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Solutions to I.E. Irodov's problems in general physics

Abhay Kumar Singh; I E Irodov

Chapter 1

Physical Fundamentals Of Mdchanics - all with Video Answers

Educators

Section 1

Kinematics

02:13

Problem 1

Let $v_{0}$ be the stream velocity and $v^{\prime}$ the velocity of motorboat with respect to water. The motorboat reached point $B$ while going downstream with velocity $\left(v_{0}+v^{\prime}\right)$ and then returned with velocity $\left(v^{\prime}-v_{0}\right)$ and passed the raft at point $C .$ Let $t$ be the time for the raft (which flows with stream with velocity $v_{0}$ ) to move from point $A$ to $C$, during which the motorboat moves from $A$ to $B$ and then from $B$ to $C$. Therefore
On solving we get $v_{0}=\frac{l}{2 \tau}$

Narayan Hari
Narayan Hari
Numerade Educator
01:05

Problem 2

Let $s$ be the total distance traversed by the point and $t_{1}$ the time taken to cover half the distance. Further let $2 t$ be the time to cover the rest half of the distance.
Therefore $\quad \frac{s}{2}=v_{0} t_{1} \quad$ or $\quad t_{1}=\frac{s}{2 v_{0}}$
and $\quad \frac{s}{2}=\left(v_{1}+v_{2}\right) t \quad$ or $\quad 2 t=\frac{s}{v_{1}+v_{2}}$
Hence the sought average velocity
$$
\left\langle v>=\frac{s}{t_{1}+2 t}=\frac{s}{\left[s / 2 v_{0}\right]+\left[s /\left(v_{1}+v_{2}\right)\right]}=\frac{2 v_{0}\left(v_{1}+v_{2}\right)}{v_{1}+v_{2}+2 v_{0}}\right.
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:41

Problem 3

As the car starts from rest and finally comes to a stop, and the rate of acceleration and deceleration are equal, the distances as well as the times taken are same in these phases of motion.

Let $\Delta t$ be the time for which the car moves uniformly. Then the acceleration / deceleration time is $\frac{\tau-\Delta t}{2}$ each. So,

Narayan Hari
Narayan Hari
Numerade Educator
01:59

Problem 4

(a) Sought average velocity $\left\langle\nu>=\frac{s}{t}=\frac{200 \mathrm{~cm}}{20 \mathrm{~s}}=10 \mathrm{~cm} / \mathrm{s}\right.$
(b) For the maximum velocity, $\frac{d s}{d t}$ should be maximum. From the figure $\frac{d s}{d t}$ is maximum for all points on the line $a c$, thus the sought maximum velocity becomes average velocity for the line $a c$ and is equal to :
(c) Time $t_{0}$ should be such that corresponding to it the slope $\frac{d s}{d t}$ should pass through the point $O$ (origin), to satisfy the relationship $\frac{d s}{d t}=\frac{s}{t_{0}} .$ From figure the tangent at point $d$ passes through the origin and thus corresponding time $t=t_{0}=16 \mathrm{~s}$.

Narayan Hari
Narayan Hari
Numerade Educator
01:41

Problem 5

.Let the particles collide at the point $A$ (Fig.), whose position vector is $\overrightarrow{r_{3}}$ (say). If $t$ be the time taken by each particle to reach at point $A$, from triangle law of vector addition :
$\overrightarrow{r_{3}}=\overrightarrow{r_{1}}+\overrightarrow{v_{1}} t=\overrightarrow{r_{2}}+\overrightarrow{v_{2}} t$
so, $\overrightarrow{r_{1}}-\overrightarrow{r_{2}}=\left(\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right) t$
therefore, $t=\frac{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|}{\left|\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right|}$
From Eqs. (1) and (2) $\overrightarrow{r_{1}}=\overrightarrow{r_{2}}-\dot{\left(\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right) \frac{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|}{\left|\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right|}}$
or, $\frac{\overrightarrow{r_{1}}-\overrightarrow{r_{2}}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|}=\frac{\overrightarrow{v_{2}}-\overrightarrow{v_{1}}}{\left|\overrightarrow{v_{2}}-\overrightarrow{v_{1}}\right|}$, which is the sought relationship.

Narayan Hari
Narayan Hari
Numerade Educator
02:10

Problem 6

We have $\vec{v}=\vec{v}-\vec{v}_{0}$
From the vector diagram [of Eq. (1)] and using properties of triangle
-
$v^{\prime}=\sqrt{v_{0}^{2}+v^{2}+2 v_{0} v \cos \varphi}=39.7 \mathrm{~km} / \mathrm{hr}$
and $\frac{v^{\prime}}{\sin (\pi-\varphi)}=\frac{v}{\sin \theta}$ or, $\sin \theta=\frac{\nu \sin \varphi}{v^{\prime}}$
$\theta=\sin ^{-1}\left(\frac{\nu \sin \varphi}{v^{\prime}}\right)$
or
Using (2) and putting the values of $v$ and $d$ $\theta=19.1^{\circ}$

Narayan Hari
Narayan Hari
Numerade Educator
02:31

Problem 7

Let one of the swimmer (say 1) cross the river along $A B$, which is obviously the shortest path. Time taken to cross the river by the swimmer $1 .$ $t_{1}=\frac{d}{\sqrt{v^{\prime 2}-v_{0}^{2}}}$, (where $A B=d$ is the width of the river)
(1)
For the other swimmer (say 2), which follows the quickest path, the time taken to cross the river.
In the time $t_{2}$, drifting of the swimmer 2 , becomes $x=v_{0} t_{2}=\frac{v_{0}}{v^{\prime}} d$, (using Eq. 2)
If $t_{3}$ be the time for swimmer 2 to walk the distance $x$ to come from $C$ to $B$ (Fig.), then $t_{3}=\frac{x}{u}=\frac{v_{0} d}{v^{\prime} u} \quad$ (using Eq. 3)
(4)
According to the problem $t_{1}=t_{2}+t_{3}$
or, $\frac{d}{\sqrt{v^{\prime 2}-v_{0}^{2}}}=\frac{d}{v^{\prime}}+\frac{v_{0} d}{v^{\prime} u}$
On solving we get

Narayan Hari
Narayan Hari
Numerade Educator
02:17

Problem 8

Let $l$ be the distance covered by the boat $A$ along the river as well as by the boat $B$ acrc the river. Let $v_{0}$ be the stream velocity and $v^{\prime}$ the velocity of each boat with respect water. Therefore time taken by the boat $A$ in its journey
$$
t_{A}=\frac{l}{v^{\prime}+v_{0}}+\frac{l}{v^{\prime}-v_{0}}
$$
and for the boat $B$
$$
t_{B}=\frac{l}{\sqrt{v^{\prime 2}-v_{0}^{2}}}+\frac{l}{\sqrt{v^{\prime 2}-v_{0}^{2}}}=\frac{2 l}{\sqrt{v^{\prime 2}-v_{0}^{2}}}
$$
Hence,
$$
\frac{t_{A}}{t_{B}}=\frac{v^{\prime}}{\sqrt{v^{\prime 2}-v_{0}^{2}}}=\frac{\eta}{\sqrt{\eta^{2}-1}}\left(\text { where } \eta=\frac{v^{\prime}}{v}\right)
$$
On substitution $\quad t_{A} / t_{B}=1 \cdot 8$

Narayan Hari
Narayan Hari
Numerade Educator
02:13

Problem 9

Let $v_{0}$ be the stream velocity and $v^{\prime}$ the velocity of boat with respect to water. A $\frac{v_{0}}{v^{\prime}}=\eta=2>0$, soine drifting of boat is inevitable. Let $\vec{v}$ make an angle $\theta$ with flow direction. (Fig.), then the time taken to cross the rive $t=\frac{d}{v^{\prime} \sin \theta}$ (where $d$ is the width of the river)
In this time interval, the drifting of the boat $x=\left(v^{\prime} \cos \theta+v_{0}\right) t$
$=\left(v^{\prime} \cos \theta+v_{0}\right) \frac{d}{v^{\prime} \sin \theta}=$
For $x_{\min }$ (minimum $\frac{d}{d \theta}(\cot \theta+\eta$
Hence,

Narayan Hari
Narayan Hari
Numerade Educator
01:23

Problem 10

The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2 , then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration :
$$
\overrightarrow{r_{12}}=\overrightarrow{r_{0}(12)}+\overrightarrow{v_{0}(12)} t+\frac{1}{2} \vec{w}_{12} t^{2}
$$
So, $\vec{r}_{12}=\vec{v}_{0(12)} t$, (because $\vec{w}_{12}=0$ and $\left.\overrightarrow{r_{o(12)}}=0\right)$
or, $\quad\left|\overrightarrow{r_{12}}\right|=\left|\vec{v}_{(12)}\right| t$
(1)
But $\left|\overrightarrow{v_{01}}\right|=\left|\overrightarrow{v_{02}}\right|=v_{0}$
So, from properties of triangle
$$
v_{\alpha(12)}=\sqrt{v_{0}^{2}+v_{0}^{2}-2 v_{0} v_{0} \cos \left(\pi / 2-\theta_{0}\right)}
$$
Hence, the sought distance
$$
\left|\overrightarrow{r_{12}}\right|=v_{0} \sqrt{2(1-\sin \theta)} t=22 \mathrm{~m}
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:11

Problem 11

Let the velocities of the paricles (say $\vec{v}_{1}$ and $\vec{v}_{2}^{\prime}$ ) becomes mutually perpendicuiar after time $t$. Then their velocitis become $\overrightarrow{v_{1}}^{\prime}=\overrightarrow{v_{1}}+\overrightarrow{g t} ; \overrightarrow{v_{2}}=\overrightarrow{v_{2}}+\overrightarrow{g t}$
(1)
As $\overrightarrow{v_{1}}^{\prime} \perp \overrightarrow{v_{2}}^{\circ} \quad$ so, $\overrightarrow{v_{1}}^{\prime} \cdot \overrightarrow{v_{2}}^{\prime}=0$
or, $\quad\left(\overrightarrow{v_{1}}+\overrightarrow{g t}\right) \cdot\left(\overrightarrow{v_{2}}+\overrightarrow{g t}\right)=0$
or $\quad-v_{1} v_{2}+g^{2} t^{2}=0$
Hence, $t=\frac{\sqrt{v_{1} v_{2}}}{g}$
Now form the Eq. $\overrightarrow{r_{12}}=\overrightarrow{r_{0}(12)}+\overrightarrow{v_{0(12)}} t+\frac{1}{2} \vec{w}_{12} t^{2}$
$\left|\overrightarrow{r_{12}}\right|=\left|\overrightarrow{v_{(12)}}\right| t$, (because here $\vec{w}_{12}=0$ and $\left.\overrightarrow{r_{0(12)}}=0\right)$
Hence the sought distance

