Question

A continuous conveyor belt passes over two discs A and B of radii $R$ and $4R$ respectively. They start from rest. A has an angular acceleration of $\frac{\pi }{4}rad/s^2$ What is the angular velocity(${\omega }_2$) of disc B at the end of $32sec$?

Answer 1

Step 1: Given data-

Radius of disc A, $R_1=R$

The radius of disc B, $R_2=4R$

Angular acceleration of A, $\alpha =\frac{\pi }{4}rad/s^2$

Time, $t=32s$

Step 2: To find

The angular velocity(${\omega }_2$) of disc B at the end of $32sec$

Step 3: Calculation

According to equations of kinematics,

$$\begin{gathered} \omega =\alpha t \\ \omega =Angularvelocity \\ \alpha =Angularacceleration \\ t=time \end{gathered}$$

${\omega }_1=\frac{\pi }{4}\times 32=8\pi rad/s$

The linear velocity of the belt at A,

$$\begin{gathered} v=r\omega \\ v=linearvelocity \\ \omega =angularvelocity \\ r=radiusofdisc \end{gathered}$$

$v=R_1{\omega }_1=8\pi R$

Since the velocity of the belt is constant as both the disc are moving with the same linear velocity as the velocity of the belt.

Velocity at disc B will be equal to the velocity at disc A

$v=R_2{\omega }_2$

${\omega }_2=\frac{v}{R_2}=\frac{8\pi R}{4R}=2\pi rad/s^{}$

Thus, the angular velocity of disc B at the end of $32$ seconds is $2\pi rad/s$.