Disk A has a mass mA=4 kg, a radius rA=300 mm, and an initial angular velocity ω0=300 rpm clockw

step 1 Hello everyone here it is given. This A having the mass 4 KG and radius 300 millimeter that is 0

Disk A has a mass mA=4 kg, a radius rA=300 mm, and an...

Disk $A$ has a mass $m_{A}=4 \mathrm{kg}$, a radius $r_{A}=300 \mathrm{mm},$ and an initial angular velocity $\omega_{0}=300 \mathrm{rpm}$ clockwise. Disk $B$ has a mitial $m_{B}=1.6 \mathrm{kg}$, a radius $r_{B}=180 \mathrm{mm},$ and is at rest when it is brought into contact with disk $A .$ Knowing that $\mu_{k}=0.35$ between the disks and neglecting bearing friction, determine $(a)$ the angular acceleration of each disk, $(b)$ the reaction at the support $C .$

Disk $A$ has a mass $m_{A}=4 \mathrm{kg}$, a radius $r_{A}=300 \mathrm{mm},$ and an initial angular velocity $\omega_{0}=300 \mathrm{rpm}$ clockwise. Disk $B$ has a mitial $m_{B}=1.6 \mathrm{kg}$, a radius $r_{B}=180 \mathrm{mm},$ and is at rest when it is brought
into contact with disk $A .$ Knowing that $\mu_{k}=0.35$ between the disks
and neglecting bearing friction, determine $(a)$ the angular acceleration
of each disk, $(b)$ the reaction at the support $C .$

Key Concepts

Angular Acceleration

Angular acceleration is the rate at which the angular velocity of a body changes with time. It is directly related to the net torque acting on the system and inversely related to the moment of inertia. Understanding angular acceleration is crucial for predicting how quickly rotating bodies speed up or slow down under the influence of applied torques.

Reaction Forces in Rotational Systems

Reaction forces at supports or bearings are the forces exerted by constraints or supports to maintain the prescribed rotational motion of a body. In rotating systems, these reaction forces balance the net external forces acting perpendicular to the axis of rotation. Analyzing these forces is essential for understanding the stability and equilibrium of the system.

Frictional Torque

Frictional torque arises from frictional forces acting at a distance from the rotational axis. When two surfaces in relative motion contact each other, kinetic friction produces a torque that can either accelerate or decelerate the rotational motion of the bodies in contact. The magnitude of this torque is determined by the friction coefficient and the normal force, as well as the effective lever arm.

Rotational Dynamics

Rotational dynamics involves analyzing the motion of bodies rotating about an axis. It encompasses the study of how torques cause changes in angular velocity, paralleling Newton's second law for translational motion but in a rotational context. This includes linking net applied torques to angular accelerations based on the moment of inertia of the rotating body.

Moment of Inertia

The moment of inertia is a measure of a body's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. For common geometries, such as disks, standard formulas are used to calculate the moment of inertia which then informs how much torque is required to achieve a given angular acceleration.

Transcript

00:02 Hello everyone here it is given.

00:04 This a having the mass 4 kg and radius 300 millimeter that is 0 .3 meter and initial angular velocity 300 rpm in clockwise direction.

00:53 This be having the mass 1 .6 kg radius 180 millimeter and at rest.

01:26 And is test when it is brought into contact with the disk a knowing that coefficient of friction is 0 .35 between the disc, neglecting, wearing friction, calculate angular acceleration of each disk, alpha a and alpha v we have to calculate and the reaction at the support c.

02:04 This is the support.

02:05 So reaction at c let us see here we will draw the free body diagram for disk b and disc a let us here is the disc b the weight b b downward friction mu k into n and reaction moment submission of force along by axis to be zero there is no motion along that so y -axis we consider to be positive summation of f5 must be 0.

03:55 So n minus weight.

04:03 So normal reaction is called to weight of the disk v.

04:07 That is mass of b into g.

04:10 So friction you will get mu k into n, that is mu k m b into g.

04:45 In counter -clockwise we are taking to be positive.

04:55 Let us see here.

04:59 Friction is tangential, so due to which torque is developed.

05:02 All other forces are acting at a point of rotation.

05:05 So torque due to them will be 0.

05:09 And that will be iv into alpha v.

05:16 Force, mu k, mass of v into g, rb.

05:25 Moment of inertia of the disk, half mr square.

05:32 On solving it angular velocity of v disk you will get twice of mkg upon rv say this is equation 1 and this will be in counterclockwise direction now by using the given values you can calculate this value coefficient of friction is given 0 .35 acceleration due to gravity 9 .8 and its radius is given 180 mm that is 0 .18 meter so it would be 38 .15 radiant per second square so this is the one of the answer of part a now summation of fx to be zero because there is no tangential motion along right we are taking positive so it can be written as f minus reaction b will be zero so reaction on b you will get is called to f and f is called to mu k m b into g mu k is 0 .35 weight of the disc b is given sorry mass is given 1 .6 so on solving it reaction b on b you will get 5 .4961 in left direction now for the disk a, this is the disk a, f will be in this direction, this is normal reaction, r a and it will move in counter -clock by direction.

09:02 This is the center a...

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