) P.E. $-M_{g} \frac{L}{2}-0.5 \mathrm{MgL}$
(b) P.E. $-\mathrm{Mg} \frac{L}{2}-0.5 \mathrm{MgL}$
(c) $P$.E. $-M g \frac{2 R}{\pi}-\frac{2 M g}{\pi} \frac{L}{\pi}-\frac{2}{\pi^{2}} M g L \approx 0.2 \mathrm{MgL}$
(d) $P . E .-\frac{4}{\pi^{2}} M g L \approx 0.4 M g L$
(c) P.E. $-M g\left(R \quad \frac{2 R}{\pi}\right)-\left(\frac{\pi-2}{\pi}\right) M g R=\frac{\pi-2}{\pi^{2}} M g L-\frac{1.14}{\pi^{2}} M g L-0.11 M g L$
(c) $<(\mathrm{c})<(\mathrm{d})<(\mathrm{a})-(\mathrm{b})$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:14

Problem 55

A pile of loose link chain, mass per unit length $\lambda$ lies on a rough surface with coefficient o kinctic friction $\mu_{k}$, One end of chain is being pulled horizontally along the surface by a constant force $P$. Determine the acceleration of chain in terms of $x$ and $\frac{d x}{d t}=v$. Solution
We have $\quad \vec{F}_{\text {sute } 4}+\vec{v} \frac{d m}{d t}-m \frac{d \vec{v}}{d t} \quad \ldots$ (1)
Here $F_{\text {exwrall }}=P-f=P-\mu_{k} \lambda x g$
and $\quad v_{t}=v-0=v ; m=\lambda x$
$\frac{d m}{d t}--\frac{d m}{d x} \cdot \frac{d x}{d t}--\lambda v \quad$ [As mass of system decreasing?
Substituting these values in cquation (1), we get
$\left[P \quad \mu_{k} \lambda g x\right] \mid v(\lambda v)-(\lambda x) \frac{d y}{d t}$
After rearranging, we get

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:50

Problem 56

A chain of length $L$ and mass per unit length $\rho$ is pilled on a horizontal surface. One end of the chain is lifted vertically with a constant velocity $v$ by a variable force $P$. Determine:
(a) $P$ as a function of the height $x$ of the end above the surface.
(b) the energy lost during the lifting of the chain. Solution
(a) Let $x$ be the displacement of the end of the chain above the surface.
$$
F_{\mathrm{cxt}}-P-\rho g x \text { and } v_{\mathrm{x}}-0-v ; \quad \frac{d M}{d t}-\rho v \quad \text { and } \quad \frac{d}{d t}-0
$$
$$
\text { or } \quad 0=(P-\rho g x)+(0-v) \rho v \quad \Rightarrow P=\rho\left(g x+v^{2}\right)
$$
(b) From work-energy theorem, $\int \rho d x-A E=A K+A U$
where $\int P d x$ is work done by external force $P, \wedge E$ is loss in energy.
$$
\int P d x-\int_{0}^{L}\left(\rho g x+\rho v^{2}\right) d x-\frac{1}{2} \rho g L^{2}+\rho v^{2} L
$$
On substituting in the work-energy equation, we gel
$$
\frac{1}{2} \rho g L^{2} \mid \rho v^{2} L \quad \Delta E-\frac{1}{2} \rho L y^{2}+\frac{1}{2} \rho g L^{2} \quad \Delta E-\frac{1}{2} \rho L v^{2}
$$

Hast Aggarwal
Hast Aggarwal
Numerade Educator
01:42

Problem 57

Figure shows a cart that carries a pile of open link chain of mass per unit length $\rho$. The chain can pass freely through a hole in the cart and is brought to rest link by link, by the tension $T$ in the portion of the chain resting on the ground and fixed at its end $A$. A constant force $P$ moves the cart and the initial mass and velocity (when $x=0$ ) are $m_{\circ}$ and $V_{0}$ respectively. Determine the expression for the acceleration $a$ and the velocity $v$ of the cart in terms of $x$ if all friction is neglected. Also find $T$.
Solution Consider threc links: link 1, the last horizontal link; link 2 , the transition link that is decelerated from the velocity $v$ to zero by tension transmitied by link $1 ;$ link 3 , that follows $\operatorname{link} 2$; link 2 cxerts no force on the following link 3 during the transition. Mass of the carl al position $x=m_{\circ}-\rho x, \quad v_{\text {rel }}=0$ and $F_{\text {cxl }}=P$ From the equation, $M \frac{d y}{d t}-F_{\text {et }}+v_{\text {nd }} \frac{d M}{d t} \Rightarrow(m g-\rho x) a-P+0 \quad \therefore a-\frac{P}{m_{\infty} \rho x}$
as $a-v \frac{d y}{d x}-\frac{P}{m_{0} \rho x} \quad$ or $\int_{v_{n}}^{r} v d y-\int_{10}^{x} \frac{P d x}{m_{0} \quad \rho x}$
$\Rightarrow \frac{v^{2}}{2}-\frac{v_{n}^{2}}{2}-\frac{P}{\rho}\left|\ln \left(m_{\circ}-\rho x\right)\right|_{0}^{x} \quad$ or $v^{2}-v_{o}^{2}+\frac{2 P}{\rho} \ln \left(\frac{m_{n}}{m_{0}-\rho x}\right)$
$\Rightarrow \quad v-\sqrt{v_{0}^{2}+\frac{2 P}{\rho} \ln \left(\frac{m_{0}}{m_{v}-\rho x}\right)}$

Hast Aggarwal
Hast Aggarwal
Numerade Educator