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 $\dfrac{t}{4}$. If the equal torques act the discs A and B, initially both being at rest. At a later instant, the angular speeds of a point on the rim of B are ${\omega _A}$ and ${\omega _B}$ respectively. Then we have:
A) ${\omega _A} > {\omega _B}$.
B) ${\omega _A} = {\omega _B}$.
C) ${\omega _A} < {\omega _B}$.
D) The relation depends on the actual magnitude of the torques.

Answer 1

Hint: The torque is a kind of twisting force which results in rotation of anybody. The angular velocity is the rate of revolving of the body around a fixed point. Angular acceleration is the acceleration of the body revolving around a fixed point.

Formula used:
The formula of the torque is given by,
$ \Rightarrow T = I \cdot \alpha $
Where torque is $T$ the moment of inertia is $I$ and the angular acceleration is $\alpha $.
The formula of the angular acceleration is given by,
$ \Rightarrow \omega = \alpha \cdot \left( {time} \right)$
Where $\omega $ is angular velocity the angular acceleration is$\alpha $.

Complete step by step solution:
It is given in the problem that 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 $\dfrac{t}{4}$ if the equal torques act the discs A and B, initially both being at rest at a later instant, the angular speeds of a point on the rim of B are ${\omega _A}$ and ${\omega _B}$ respectively then we need to find the relation between the angular speeds of discs A and B.
The torque on both the discs is same therefore we get,
$ \Rightarrow {T_A} = {T_B}$
$ \Rightarrow {I_A} \cdot {\alpha _A} = {I_B} \cdot {\alpha _B}$………eq. (1)
The moment of inertia of the disc A is equal to,
$ \Rightarrow {I_A} = {M_A}{r^2}$
$ \Rightarrow {I_A} = \pi \rho t{r^4}$………eq. (2)
The moment of inertia of the disc B is equal to,
$ \Rightarrow {I_B} = {M_B}{\left( {4r} \right)^2}$
$ \Rightarrow {I_B} = \pi \rho t \cdot \left( {64{r^4}} \right)$………eq. (3)
Replace the value of ${I_A}$ and ${I_B}$ from equation (2) and equation (3) in equation (1).
$ \Rightarrow {I_A} \cdot {\alpha _A} = {I_B} \cdot {\alpha _B}$
$ \Rightarrow \left( {\pi \rho t{r^4}} \right) \cdot {\alpha _A} = \left[ {\pi \rho t \cdot \left( {64{r^4}} \right)} \right] \cdot {\alpha _B}$
$ \Rightarrow {\alpha _A} = 64 \cdot {\alpha _B}$………eq. (4)
Since the formula of the angular acceleration is given by,
$ \Rightarrow \omega = \alpha \cdot \left( {time} \right)$
Where $\omega $ is angular velocity the angular acceleration is$\alpha $.
\[ \Rightarrow \dfrac{{{\omega _A}}}{{{\alpha _A}}} = \dfrac{{{\omega _B}}}{{{\alpha _B}}}\]
Replacing the value of the ${\alpha _A}$ in the above relation we get,
\[ \Rightarrow \dfrac{{{\omega _A}}}{{{\alpha _A}}} = \dfrac{{{\omega _B}}}{{{\alpha _B}}}\]
\[ \Rightarrow \dfrac{{{\omega _A}}}{{64 \cdot {\alpha _B}}} = \dfrac{{{\omega _B}}}{{{\alpha _B}}}\]
\[ \Rightarrow \dfrac{{{\omega _A}}}{{64}} = {\omega _B}\]
\[ \Rightarrow \dfrac{{{\omega _A}}}{{{\omega _B}}} = 64\]
$ \Rightarrow {\omega _A} > {\omega _B}$.
The relation between the angular velocity of the discs A and B is equal to ${\omega _A} > {\omega _B}$.

The correct answer for this problem is option (A).

Note: It is advisable for students to remember the formula of the torque also the formula of the moment of inertia of the disc. The ratio of the angular acceleration of the disc is proportional to the ratio of the angular velocities.