Narayan Hari
Narayan Hari
Numerade Educator
02:27

Problem 12

From the symmetry of the problem all the three points are always located at the vertices of equilateral triangles of varying side length and finally meet at the centriod of the initial equilateral triangle whose side length is $a$, in the sought time interval (say $t)$.
Let us consider an arbitrary equilateral triangie of edge length $l$ (say). Then the rate by which 1 approaches 2,2 approches 3 , and 3 approches 1 , becomes :
On integrating :
$-\int_{a}^{0} d l=\frac{3 v}{2} \int_{0}^{t} d t$
$a=\frac{3}{2} v t$ so $t=\frac{2 a}{3 v}$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:31

Problem 13

Let us locate the points $A$ and $B$ at an arbitrary instant of time (Fig.). If $A$ and $B$ are separated by the distance $s$ at this moment, then the points converge or point $A$ approaches $B$ with velocity $\frac{-d s}{d t}=v-u \cos \alpha$ where angle $\alpha$ varies with time. On intergating,
(where $T$ is the sought time.)
or $\quad l=\int_{0}^{T}(v-u \cos \alpha) d t$
(1)
As both $A$ and $B$ cover the same distance in $x$ -direction during the sought time interval, so the other condition which is required, can be obtained by the equation $\Delta x=\int v_{x} d t$
So,
(2)
Solving (1) and (2), we get $T=\frac{u l}{v^{2}-u^{2}}$ One can see that if $u=v$, or $u<v$, point $A$ cannot catch $B$.

Narayan Hari
Narayan Hari
Numerade Educator
08:17

Problem 14

In the reference frame fixed to the train, the distance between the two events is obviously equal to $l$. Suppose the train starts moving at time $t=0$ in the positive $x$ direction and take the origin $(x=0)$ at the head-light of the train at $t=0 .$ Then the coordinate of first event in the earth's frame is
$$
x_{1}=\frac{1}{2} w t^{2}
$$
and similarly the coordinate of the second event is
$$
x_{2}=\frac{1}{2} w(t+\tau)^{2}-l
$$
The distance between the two events is obviously.
$$
x_{1}-x_{2}=l-w \tau(t+\tau / 2)=0.242 \mathrm{~km}
$$
in the reference frame fixed on the earth.. For the two events to occur at the same point in the reference frame $K$, moving with constant velocity $V$ relative to the earth, the distance travelled by the frame in the time interval $T$ must be equal to the above distance. Thus
$$
\begin{gathered}
V \tau=l-w \tau(t+\tau / 2) \\
V=\frac{l}{\tau}-w(t+\tau / 2)=4.03 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$
So,
The frame $K$ must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point $x_{1}$ on the earth at time $t$ it coincides with the point $x_{2}$ at time $t+\tau$.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
06:49

Problem 15

(a) One good way to solve the problem is to work in the elevator's frame having the observer at its bottom (Fig.). Let us denote the separation between floor and celing by $h=2 \cdot 7 \mathrm{~m}$. and the acceleration of the elevator by $w=1.2 \mathrm{~m} / \mathrm{s}^{2}$ From the kinematical formula
$$
y=y_{0}+v_{0 y} t+\frac{1}{2} w_{y} t^{2}
$$
Here $y=0, y_{0}=+h, v_{0 y}=0$
and $\quad w_{y}=w_{\text {bott }(y)}-w_{\text {de }(y)}$
$=(-g)-(w)=-(g+w)$
So, $\quad 0=h+\frac{1}{2}\{-(g+w)\} t^{2}$
or, $t=\sqrt{\frac{2 h}{g+w}}=0 \cdot 7 \mathrm{~s}$
(b) At the moment the bolt loses contact with the elevator, it has already aquired the velocity equal to elevator, given by :
$v_{0}=(1 \cdot 2)(2)=2 \cdot 4 \mathrm{~m} / \mathrm{s}$
In the reference frame attached with the elevator shaft (ground) and pointing the $y$ -axis upward, we have for the displacement of the bolt,
$$
\begin{aligned}
\Delta y &=v_{0 y} t+\frac{1}{2} w_{y} t^{2} \\
&=v_{0} t+\frac{1}{2}(-g) t^{2}
\end{aligned}
$$
er, $\Delta y=(2 \cdot 4)(0 \cdot 7)+\frac{1}{2}(-9 \cdot 8)(0 \cdot 7)^{2}=-0 \cdot 7 \mathrm{~m}$
Hence the bolt comes down or displaces downward relative to the point, when it loses contact with the elevator by the amount $0.7 \mathrm{~m}$ (Fig.). Obviously the total distance covered by the bolt during its free fall time
$$
s=|\Delta y|+2\left(\frac{v_{0}^{2}}{2 g}\right)=0 \cdot 7 \mathrm{~m}+\frac{(2 \cdot 4)^{2}}{(9 \cdot 8)} \mathrm{m}=1 \cdot 3 \mathrm{~m}
$$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:06

Problem 16

Let the particle 1 and 2 be at points $B$ and $A$ at $t=0$ at the distances $l_{1}$ and $l_{2}$ from intersection point $O$. Let us fix the inertial frame with the particle $2 .$ Now the particle 1 moves in relative to this reference frame with a relative velocity $\vec{v}_{12}=\overrightarrow{v_{1}}-\overrightarrow{v_{2}}$ and its trajectory is the straight line $B P$. Obviously, the minimum distance between the particles is equal to the length of the perpendicular $A P$ dropped from point $A$ on to the straight line $B P$ (Fig.).
$v_{12}=\sqrt{v_{1}^{2}+v_{2}^{2}}$, and $\tan \theta=\frac{v_{1}}{v_{2}}$
From Fig. (b), $\quad v_{12}=\sqrt{v_{1}^{2}+v_{2}^{2}}$, and $\tan \theta=\frac{1}{v}$
The shortest distanice
$A P=A M \sin \theta=(O A-O M) \sin \theta=\left(l_{2}-l_{1} \cot \theta\right) \sin \theta$
or $\quad A P=\left(l_{2}-l_{1} \frac{v_{2}}{v_{1}}\right) \frac{v_{1}}{\sqrt{v_{1}^{2}+v_{2}^{2}}}=\frac{v_{1} l_{2}-v_{2} l_{1}}{\sqrt{v_{1}^{2}+v_{2}^{2}}}$
(using 1)
The sought time can be obtained directly from the condition that $\left(l_{1}-v_{1} t\right)^{2}+\left(l_{2}-v_{2} t\right)^{2}$ is minimum. This gives $t=\frac{l_{1} v_{1}+l_{2} v_{2}}{v_{1}^{2}+v_{2}^{2}}$.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
01:49

Problem 17

Let the car turn off the highway at a distance $x$ from the point $D$. So, $C D=x$, and if the speed of the car in the field is $v$, then the time taken by the car to cover the distance $A C=A D-x$ on the highway
$$
t_{1}=\frac{A D-x}{\eta v}
$$
and the time taken to travel the distance $C B$ in the field
$$
t_{2}=\frac{\sqrt{l^{2}+x^{2}}}{v}
$$
So, the total time elapsed to move the car from
$$
t=t_{1}+t_{2}=\frac{A D-x}{\eta v}+\frac{\sqrt{l^{2}+x^{2}}}{v}
$$
For $t$ to be minimum
$$
\frac{d t}{d x}=0 \text { or } \frac{1}{v}\left[-\frac{1}{\eta}+\frac{x}{\sqrt{l^{2}+x^{2}}}\right]=0
$$
or $\quad \eta^{2} x^{2}=l^{2}+x^{2} \quad$ or $\quad x=\frac{l}{\sqrt{\eta^{2}-1}}$

Narayan Hari
Narayan Hari
Numerade Educator
08:47

Problem 18

To plot $x(t), s(t)$ and $w_{x}(t)$ let us partion the given plot $v_{x}(t)$ into five segments (for detailed analysis) as shown in the figure. For the part $o a: w_{x}=1$ and $v_{x}=t=v$ Thus, $\Delta x_{1}(t)=\int v_{x} d t=\int_{0}^{t} d t=\frac{t^{2}}{2}=s_{1}(t)$
Putting $t=1$, we get, $\Delta x_{1}=s=\frac{1}{2}$ unit For the part ab :
$w_{x}=0$ and $v_{x}=v=$ constant $=1$
Thus $\Delta x_{2}(t)=\int v_{x} d t=\int_{1}^{t} d t=(t-1)=s_{2}(t)$
Putting $\quad t=3, \Delta x_{2}=s_{2}=2$ unit
For the part $b 4: w_{x}=1$ and $\left.v_{x}=1-(t-3)=4-t\right)=v$ Thus $\quad \Delta x_{3}(t)=\int_{3}^{t}(4-t) d t=4 t-\frac{t^{2}}{2}-\frac{15}{2}=s_{3}(t)$
Putting $t=4, \Delta x_{3}=x_{3}=\frac{1}{2}$ unit
For the part $4 d: \quad v_{x}=-1$ and $v_{x}=-(1-4)=4-1$
So, $\quad v=\left|v_{x}\right|=t-4$ for $t>4$
Thus $\Delta x_{4}(t)=\int_{4}^{t}(1-t) d t=4 t-\frac{t^{2}}{2}-8$
Putting $\quad t=6, \Delta x_{4}=-1$
Similarly $\quad s_{4}(t)=\int\left|v_{x}\right| d t=\int_{4}^{t}(t-4) d t=\frac{t^{2}}{2}-4 t+8$
Putting $\quad t=6, s_{4}=2$ unit For the part $d 7: \quad w_{x}=2$ and $v_{x}=-2+2(t-6)=2(t-7)$ $v=\left|v_{x}\right|=2(7-t)$ for $t \leftarrow 7$Now, $\quad \Delta x(t)=\int_{t} 2(t-7) d t=t^{2}-14 t+48$
Putting $t=4, \Delta x_{5}=-1$
Similarly $s_{5}(t)=\int_{t}^{6} 2(7-t) d t=14 t-t^{2}-48$
Putting $\quad t=7, s_{5}=1$
On the basis of these obtained expressions $w_{x}(t), x(t)$ and $s(t)$ plots can be easily plotted as shown in the figure of answersheet.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:00

