Question

A uniform circular disc $A$ of radius $r$ is made from a metal plate of thickness $t$ and another uniform circular disc $B$ of radius $4r$ is made from the same metal plate of thickness $t/4$. Equal torques act on the discs $A$ and $B$, initially both being at rest. At a later instant, the angular speeds of a point on rim $A$ and another point on rim $B$ are ${\omega }_A$ and ${\omega }_B$ respectively. Find the relation between ${\omega }_A$ and ${\omega }_B$ .

Answer 1

Uniform circular disc A of radius $=r$

Thickness$=5$

Another uniform circular disc B of radius $=4r$

Thickness $=t/4$

Torque will be equal to same disc A and B initially

Angular speed of a point on rim

$A=w_A$

Angular speed of a point on

Rin B$=w_B$

${\tau }_A=Fl\sin \theta$

$\frac{P}{w_A}=P_{pressure}\times \pi r^2\times t=Pressure\times \pi r^2\times t\to \left(1\right){\tau }_B=Pressure\times \pi \left(16r^2\right)\times \frac{5}{4}\to \left(2\right)$

Now modifying equation $\left(1\right)$ and $\left(2\right)$ and we get,

$P_{pressure}\times \pi r^2\times t{w}_A=P\to \left(3\right)Pressure\times \pi 4r^{2}t{w}_B=P\to \left(4\right)$

Now equating this equation $\left(3\right)$ and $\left(4\right)$ and we will get,

$\pi r^{2}t{w}_a=\pi 4r^{2}t{w}_b\frac{w_a}{w_b}=\frac{4r^2{t}^2}{r^{2}t}=4:1w_A:w_B=4:1$