Problem 19

(a) Mean velocity $<v>=\frac{\text { Total distance covered }}{\text { Time elapsed }}$ $=\frac{s}{t}=\frac{\pi R}{\tau}=50 \mathrm{~cm} / \mathrm{s}$
(b) Modulus of mean velocity vector $\mid\langle\vec{v}>|=\frac{|\Delta \vec{r}|}{\Delta t}=\frac{2 R}{\tau}=32 \mathrm{~cm} / \mathrm{s}$
(c) Let the point moves from $i$ to $f$ along the half circle (Fig.) and $v_{0}$ and $v$ be the spe
at the points respectively. We have $\quad \frac{d v}{d t}=w_{t}$
or, $v=v_{0}+w_{t} t$ (as $w_{t}$ is constant, according to the problem)
So,
So, from (1) and (3) $\frac{v_{0}+v}{2}=\frac{\pi R}{\tau}$
Now the modulus of the mean vector of total acceleration $|<\vec{w}>|=\frac{|\Delta \vec{v}|}{\Delta t}=\frac{\left|\vec{v}-\vec{v}_{0}\right|}{\tau}=\frac{v_{0}+v}{\tau}$ (see Fig.)
Using (4) in (5), we get :
$$
\begin{aligned} * \vec{w}>\mid=\frac{2 \pi R}{\tau^{2}} \end{aligned}
$$

Narayan Hari
Narayan Hari
Numerade Educator
02:03

Problem 20

(a) we have $\vec{r}=\overrightarrow{a t}(1-\alpha t)$
So, $\vec{v}=\frac{d r}{d t}=\vec{a}(1-2 \alpha t)$
and $\vec{w}=\frac{d \vec{v}}{d t}=-2 \alpha \vec{a}$
(b) From the equation $\quad \vec{r}=\overrightarrow{a t}(1-\alpha t)$,
So, the sought time $\Delta t=\frac{1}{\alpha}$ As $\vec{v}=\vec{a}(1-2 \alpha t)$
So, $\left.v=\mid \overrightarrow{v\rceil}=\begin{array}{l|l}a(1-2 \alpha t) & \text { for } t \leq \frac{1}{2 \alpha} \\ & a(2 \alpha t-1)\end{array}\right\}$ for $t>\frac{1}{2 \alpha}$Hence, the sought distance
$$
s=\int v d t=\int_{0}^{1 / 2 \alpha} a(1-2 \alpha t) d t+\int_{1 / 2 a}^{1 / \alpha} a(2 \alpha t-1) d t
$$
Simplifying, we get, $s=\frac{a}{2 \alpha}$

Narayan Hari
Narayan Hari
Numerade Educator
07:24

Problem 21

(a) As the particle leaves the origin at $t=0$
So, $\Delta x=x=\int v_{x} d t$
As $\quad \vec{v}=\overrightarrow{v_{0}}\left(1-\frac{t}{\tau}\right)$,
where $\overrightarrow{v_{0}}$ is directed towards the +ve $x$ -axis
So,
$$
v_{x}=v_{0}\left(1-\frac{t}{\tau}\right)
$$
From (1) and (2),
$$
x=\int_{0}^{t} v_{0}\left(1-\frac{t}{\tau}\right) d t=v_{0} t\left(1-\frac{t}{2 \tau}\right)
$$
Hence $x$ coordinate of the particle at $t=6 \mathrm{~s}$
$$
x=10 \times 6\left(1-\frac{6}{2 \times 5}\right)=24 \mathrm{~cm}=0.24 \mathrm{~m}
$$
Similarly at $t=10 \mathrm{~s}$
$$
x=10 \times 10\left(1-\frac{10}{2 \times 5}\right)=0
$$
and at $t=20 \mathrm{~s}$
$$
x=10 \times 20\left(1-\frac{20}{2 \times 5}\right)=-200 \mathrm{~cm}=-2 \mathrm{~m}
$$
(b) At the moments the particle is at a distance of $10 \mathrm{~cm}$ from the origin, $x=\pm 10 \mathrm{~cm}$. Putting $x=+10$ in Eq. (3)
So,
$$
\begin{gathered}
10=10 t\left(1-\frac{t}{10}\right) \text { or, } t^{2}-10 t+10=0 \\
t=t=\frac{10 \pm \sqrt{100-40}}{2}=5 \pm \sqrt{15} \mathrm{~s}
\end{gathered}
$$
Now putting $\quad x=-10$ in Eqn (3)
$$
-10=10\left(1-\frac{t}{10}\right),
$$
On solving, $t=5 \pm \sqrt{35} \mathrm{~s}$On solving, $t=5 \pm \sqrt{35}$
As $t$ cannot be negative, so,
$$
t=(5+\sqrt{35}) \mathrm{s}
$$Hence the particle is at a distance of $10 \mathrm{~cm}$ from the origin at three moments of time :
$$
t=5 \pm \sqrt{15} \quad \mathrm{~s}, 5+\sqrt{35} \mathrm{~s}
$$
(c) We have
$$
\vec{v}=\overrightarrow{v_{0}}\left(1-\frac{t}{\tau}\right)
$$
So,
$$
\left.v=\mid \overrightarrow{v\rceil}=\begin{array}{l}
v_{0}\left(1-\frac{t}{\tau}\right) \\
v_{0}\left(\frac{t}{\tau}-1\right)
\end{array}\right\} \text { for } t \leq \tau
$$
So
$$
\begin{aligned}
s &=\int_{0}^{t} v_{0}\left(1-\frac{t}{\tau}\right) d t \text { for } t \leq \tau=v_{0} t(1-t 2 \tau) \\
s &=\int_{0} v_{0}\left(1-\frac{t}{2}\right) d t+\int_{\tau} v_{0}\left(\frac{t}{\tau}-1\right) d t \text { for } t>\tau
\end{aligned}
$$
and
$$
\begin{gathered}
=v_{0} \tau\left[1+(1-t / \tau)^{2}\right] / 2 \text { for } t>\tau \\
s=\int_{0}^{4} v_{0}\left(1-\frac{t}{\tau}\right) d t=\int_{0} 10\left(1-\frac{t}{5}\right) d t=24 \mathrm{~cm}
\end{gathered}
$$
And for $t=8 \mathrm{~s}$
$$
s=\int_{0}^{5} 10\left(1-\frac{t}{5}\right) d t+\int_{5}^{8} 10\left(\frac{t}{5}-1\right) d t
$$
On integrating and simplifying, we get
$$
s=34 \mathrm{~cm}
$$
On the basis of Eqs. (3) and $(4), x(t)$ and $s(t)$ plots can be drawn as shown in the answer sheet.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
01:37

Problem 22

As particle is in unidirectional motion it is directed along the $x$ -axis all the time. As at $t=0, x=0$
So,
$$
\Delta x=x=s, \text { and } \frac{d v}{d t}=w
$$
Therefore, $v=\alpha \sqrt{x}=\alpha \sqrt{s}$
or, $w=\frac{d v}{d t}=\frac{\alpha}{2 \sqrt{s}} \frac{d s}{d t}=\frac{\alpha}{2 \sqrt{s}}$
$$
\begin{aligned}
&=\frac{\alpha v}{2 \sqrt{s}}=\frac{\alpha \alpha \sqrt{s}}{2 \sqrt{s}}=\frac{\alpha^{2}}{2} \\
&w=\frac{d v}{d t}=\frac{\alpha^{2}}{2}
\end{aligned}
$$
As,
On integrating, $\int_{0}^{v} d v=\int_{0}^{t} \frac{\alpha^{2}}{2} d t$ or, $v=\frac{\alpha^{2}}{2} t$
13
(b) Let $s$ be the time to cover first $s \mathrm{~m}$ of the path. From the Eq.
$$
\begin{aligned}
s &=\int v d t \\
s &=\int_{0} \frac{\alpha^{2}}{2} d t=\frac{\alpha^{2}}{2} \frac{t^{2}}{2}
\end{aligned}
$$
(using 2)
or $t=\frac{2}{\alpha} \sqrt{s}$The mean velocity of particle
$$
\left\langle V>=\frac{\int v(t) d t}{\int d t}=\frac{\int_{0}^{2} \frac{\alpha^{2}}{2} t d t}{2 \sqrt{s} / \alpha}=\frac{\alpha \sqrt{s}}{2}\right.
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:17

Problem 23

According to the problem $-\frac{v d v}{d s}=a \sqrt{v}($ as $v$ decreases with time)
or,
$$
-\int_{v_{0}}^{0} \sqrt{v} d v=a \int_{0}^{s} d s
$$
On integrating we get $s=\frac{2}{3 a} v_{0}^{3 / 2}$
Again according to the problem
$-\frac{d v}{d t}=a \sqrt{v}$ or $-\frac{d v}{\sqrt{v}}=a d t$
or,
$$
\int_{v_{0}}^{0} \frac{d v}{\sqrt{v}}=a \int_{0}^{t} d t
$$
Thus $t=\frac{2 \sqrt{v_{0}}}{a}$

Narayan Hari
Narayan Hari
Numerade Educator
02:40

Problem 24

$\begin{array}{lrl}\text { (a) As } & \vec{r}= & a t \vec{i}-b t^{2} \vec{j} \\ \text { So, } & & x=a t, y=-b t^{2}\end{array}$
and therefore $y=\frac{-b x^{2}}{a^{2}}$
which is Eq. of a parabola, whose graph is shown in the Fig.
(b) As $\vec{r}=a t \vec{i}-b t^{2} \vec{j}$
$\vec{v}=\frac{d \vec{r}}{d t}=a \vec{i}-2 b t \vec{j}$
So, $v=\sqrt{a^{2}(-2 b t)^{2}}=\sqrt{a^{2}+4 b^{2} t^{2}}$
Diff. Eq. (1) w.r.t. time, we get $\vec{w}=\frac{d \vec{v}}{d t}=-2 b \vec{j}$
So, $\quad|\vec{w}|=w=2 b$
(c) $\quad \cos \alpha=\frac{\vec{v} \cdot \vec{w}}{v w}=\frac{(a \vec{i}-2 b t \vec{j}) \cdot(-2 b \vec{j}}{\left(\sqrt{a^{2}+4 b^{2} t^{2}}\right) 2 b}$
or, $\cos \alpha=\frac{2 b t}{\sqrt{a^{2}+4 b^{2} t^{2}}}$
so, $\tan \alpha=\frac{a}{2 b t}$
or, $\quad \alpha=\tan ^{-1}\left(\frac{a}{2 b t}\right)$
(d) The mean velocity vector

Narayan Hari
Narayan Hari
Numerade Educator
02:01

Problem 25

(a) We have
$$
x=a t \text { and } y=a t(1-\alpha t)
$$
Hence, $y(x)$ becomes,
$$
y=\frac{a x}{a}\left(1-\frac{\alpha x}{a}\right)=x-\frac{\alpha}{a} x^{2} \text { (parabola) }
$$
(b) Dixferentiating Eq. (1) we get
So,
$$
\begin{gathered}
v_{x}=a \text { and } v_{y}=a(1-2 \alpha t) \\
v=\sqrt{v_{x}^{2}+v_{y}^{2}}=a \sqrt{1+(1-2 \alpha t)^{2}}
\end{gathered}
$$
Diff. Eq. (2) with respect to time
So,
$$
\begin{aligned}
&w_{x}=0 \text { and } w_{y}=-2 a \alpha \\
&w=\sqrt{w_{x}^{2}+w_{y}^{2}}=2 a \alpha
\end{aligned}
$$
(c) From Eqs. (2) and (3)
We have
$$
\vec{v}=a \vec{i}+a(1-2 \alpha t) \vec{j} \text { and } \vec{w}=2 a \alpha \vec{j}
$$
So, $\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}=\frac{\vec{v} \cdot \vec{w}}{v w}=\frac{-a\left(1-2 \alpha t_{0}\right) 2 a \alpha}{a \sqrt{1+\left(1-2 \alpha t_{0}\right)^{2}} 2 a \alpha}$
On simplifying. $1-2 \alpha t_{0}=\pm 1$
As, $t_{0} \approx 0, t_{0}=\frac{1}{\alpha}$

Narayan Hari
Narayan Hari
Numerade Educator
01:30

Problem 26

Differentiating motion law : $x=a \sin \omega t, y=a(1-\cos \omega t)$, with respect to time, $v_{x}=a \omega \cos \omega t, v_{y}=a \omega \sin \omega t$
So, $\vec{v}=a \omega \cos \omega t \vec{i}+a \omega \sin \omega t \vec{j}$
and $\quad v=a \omega=$ Const. Differentiating Eq. (1) with respect to time
$$
\vec{w}=\frac{d \vec{v}}{d t}=-a \omega^{2} \sin \omega t \vec{i}+a \omega^{2} \cos \omega t \vec{j}
$$
(a) The distance $s$ traversed by the point during the time $\tau$ is given by
$$
s=\int_{0}^{\tau} v d t=\int_{0}^{T} a \omega d t=a \omega \tau \quad(\text { using } 2)
$$
(b) Taking inner product of $\vec{v}$ and $\vec{w}$ We get, $\vec{v} \cdot \vec{w}=(a \omega \cos \omega t \vec{i}+a \omega \sin \omega t \vec{j}) \cdot\left(a \omega^{2} \sin \omega t(-i)+a \omega^{2} \cos \omega t-\vec{j}\right)$
So, $\vec{v} \cdot \vec{w}=-a^{2} \omega^{2} \sin \omega t \cos \omega t+a^{2} \omega^{3} \sin \omega t \cos \omega t=0$
Thus, $\vec{v} \perp \vec{w}$, i.e., the angle between velocity vector and acceleration vector equals $\frac{\pi}{2}$.

Narayan Hari
Narayan Hari
Numerade Educator
01:35

Problem 27

Accordiing to the problem
So,
$$
\begin{gathered}
\vec{w}=w(-\vec{j}) \\
w_{x}=\frac{d v_{x}}{d t}=0 \text { and } w_{y}=\frac{d v_{y}}{d t}=-w
\end{gathered}
$$
Differentiating Eq. of trajectory, $y=a x-b x^{2}$, with respect to time
$$
\frac{d y}{d t}=\frac{a d x}{d t}-2 b x \frac{d x}{d t}
$$
So,
$$
\left.\frac{d y}{d t}\right|_{x=0}=\left.a \frac{d x}{d t}\right|_{x=0}
$$
Again differentiating with respect to time
or,
$$
\begin{gathered}
\frac{d^{2} y}{d t^{2}}=\frac{a d^{2} x}{d t^{2}}-2 b\left(\frac{d x}{d t}\right)^{2}-2 b x \frac{d^{2} x}{d t^{2}} \\
-w=a(0)-2 b\left(\frac{d x}{d t}\right)^{2}-2 b x(0)(\text { using } 1)
\end{gathered}
$$
or,
$$
\frac{d x}{d t}=\sqrt{\frac{w}{2 b}} \text { (using 1) }
$$
Using (3) in (2) $\left.\quad \frac{d y}{d t}\right|_{x=0}=a \sqrt{\frac{w}{2 b}}$
Hence, the velocity of the particle at the origin $v=\sqrt{\left(\frac{d x}{d t}\right)_{x=0}^{2}+\left(\frac{d y}{d t}\right)_{x=0}^{2}}=\sqrt{\frac{w}{2 b}+a^{2} \frac{w}{2 b}}$ (using Eqns (3) and (4))
Hence, $v=\sqrt{\frac{w}{2 b}\left(1+a^{2}\right)}$

Narayan Hari
Narayan Hari
Numerade Educator
02:00

Problem 28

As the body is under gravity of constant accelration $\vec{g}$, it's velocity vector and displacemen vectors are:
$\vec{v}=\overrightarrow{v_{0}}+\overrightarrow{g t}$
and $\Delta \vec{r}=\vec{r}=\overrightarrow{v_{0}} t+\frac{1}{2} g t^{2} \quad(\vec{r}=0$ at $t=0)$
So, $<\overrightarrow{v>}$ over the first $t$ seconds $\left\langle\overrightarrow{v>}=\frac{\Delta \vec{r}}{\Delta t}=\frac{\vec{r}}{t}=\overrightarrow{v_{0}}+\frac{\overrightarrow{g t}}{2}\right.$
Hence from Eq. (3), $\langle\vec{\nu}>$ over the first $\left\langle\overrightarrow{v>}=\overrightarrow{v_{0}}+\frac{\vec{g}}{2} \tau\right.$
For evaluating $t$, take $\overrightarrow{v v}=\left(\overrightarrow{v_{0}}+\overrightarrow{g t}\right) \cdot\left(\overrightarrow{v_{0}}+\overrightarrow{g t}\right)=v_{0}^{2}+2\left(\overrightarrow{v_{0}} \cdot \overrightarrow{g)} t+g^{2} t^{2}\right.$
or, $v^{2}=v_{0}^{2}+\left(\overrightarrow{v_{0}} \cdot g\right) t+g^{2} t^{2}$
But we have $v=v_{0}$ at $t=0$ and
Also at $t=\tau$ (Fig.) (also from energy conservation)
$\mathbf{1 7}$
Hence using this propety in Eq. (5)
$$
v_{0}^{2}=v_{0}^{2}+2\left(\vec{v}_{0} \vec{g}\right) \tau+g^{2} \tau^{2}
$$
As $\tau \neq 0$, so, $\tau=-\frac{2\left(\vec{v}_{o} \cdot \vec{g}\right)}{g^{2}}$
Putting this value of $\tau$ in Eq. (4), the average velocity over the time of flight $\left\langle\overrightarrow{v>}=\overrightarrow{v_{0}}-\vec{g} \frac{\left(\overrightarrow{v_{0}} \cdot \vec{g}\right)}{g^{2}}\right.$

Narayan Hari
Narayan Hari
Numerade Educator
04:30

Problem 29

The body thrown in air with velocity $v_{0}$ at an angle $\alpha$ from the horizontal lands at point $P$ on the Earth's surface at same horizontal level (Fig.). The point of projection is taken as origin, so, $\Delta x=x$ and $\Delta y=y$
(a) From the Eq. $\Delta y=v_{0}, t+\frac{1}{2} w_{y} t^{2}$
$0=v_{0} \sin \alpha \tau-\frac{1}{2} g \tau^{2}$
As $\tau \neq 0$, so, time of motion $\tau=\frac{2 v_{0} \sin \alpha}{g}$
(b) At the maximum height of ascent, $v_{y}=0$
so, from the Eq. $\quad v_{y}^{2}=v_{0 y}^{2}+2 w_{y} \Delta y$
$0=\left(v_{0} \sin \alpha\right)^{2}-2 g H$
Hence maximum height $H=\frac{v_{0}^{2} \sin ^{2} \alpha}{2 g}$
During the time of motion the net horizontal displacement or horizontal range, will be obtained by the equation $\Delta x=v_{0 x} t+\frac{1}{2} w_{x} \tau^{2}$
or, $R=v_{0} \cos \alpha \tau-\frac{1}{2}(0) \tau^{2}=v_{0} \cos \alpha \tau=\frac{v_{0}^{2} \sin 2 \alpha}{g}$
when $\quad R=H$
$\frac{v_{0}^{2} \sin ^{2} \alpha}{g}=\frac{v_{0}^{2} \sin ^{2} \alpha}{2 g}$ or $\tan \alpha=4$, so, $\alpha=\tan ^{-1} 4$
(c) For the body, $x(t)$ and $y(t)$ are
$x=v_{0} \cos \alpha t$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:35

Problem 30

We have, $v_{x}=v_{0} \cos \alpha, v_{y}=v_{0} \sin \alpha-g t$
As $\vec{v} \uparrow \uparrow \hat{u}_{t}$ all the moments of time.
Thus $v^{2}=v_{t}^{2}-2 g t v_{0} \sin \alpha+g^{2} t^{2}$
Now, $\quad w_{t}=\frac{d v_{t}}{d t}=\frac{1}{2 v_{t}} \frac{d}{d t}\left(v_{t}^{2}\right)=\frac{1}{v_{t}}\left(g^{2} t-g v_{0} \sin \alpha\right)$
$=-\frac{g}{v_{t}}\left(v_{0} \sin \alpha-g t\right)=-g \frac{v_{y}}{v_{t}}$
Hence $\quad\left|w_{t}\right|=g \frac{\left|v_{y}\right|}{v}$
Now $w_{n}=\sqrt{w^{2}-w_{t}^{2}}=\sqrt{g^{2}-g^{2} \frac{v_{y}^{2}}{v_{t}^{2}}}$
or $\quad w_{n}=g \frac{v_{x}}{v_{t}}\left(\right.$ where $\left.v_{x}=\sqrt{v_{t}^{2}-v_{y}^{2}}\right)$
As $\quad \vec{v} \uparrow \uparrow \hat{v}_{t}$, during time of motion
$w_{v}=w_{t}=-g \frac{v_{y}}{v}$
On the basis of obtained expressions or facts the sought plots can be drawn as shown in the figure of answer sheet.

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:03

Problem 31

The ball strikes the inclined plane $(O x)$ at point $O$ (origin) with velocity $v_{0}=\sqrt{2 g h}$ (1) As the ball elastically rebounds, it recalls with same velocity $v_{0}$, at the same angle $\alpha$ from the normal or $y$ axis (Fig.). Let the ball strikes the incline second time at $P$, which is at a distance $l$ (say) from the point $O$, along the incline. From the equation $y=v_{0 y} t+\frac{1}{2} w_{y} t^{2}$
where $\tau$ is the time while moving from As $\tau \neq 0$, so, $\tau=\frac{2 v_{0}}{g}$
Now from the equation.
$x=v_{0 x} t+\frac{1}{2} w_{x} t^{2}$
so, $l=v_{0} \sin \alpha \tau+\frac{1}{2} g \sin \alpha \tau^{2}$
$l=v_{0} \sin \alpha\left(\frac{2 v_{0}}{ }\right)$
$=\frac{4 v_{0}^{2} \sin \alpha}{g}($ using 2$)$
Hence the sought distance, $l=\frac{4(2 g h) \sin \alpha}{g}=8 h \sin \alpha$ (Using Eq. 1)

Narayan Hari
Narayan Hari
Numerade Educator
02:07

Problem 32

Total time of motion
$$
\tau=\frac{2 v_{0} \sin \alpha}{g} \text { or } \sin \alpha=\frac{\tau g}{2 v_{0}}=\frac{9 \cdot 8 \tau}{2 \times 240}
$$
and horizontal range
$$
R=v_{0} \cos \alpha \tau \text { or } \cos \alpha=\frac{R}{v_{0} \tau}=\frac{5100}{240 \tau}=\frac{85}{4 \tau}
$$
From Eqs. (1) and (2)
$$
\frac{(9 \cdot 8)^{2} \tau^{2}}{(480)^{2}}+\frac{(85)^{2}}{\left(4 \tau^{2}\right)^{2}}=1
$$
On simplifying $\tau^{4}-2400 \tau^{2}+1083750=0$
Solving for $\tau^{2}$ we get :
$$
\tau^{2}=\frac{2400 \pm \sqrt{1425000}}{2}=\frac{2400 \pm 1194}{2}
$$
Thus $\tau=42 \cdot 39 \mathrm{~s}=0 \cdot 71 \mathrm{~min}$ and
$\tau=24 \cdot 55 \mathrm{~s}=0.41 \mathrm{~min}$ depending on the angle $\alpha .$

Narayan Hari
Narayan Hari
Numerade Educator
02:37

Problem 33

Let the shells collide at the point $P(x, y)$. If the first shell takes $t$ s to collide with second and $\Delta t$ be the time interval between the firings, then $x=v_{0} \cos \theta_{1} t=v_{0} \cos \theta_{2}(t-\Delta t)$
(1) and $y=v_{0} \sin \theta_{1} t-\frac{1}{2} g t^{2}$
$=v_{0} \sin \theta_{2}(t-\Delta t)-\frac{1}{2} g(t-\Delta t)^{2}$
From Eq. (1) $t=\frac{\Delta t \cos \theta_{2}}{\cos \theta_{2}-\cos \theta_{1}} \quad$ (3)
From Eqs. $\Delta t=\frac{2 v_{0} \sin \left(\theta_{1}-\theta_{2}\right)}{g\left(\cos \theta_{2}+\cos \theta_{1}\right)}$ as $\Delta t \neq 0$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:48

Problem 34

According to the problem
(a) $\frac{d y}{d t}=v_{0}$ or $d y=v_{0} d t$
Integrating
$$
\int_{0}^{y} d y=v_{0} \int_{0}^{t} d t \text { or } y=v_{0} t
$$
And also we have $\frac{d x}{d t}=a y$ or $d x=a y d t=a v_{0} t d t \quad$ (using 1$)$
So,
$$
\int_{0}^{x} d x=a v_{0} \int_{0}^{t} t d t, \text { or, } x=\frac{1}{2} a v_{0} t^{2}=\frac{1}{2} \frac{a y^{2}}{v_{0}} \text { (using 1) }
$$
(b) According to the problem
So,
$$
\begin{gathered}
v_{y}=v_{0} \text { and } v_{x}=a y \\
v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{v_{0}^{2}+a^{2} y^{2}}
\end{gathered}
$$
Therefore $\quad w_{t}=\frac{d v}{d t}=\frac{a^{2} y}{\sqrt{v_{0}^{2}+a y^{2}}} \frac{d y}{d t}=\frac{a^{2} y}{\sqrt{1+\left(a y / v_{0}\right)}}$
Diff. Eq. (2) with respect to time.
$$
\frac{d v_{y}}{d t}=w_{y}=0 \text { and } \frac{d v_{x}}{d t}=w_{x}=a \frac{d y}{d t}=a v_{0}
$$
So,
$$
w=\left|w_{x}\right|=a v_{0}
$$Hence $w_{n}=\sqrt{w^{2}-w_{t}^{2}}=\sqrt{a^{2} v_{0}^{2}-\frac{a^{4} y^{2}}{1+\left(a y / v_{0}\right)^{2}}}=\frac{a v_{0}}{\sqrt{1+\left(a y / v_{0}\right)^{2}}}$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:18

Problem 35

(a) The velocity vector of the particle $\vec{v}=a \overrightarrow{i+b x} \vec{j}$
So,
$$
\frac{d x}{d t}=a: \frac{d y}{d t}=b x
$$
From (1)
$$
\int_{0}^{x} d x=a \int_{0}^{t} d t \text { or, } x=a t
$$
And $d y=b x d t=b a t d t$
Integrating
$$
\int_{0}^{y} d y=a b \int_{0}^{t} t d t \text { or, } y=\frac{1}{2} a b t^{2}
$$
From Eqs. (2) and (3), we get, $y=\frac{b}{2 a} x^{2}$
(b) The curvature radius of trajectory $y(x)$ is :
$$
R=\frac{\left[1+(d y / d x)^{2}\right]^{\frac{3}{2}}}{\left|d^{2} y / d x^{2}\right|}
$$
Let us differentiate the path Eq. $y=\frac{b}{2 a} x^{2}$ with respect to $x$,
$$
\frac{d y}{d x}=\frac{b}{a} x \text { and } \frac{d^{2} y}{d x^{2}}=\frac{b}{a}
$$
From Eqs. (5) and (6), the sought curvature radius :
$$
R=\frac{a}{b}\left[1+\left(\frac{b}{a} x\right)^{2}\right]^{\frac{3}{2}}
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:19

Problem 36

In accordance with the problem $w_{t}=\vec{a} \cdot \vec{\tau}$
But $w_{t}=\frac{v d v}{d s}$ or $v d v=w_{t} d s$
So, $v d v=(\vec{a} \cdot \vec{\tau}) d s=\vec{a} \cdot d \vec{r}$
or, $v d v=a \vec{i} \cdot d \vec{r}=a d x$ (because $\vec{a}$ is directed towards the $x$ -axis)
So, $\int_{0}^{v} v d v=a \int_{0}^{x} d x$
Hence $v^{2}=2 a x$ or, $v=\sqrt{2 a x}$

Narayan Hari
Narayan Hari
Numerade Educator
02:32

Problem 37

The velocity of the particle $v=$ at
So,
$$
\begin{aligned}
&\frac{d v}{d t}=w_{t}=a \\
&\left.w_{n}=\frac{v^{2}}{R}=\frac{a^{2} t^{2}}{R} \text { (using } v=a t\right)
\end{aligned}
$$
And
From
$$
\begin{aligned}
s &=\int v d t \\
-2 \pi R \eta &=a \int_{0} v d t=\frac{1}{2} a t^{2} \\
\frac{4 \pi \eta}{a} &=\frac{t^{2}}{R}
\end{aligned}
$$
So,
From Eqs. (2) and (3) $w_{n}=4 \pi a \eta$
$$
\text { Hence } \begin{aligned}
w=& \sqrt{w_{t}^{2}+w_{n}^{2}} \\
&=\sqrt{a^{2}+(4 \pi a \eta)^{2}}=a \sqrt{1+16 \pi^{2} \eta^{2}}=0 \cdot 8 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
$$

Narayan Hari
Narayan Hari
Numerade Educator
03:05

Problem 38

According to the problem
For $v(t)$
$$
\begin{aligned}
&\left|w_{t}\right|=\left|w_{n}\right| \\
&\frac{-d v}{d t}=\frac{v^{2}}{R}
\end{aligned}
$$
Integrating this equation from $v_{0} \leq v \leq v$ and $0 \leq t \leq t$
$$
-\int_{v_{0}}^{v} \frac{d v}{v^{2}}=\frac{1}{R} \int_{0}^{t} d t \text { or, } v=\frac{v_{0}}{\left(1+\frac{v_{0} t}{R}\right)}
$$
Njw for $v(s),-\frac{v d v}{d s}=\frac{v^{2}}{R}$, Integrating this equation from $v_{0} \leq v \leq v$ and $0 \leq s \leq s$
So,
$$
\int_{v_{0}}^{v} \frac{d v}{v}=-\frac{1}{R} \int_{0}^{s} d s \text { or, } \ln \frac{v}{v_{0}}=-\frac{s}{R}
$$
Hence
$$
v=v_{0} e^{-s / R}
$$
(b) The normal acceleration of the point
$$
w_{n}=\frac{v^{2}}{R}=\frac{v^{2} e^{-2 s / R}}{R}(\text { using } 2)
$$
And as accordance with the problem
$$
\left|w_{t}\right|=\left|w_{n}\right| \text { and } w_{t} \hat{u}_{t} \perp w_{n} \hat{u}_{n}
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:01

Problem 39

From the equation $v=a \sqrt{s}$
$$
\begin{aligned}
&w_{t}=\frac{d \nu}{d t}=\frac{a}{2 \sqrt{s}} \frac{d s}{d t}=\frac{a}{2 \sqrt{s}} a \sqrt{s}=\frac{a^{2}}{2}, \text { and } \\
&w_{n}=\frac{v^{2}}{R}=\frac{a^{2} s}{R}
\end{aligned}
$$
As $w_{t}$ is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity vector coincides in direction. Hence the angle between $\vec{v}$ and $\vec{w}$ is equal to between $w_{t} \hat{v}_{t}$ an $\vec{w}$, and $\alpha$ can be found by means of the formula : $\tan \alpha=\frac{\left|w_{n}\right|}{\left|w_{t}\right|}=\frac{a^{2} s / R}{a^{2} / 2}=\frac{2 s}{R}$

Narayan Hari
Narayan Hari
Numerade Educator
03:34

Problem 40

From the equation $l=a \sin \omega t$
$$
\begin{aligned}
&\frac{d l}{d t}=v=a \omega \cos \omega t \\
&\text { So, } \quad w_{t}=\frac{d \nu}{d t}=-a \omega^{2} \sin \omega t, \text { and } \\
&w_{n}=\frac{\nu^{2}}{R}=\frac{a^{2} \omega^{2} \cos ^{2} \omega t}{R}
\end{aligned}
$$
(a) At the point $l=0, \sin \omega t=0$ and $\cos \omega t=\pm 1$ so, $\omega t=0, \pi$ etc.
Hence
$$
w=w_{n}=\frac{a^{2} \omega^{2}}{R}
$$
Similarly at $l=\pm a, \sin \omega t=\pm 1$ and $\cos \omega t=0$, so, $w_{n}=0$
Hence $w=\left|w_{t}\right|=a \omega^{2}$

Narayan Hari
Narayan Hari
Numerade Educator
01:35

Problem 41

As $w_{t}=a$ and at $t=0$, the point is at rest
So,
$v(t)$ and $s(t)$ are, $v=a t$ and $s=\frac{1}{2} a t^{2}$
Let $R$ be the curvature radius, then
$$
w_{n}=\frac{v^{2}}{R}=\frac{a^{2} t^{2}}{R}=\frac{2 a s}{R}(\text { using } 1)
$$
But according to the problem
$$
w_{n}=b t^{4}
$$
So, $b t^{4}=\frac{a^{2} t^{2}}{R}$
or, $R=\frac{a^{2}}{b t^{2}}=\frac{a^{2}}{2 b s}$ (using 1)
Therefore $w=\sqrt{w_{t}^{2}+w_{n}^{2}}=\sqrt{a^{2}+(2 a s / R)^{2}}=\sqrt{a^{2}+\left(4 b s^{2} / a^{2}\right)^{2}}$ (using 2)
Hence $w=a \sqrt{1+\left(4 b s^{2} / a^{3}\right)^{2}}$

Narayan Hari
Narayan Hari
Numerade Educator
03:02

Problem 42

(a) Let us differentiate twice the path equation $y(x)$ with respect to time.
$$
\frac{d y}{d t}=2 a x \frac{d x}{d t} ; \frac{d^{2} y}{d t^{2}}=2 a\left[\left(\frac{d x}{d t}\right)^{2}+x \frac{d^{2} x}{d t^{2}}\right]
$$
Since the particle moves uniformly, its acceleration at all points of the path is normal and at the point $x=0$ it coincides with the direction of derivative $d^{2} y / d t^{2}$. Keeping in mind that at the point $x=0,\left|\frac{d x}{d t}\right|=v$
We get
$$
w=\left|\frac{d^{2} y}{d t^{2}}\right|_{x=0}=2 a v^{2}=w_{n}
$$
So, $w_{n}=2 a v^{2}=\frac{v^{2}}{R}$, or $R=\frac{1}{2 a}$
Note that we can also calculate it from the formula of problem $(1.35 b)$
(b) Differentiating the equation of the trajectory with respect to time we see that
$$
b^{2} x \frac{d x}{d t}+a^{2} y \frac{d y}{d t}=0
$$
which implies that the vector $\left(b^{2} x \vec{i}+a^{2} y \vec{j}\right)$ is normal to the velocity vector $\vec{v}=\frac{d x}{d t} i+\frac{d y}{d t} \vec{j}$ which, of course, is along the tangent. Thus the former vactor is along the normal and the normal component of acceleration is clearly
$$
w_{n}=\frac{b^{2} x \frac{d^{2} x}{d t^{2}}+a^{2} y \frac{d^{2} y}{d t^{2}}}{\left(b^{4} x^{2}+a^{4} y^{2}\right)^{1 / 2}}
$$

Narayan Hari
Narayan Hari
Numerade Educator
02:23

Problem 43

Let us fix the co-ordinate system at the point $O$ as shown in the figure, such that the radius vector $\vec{r}$ of point $A$ makes an angle $\theta$ with $x$ axis at the moment shown. Note that the radius vector of the particle $A$ rotates clockwise and we here take line $o x$ as reference line, so in this case obviously the angular velocity $\omega=\left(-\frac{d \theta}{d t}\right)$ taking anticlockwise sense of angular displacement as positive. Also from the geometry of the triangle $O A C$ $\frac{R}{\sin \theta}=\frac{r}{\sin (\pi-2 \theta)}$ or, $r=2 R \cos \theta$
Let us write, $\vec{r}=r \cos \theta \vec{i}+r \sin \theta \vec{j}=2 R \cos ^{2} \theta \vec{i}+R \sin 2 \theta \vec{j}$
Differentiating with respect to time. $\frac{\overrightarrow{d r}}{d t}$ or $\vec{v}=2 R 2 \cos \theta(-\sin \theta) \frac{d \theta}{d t} \vec{i}+2 R \cos 2 \theta \frac{d \theta}{d t} \vec{j}$
or, $\vec{v}=2 R\left(\frac{-d \theta}{d t}\right)[\sin 2 \theta \vec{i}-\cos 2 \theta \vec{j}]$
or, $\vec{v}=2 R \omega\left(\sin 2 \theta \vec{i}-\cos ^{2} \theta \vec{j}\right)$
So, $|\vec{v}|$ or $v=2 \omega R=0 \cdot 4 \mathrm{~m} / \mathrm{s}$
As $\omega$ is constant, $v$ is also constant and $w_{t}=\frac{d v}{d t}=0$,
So, $w=w_{n}=\frac{v^{2}}{R}=\frac{(2 \omega R)^{2}}{R}=4 \omega^{2} R=0.32 \mathrm{~m} / \mathrm{s}^{2}$
Alternate : From the Fig. the angular velocity of the point $A$, with respect to centre of the circle $C$ becomesThus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius $R$ with constant angular velocity $2 \omega$.
Hence $\quad v=2 \omega R$ and
$$
w=w_{n}=\frac{v^{2}}{R}=\frac{(2 \omega R)^{2}}{R}=4 \omega^{2} R
$$

Narayan Hari
Narayan Hari
Numerade Educator
02:09

Problem 44

Differentiating $\varphi(t)$ with respect to time
$$
\frac{d \varphi}{d t}=\omega_{z}=2 a t
$$
For fixed axis rotation, the speed of the point A:
$$
v=\omega R=2 a t R \text { or } R=\frac{v}{2 a t}
$$
Differentiating with respect to time
But
$$
\begin{aligned}
&w_{t}=\frac{d \nu}{d t}=2 a R=\frac{v}{t}, \text { (using 1) } \\
&w_{n}=\frac{v^{2}}{R}=\frac{v^{2}}{v / 2 a t}=2 a t v \quad \text { (using 2) }
\end{aligned}
$$
So,
$$
w=\sqrt{w_{t}^{2}+w_{n}^{2}}=\sqrt{(v / t)^{2}+(2 a t v)^{2}}
$$
$$
=\frac{v}{t} \sqrt{1+4 a^{2} t^{4}}
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:12

Problem 45

The shell acquires a constant angular acceleration at the same time as it accelerates linearly. The two are related by (assuming both are constant)
$$
\frac{w}{l}=\frac{\beta}{2 \pi n}
$$
Where $w=$ linear acceleration and $\beta=$ angular acceleration Then, $\omega=\sqrt{2 \cdot \beta 2 \pi n}=\sqrt{2 \cdot \frac{w}{l}(2 \pi n)^{2}}$
But $v^{2}=2 w l$, hence finally $\omega=\frac{2 \pi n v}{l}$

Narayan Hari
Narayan Hari
Numerade Educator
02:28

Problem 46

Let us take the rotation axis as $z$ -axis whose positive direction is associated with the positive direction of the cordinate $\varphi$, the rotation angle, in accordance with the right-hand screw rule (Fig.)
(a) Defferentiating $\varphi(t)$ with respect to time.
$$
\frac{d \varphi}{d t}=a-3 b t^{2}=\omega_{z}
$$
(1) and
$$
\frac{d^{2} \varphi}{d t^{2}}=\frac{d \omega_{z}}{d t}=\beta_{z}=-6 b t
$$
$(109$
From (1) the solid comes to stop at $\Delta t=t=\sqrt{\frac{a}{3 b}}$ The angular velocity $\omega=a-3 b t^{2}$, for $0 \leq t \leq \sqrt{a / 3 b}$
$$
\text { So, }<\omega>=\frac{\int \omega d t}{\int d t}=\frac{\sqrt{a / 3 b}\left(a-3 b t^{2}\right) d t}{\sqrt{a / 3 b}}=\left[a t-b t^{3}\right]_{0}^{\sqrt{a / 3 b}} / \sqrt{a / 3 b}=2 a / 3
$$
Similarly $\beta=\left|\beta_{z}\right|=6 b t \quad$ for all values of $t$So, $\quad<\beta>=\frac{\int \beta d t}{\int d t}=\frac{\sqrt{a / 3 b} 6 b t d t}{\sqrt{a / 3 b} d t}=\sqrt{3 a b}$
(b) From Eq. (2) $\beta_{z}=-6 b t$
So, $\left(\beta_{z}\right)_{t}=\sqrt{a / 3 b}=-6 b \sqrt{\frac{a}{3 b}}=-2 \sqrt{a b}$
Hence $\quad \beta=\left|\left(\beta_{z}\right)_{t-\sqrt{a / 3 b}}\right|=2 \sqrt{3 a b}$

Narayan Hari
Narayan Hari
Numerade Educator
02:01

Problem 47

Angle $\alpha$ is related with $\left|w_{t}\right|$ and $w_{n}$ by means of the fomula :
$$
\tan \alpha=\frac{w_{n}}{\left|w_{t}\right|}, \text { where } w_{n}=\omega^{2} R \text { and }\left|w_{t}\right|=\beta R
$$
where $R$ is the radius of the circle which an arbitrary point of the body circumscribes. From the given equation $\beta=\frac{d \omega}{d t}=$ at (here $\beta=\frac{d \omega}{d t}$, as $\beta$ is positive for all values of $t$ )
Integrating within the limit $\int_{0}^{\infty} d \omega=a \int_{0}^{t} t d t$ or, $\omega=\frac{1}{2} a t^{2}$
So,
$$
w_{n}=\omega^{2} R=\left(\frac{a t^{2}}{2}\right)^{2} R=\frac{a^{2} t^{4}}{4} R
$$
and
$$
\left|w_{t}\right|=\beta R=a t R
$$
Putting the values of $\left|w_{t}\right|$ and $w_{n}$ in Eq. (1), we get,
$$
\tan \alpha=\frac{a^{2} t^{4} R / 4}{a t R}=\frac{a t^{3}}{4} \text { or, } t=\left[\left(\frac{4}{a}\right) \tan \alpha\right]^{1 / 3}
$$

Narayan Hari
Narayan Hari
Numerade Educator
02:01

Problem 48

In accordance with the problem, $\beta_{z}<0$
Thus $-\frac{d \omega}{d t}=k \sqrt{\omega}$, where $k$ is proportionality constant
or, $-\int_{\infty_{0}}^{\infty} \frac{d \omega}{\sqrt{\omega}}=k \int_{0}^{t} d t$ or, $\sqrt{\omega}=\sqrt{\omega_{0}}-\frac{k t}{2}$
When $\omega=0$, total time of rotation $t=\tau=\frac{2 \sqrt{\omega_{0}}}{k}$

Narayan Hari
Narayan Hari
Numerade Educator
01:39

Problem 49

We have $\omega=\omega_{0}-a \varphi=\frac{d \varphi}{d t}$
Integratin this Eq. within its limit for $(\varphi) t$
$$
\int_{0}^{\varphi} \frac{d \varphi}{\omega_{0}-k \varphi}=\int_{0}^{t} d t \text { or }, \ln \frac{\omega_{0}-k \varphi}{\omega_{0}}=-k t
$$
Hence
$$
\varphi=\frac{\omega_{0}}{k}\left(1-e^{-k t}\right)
$$
(b) From the Eq., $\omega=\omega_{0}-k \varphi$ and Eq. (1) or by differentiating Eq. (1)
$$
\omega=\omega_{0} e^{-k t}
$$

Narayan Hari
Narayan Hari
Numerade Educator
01:25

Problem 50

Let us choose the positive direction of $z$ -axis (stationary rotation axis) along the vector $\beta_{0} .$ In accordance with the equation
or, $\omega_{z} d \omega_{z}=\beta_{z} d \varphi=\beta \cos \varphi d \varphi$,
Integrating this Eq. within its limit for
$\omega_{z}(\varphi)$
or, $\int_{0}^{\omega_{t}} d \omega_{z}=\beta_{0} \int_{0}^{\varphi} \cos \varphi d \varphi$
or, $\frac{\omega_{z}^{2}}{2}=\beta_{0} \sin \varphi$
Hence $\quad \omega_{z}=\pm \sqrt{2 \beta_{0} \sin \varphi}$
The plot $\omega_{z}(\varphi)$ is shown in the Fig. It can be seen that as the angle $\varphi$ grows, the vector $\vec{\omega}$ first increases, coinciding with the direction of the vector $\overrightarrow{\beta_{0}}\left(\omega_{z}>0\right)$, reaches the maximum at $\varphi=\varphi / 2$, then starts decreasing and finally turns into zero at $\varphi=\pi .$ After that the body starts rotating in the opposite direction in a similar fashion $\left(\omega_{z}<0\right) .$ As a result, the body will oscillate about the position $\varphi=\varphi / 2$ with an amplitude equal to $\pi / 2$.

Narayan Hari
Narayan Hari
Numerade Educator
02:51

Problem 51

Rotating disc moves along the $x$ -axis, in plane motion in $x-y$ plane. Plane motion of a solid can be imagined to be in pure rotation about a point (say $I$ ) at a certain instant known as instantaneous centre of rotation. The instantaneous axis whose positive sense is
directed along $\vec{\omega}$ of the solid and which passes through the point $I$, is known as instantaneous axis of rotation. Therefore the velocity vector of an arbitrary point $(P)$ of the solid can be represented as :
$\overrightarrow{v_{p}}=\vec{\omega} \times \overrightarrow{r_{P I}}$
On the basis of Eq. (1) for the C. M. (C) of the disc $\overrightarrow{v_{c}}=\vec{\omega} \times \overrightarrow{r_{d}}$
According to the problem $\vec{\omega} \uparrow \uparrow \vec{k}$ i.e. $\vec{\omega} \perp x-y$ plane, so to satisy the Eqn. (2) $\vec{r}_{c I}$ is directed along $(-\vec{j})$. Hence point
$I$ is at a distance $r_{C I}=y$, above the centre of the disc along $y$ -axis. Using all these facts
in Eq. (2), we get $v_{C}=\omega y$ or $y=\frac{v_{c}}{\omega}$
(a) From the angular kinematical equation
On the other hand $x=v t$, (where $x$ is the $x$ coordinate of the C.M.)
or, $t=\frac{x}{v}$
From Eqs. (4) and $(5), \omega=\frac{\beta x}{v}$ Using this value of $\omega$ in Eq. (3) we get $\dot{y}=\frac{v_{c}}{\omega}=\frac{v}{\beta x / v}=\frac{\nu^{2}}{\beta x}$ (hyperbola)
(b) As centre $C$ moves with constant acceleration $w$, with zero initial velocitySo, $x=\frac{1}{2} w t^{2}$ and $v_{c}=w t$
Therefore, $v_{c}=w \sqrt{\frac{2 x}{w}}=\sqrt{2 x w}$
Hence $y=\frac{v_{c}}{\omega}=\frac{\sqrt{2 w x}}{\omega}$ (parahola)

Narayan Hari
Narayan Hari
Numerade Educator
03:39

Problem 52

The plane motion of a solid can be imagined as the combination of translation of the C.M. and rotation about C.M. So, we may write $\overrightarrow{v_{A}}=\overrightarrow{v_{C}}+\overrightarrow{v_{A} C}$
$=\overrightarrow{v_{C}}+\vec{\omega} \times \overrightarrow{r_{A} C} \quad$ (1) and
$\vec{w}_{A}=\vec{w}_{C}+\vec{w}_{A C}$
$=\vec{w}_{C}+\omega^{2}\left(-\overrightarrow{r_{A} c}\right)+\left(\vec{\beta} \times \overrightarrow{r_{A} C}\right)$
$\overrightarrow{r_{A C}}$ is the position of vector of $\boldsymbol{A}$ with respect to $\mathrm{C}$ In the problem $v_{C}=v=$ constant, and the rolling is without slipping i.e., $v_{C}=v=\omega R$, So, $w_{C}=0$ and $\beta=0 .$ Using these conditions in Eq. (2)
Here, $\hat{u}_{A C}$ is the unit vector directed along $\overrightarrow{r_{A} C}$. Hence $w_{A}=\frac{v^{2}}{R}$ and $\vec{w}_{A}$ is directed along $\left(-\hat{u}_{A C}\right)$ or directed toward the centre of the wheel.
(b) Let the centre of the wheel move toward right (positive $x$ -axis) then for pure tolling on the rigid horizontal surface, wheel will have to rotate in clockwise sense. If $\omega$ be the angular velocity of the wheel then $\omega=\frac{v_{C}}{R}=\frac{v}{R}$. Let the point $A$ touches the horizontal surface at $t=0$, further let us locate the point $A$ at $t=t$, When it makes $\theta=\omega t$ at the centre of the wheel. From Eqn. (1) $\overrightarrow{v_{A}}=\overrightarrow{v_{C}}+\vec{\omega} \times \overrightarrow{r_{A C}}$
$=v \overrightarrow{i+} \omega(-\vec{k}) \times[R \cos \theta(-\vec{j})+R \sin \theta(-\vec{i})]$
or, $\overrightarrow{v_{A}}=v \vec{i}+\omega R[\cos \omega t(-\overrightarrow{i)}+\sin \omega t \vec{j}]$
$=(v-\cos \omega t) \vec{i}+v \sin \omega t \vec{j} \quad($ as $v=\omega R)$So, $\quad v_{A}=\sqrt{(v-v \cos \omega t)^{2}+(v \sin \omega t)^{2}}$
$$
=v \sqrt{2(1-\cos \omega t)}=2 v \sin (\omega t / 2)
$$
Hence distance covered by the point $A$ during $T=2 \pi / \omega$
$$
s=\int v_{A} d t=\int_{0}^{2 \pi / \omega} 2 v \sin (\omega t / 2) d t=\frac{8 v}{\omega}=8 R
$$

Narayan Hari
Narayan Hari
Numerade Educator
11:25

Problem 53

Let us fix the co-ordinate axis $x y z$ as shown in the fig. As the ball rolls without slipping along the rigid surface so, on the basis of the solution of problem $1.52$ :
$\overrightarrow{v_{0}}=\overrightarrow{v_{c}}+\vec{\omega} \times \overrightarrow{r_{o c}}=0$
Thus $v_{c}=\omega R$ and $\vec{\omega} \uparrow \uparrow(-\vec{k})$ as $\left.\vec{v}_{c} \uparrow \uparrow \vec{i}\right\}$$\mathbf{3 1}$
and $w_{c}=\beta R$ and $\vec{\omega}_{c}+\vec{\beta} \times \vec{r}_{o c}=0$
At the position corresponding to that of Fig., in accordance with the problem, $w_{c}=w$, so $v_{c}=w t$
and $\quad \omega=\frac{v_{c}}{R}=\frac{w t}{R}$ and $\beta=\frac{w}{R}$ (using 1)
(a) Let us fix the co-ordinate system with the frame attached with the rigid surface as shown in the Fig. As point $O$ is the instantaneous centre of rotation of the ball at the moment shown in Fig.
so, $\overrightarrow{v_{0}}=0$
Now, $\overrightarrow{v_{A}}=\overrightarrow{v_{C}}+\vec{\omega} \times \overrightarrow{r_{A C}}$
$=v_{C} \vec{i}+\omega(-\vec{k}) \times R(\vec{j})=\left(v_{C}+\omega R\right) \vec{i}$
So, $\overrightarrow{v_{A}}=2 v_{C} \vec{i}=2 w t \vec{i}$ (using
1) Similalry $\overrightarrow{v_{B}}=\overrightarrow{v_{C}}+\vec{\omega} \times \overrightarrow{r_{B C}}=v_{C} \vec{i}+\omega(-\vec{k}) \times R(\overrightarrow{i)}$
$=v_{C} \vec{i}+\omega R(-\vec{j})=v_{C} \vec{i}+v_{C}(-\vec{j})$
So, $v_{B}=\sqrt{2} v_{c}=\sqrt{2} w t$ and $\vec{v}_{B}$ is at an angle $45^{\circ}$ from both $\vec{i}$ and $\vec{j}$ (Fig.)
(b) $\vec{w}_{0}=\vec{w}_{C}+\omega^{2}\left(-\vec{r}_{o c}\right)+\vec{\beta} \times \vec{r}_{O C}$
$=\omega^{2}\left(-\vec{r}_{O C}\right)=\frac{\nu_{C}^{2}}{R}\left(-\hat{u}_{O C}\right)$ (using 1)
where $\hat{u}_{o c}$ is the unit vector along $\vec{r}_{o c}$
so, $w_{0}=\frac{v_{0}^{2}}{R}=\frac{w^{2} t^{2}}{R}$ (using 2) and $\vec{w}_{0}$ is
directed towards the centre of the ball Now $\vec{w}_{A}=\vec{w}_{C}+\omega^{2}\left(-\vec{r}_{A C}\right)+\vec{\beta} \times \vec{r}_{A C}$
$=w \vec{i}+\omega^{2} R(-\vec{j})+\beta(-\vec{k}) \times R \vec{j}$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:11

Problem 54

Let us draw the kinematical diagram of the rolling cylinder on the basis of the solutiol of problem $1.53$.
As, an arbitrary point of the cylinder follows a curve, its normal acceleration and radius of curvature are related by the well known equation
$w_{n}=\frac{v^{2}}{R}$
so, for point $A$, $w_{A(n)}=\frac{v_{A}^{2}}{R_{A}}$
or, $R_{A}=\frac{4 v_{c}^{2}}{\omega_{r}^{2}}=4 r$ (because $v_{c}=\omega r$, for pure rolling $)$
Similarly for point $B$
$\omega^{2} r \cos 45^{\circ}=\frac{\left(\sqrt{2} v_{c}\right)^{2}}{R_{B}}$
or, $R_{B}=2 \sqrt{2} \frac{v_{C}^{2}}{\omega^{2} r}=2 \sqrt{2} r$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
07:09

Problem 55

The angular velocity is a vector as infinitesimal rotation commute. Then the relative angular velocity of the body 1 with respect to the body 2 is clearly.
$$
\vec{\omega}_{12}=\overrightarrow{\omega_{1}}-\overrightarrow{\omega_{2}}
$$
as for relative linear velocity. The relative acceleration of 1 w.r.t. 2 is
$$
\left(\frac{d \overrightarrow{\omega_{1}}}{d t}\right)_{s^{\prime}}
$$where $S^{\prime}$ is a frame corotating with the second body and $S$ is a space fixed frame with origin coinciding with the point of intersection of the two axes,
but
$$
\left(\frac{d \vec{\omega}_{1}}{d t}\right)_{s}=\left(\frac{d \vec{\omega}_{1}}{d t}\right)_{s}+\vec{\omega}_{2} \times \vec{\omega}_{1}
$$
Since $S^{\prime}$ rotates with angular velocity $\vec{w}_{2}$. However $\left(\frac{d \vec{\omega}_{1}}{d t}\right)_{s}=0$ as the first body rotates with constant angular velocity in space, thus
$$
\overrightarrow{\beta_{12}}=\overrightarrow{\omega_{1}} \times \overrightarrow{\omega_{2}}
$$
Note that for any vector $\vec{b}$, the relation in space forced frame $(k)$ and a frame $\left(k^{\prime}\right)$ rotating with angular velocity $\vec{\omega}$ is
$$
\left.\frac{d \vec{b}}{d t}\right|_{K}=\left.\frac{d \bar{b}}{d t}\right|_{K}+\vec{\omega} \times \vec{b}
$$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
01:47

Problem 56

We have $\quad \vec{\omega}=$ at $\vec{i}+b t^{2} \vec{j}$
So, $\omega=\sqrt{(a t)^{2}+\left(b t^{2}\right)^{2}}$, thus, $\left.\omega\right|_{t=10 s}=7.81 \mathrm{rad} / \mathrm{s}$
Differentiating Eq. (1) with respect to time
So,
$$
\begin{aligned}
&\vec{\beta}=\frac{d \vec{\omega}}{d t}=a \overrightarrow{i+2} b t \vec{j} \\
&\beta=\sqrt{a^{2}+(2 b t)^{2}}
\end{aligned}
$$
and
$$
\left.\beta\right|_{t=10 s}=1 \cdot 3 \mathrm{rad} / \mathrm{s}^{2}
$$
(b)
$$
\cos \alpha=\frac{\vec{\omega} \cdot \vec{\beta}}{\omega \beta}=\frac{\left(a t \vec{i}+b t^{2} \vec{j}\right) \cdot(a \vec{i}+2 b t \vec{j})}{\sqrt{(a t)^{2}+\left(b t^{2}\right)^{2}} \sqrt{a^{2}+(2 b t)^{2}}}
$$
Putting the values of $(a)$ and $(b)$ and taking $t=10 \mathrm{~s}$, we get $\alpha=17^{\circ}$

Narayan Hari
Narayan Hari
Numerade Educator
05:52

Problem 57

(a) Let the axis of the cone $(O C)$ rotates in anticlockwise sense with constant angular velocity $\vec{\omega}$ and the cone itself about it's own axis $(O C)$ in clockwise sense with angular velocity $\overrightarrow{\omega_{0}}$ (Fig.). Then the resultant angular velocity of the cone. $\vec{\omega}=\vec{\omega}+\overrightarrow{\omega_{0}}$
As the rolling is pure the magnitudes of the vectors $\vec{\omega}$ and $\overrightarrow{\omega_{0}}$ can be easily found from Fig.
(2)
As $\vec{\omega} \perp \vec{\omega}_{0}$ from Eq. (1) and (2)$$
\begin{aligned}
\omega=& \sqrt{\omega^{\prime 2}+\omega_{0}^{2}} \\
& \sqrt{\left(\frac{v}{R \cot \alpha}\right)^{2}+\left(\frac{v}{R}\right)^{2}}=\frac{v}{R \cos \alpha}=2 \cdot 3 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
(b) Vector of angular acceleration
$$
\vec{\beta}=\frac{d \vec{\omega}}{d t}=\frac{d\left(\vec{\omega}+\vec{\omega}_{0}\right)}{d t} \text { (as } \vec{\omega}=\text { constant.) }
$$
The vector $\vec{\omega}_{0}$ which rotates about the $O O^{\prime}$ axis with the angular velocity $\vec{\omega}$, retains $\mathrm{i}$ magnitude. This increment in the time interval dt is equal to
$\left|d \vec{\omega}_{0}\right|=\omega_{0} \cdot \omega^{\prime} d t$ or in vector form $d \vec{\omega}_{0}=\left(\vec{\omega} \times \vec{\omega}_{0}\right) d t .$
Thus $\vec{\beta}=\vec{\omega}^{\prime} \times \vec{\omega}_{0}$
The magnitude of the vector $\vec{\beta}$ is equal to
$$
\beta=\omega^{\prime} \omega_{0}\left(\operatorname{as} \vec{\omega}^{\prime} \perp \vec{\omega}_{0}\right)
$$
So, $\beta=\frac{v}{R \cot \alpha} \frac{v}{R}=\frac{\nu^{2}}{R^{2}} \tan \alpha=2 \cdot 3 \mathrm{rad} / \mathrm{s}$

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
02:48

Problem 58

The axis $A B$ acquired the angular velocity $\vec{\omega}=\overrightarrow{\beta_{o}} t$
Using the facts of the solution of $1.57$, the angular velocity of the body
$$
\omega=\sqrt{\omega_{o}^{2}+\omega^{\prime 2}}
$$
$=\sqrt{\omega_{0}^{2}+\beta_{0}^{2} t^{2}}=0 \cdot 6 \mathrm{rad} / \mathrm{s}$
And the angular acceleration. $\vec{\beta}=\frac{d \vec{\omega}}{d t}=\frac{d\left(\vec{\omega}+\vec{\omega}_{0}\right)}{d t}=\frac{d \vec{\omega}}{d t}+\frac{d \vec{\omega}_{0}}{d t}$
But $\frac{d \vec{\omega}_{0}}{d t}=\vec{\omega} \times \vec{\omega}_{0}$, and $\frac{d \vec{\omega}}{d t}=\overrightarrow{\beta_{0}} t$
So, $\quad \vec{\beta}=\left(\overrightarrow{\beta_{0}} t \times \overrightarrow{\omega_{0}}\right)+\overrightarrow{\beta_{0}}$
As, $\overrightarrow{\beta_{0}} \perp \overrightarrow{\omega_{0}}$ so, $\beta=\sqrt{\left(\omega_{0} \boldsymbol{\beta}_{0} t\right)^{2}+\beta_{0}^{2}}=\beta_{0} \sqrt{1+\left(\omega_{0} t\right)^{2}}=0.2 \mathrm{rad} / \mathrm{s}^{2}$

Narayan Hari
Narayan Hari
Numerade Educator