Question

The two rigid objects shown have the same mass, radius, and angular speed.  

If the same braking torque is applied to each, which takes longer to stop?

 

Choices:

- B

- more information is needed

- A

 

Explain.

Answer 1

Given that \(m, r, \omega\), and \(\tau\) are constants. Case I: (Solid disk-A)

Moment of inertia of solid disk, \(I_{1}=\frac{m r^{2}}{2}\) Initial angular velocity, \(\omega_{i}=\omega\) Final angular velocity, \(\omega_{f}=0\) Time taken to come to rest \(=t_{1}\)

Torque \(=\tau\) Torque, \(\tau=I_{1} \alpha_{1}\)

$$ \tau=\left(\frac{m r^{2}}{2}\right)\left(\frac{\omega_{f}-\omega_{i}}{t_{1}}\right) $$

But, \(\omega_{f}=0\)

$$ \tau=-\left(\frac{m r^{2}}{2}\right)\left(\frac{\omega_{i}}{t_{1}}\right) $$

Here, negative sign represent retarding torque.

Case I: (Ring - B) Moment of inertia of ring, \(I_{2}=m r^{2}\) Initial angular vel ocity, \(\omega_{i}=\omega\) Final angular velocity, \(\omega_{\rho}=0\) Time taken to come to rest \(=t_{2}\) Torque \(=\tau\) Torque, \(\tau=I_{2} \alpha_{2}\)

$$ \tau=\left(m r^{2}\right)\left(\frac{\omega_{f}-\omega_{i}}{t_{2}}\right) $$

But, \(\omega_{f}=0\)

$$ \tau=-\left(m r^{2}\right)\left(\frac{\omega_{i}}{t_{2}}\right) \ldots \ldots(2) $$

Here, negative sign represent retarding torque.

By dividing (1) with (2),

$$ \frac{\tau}{\tau}=\frac{-\left(\frac{m r^{2}}{2}\right)\left(\frac{a_{i}}{t_{1}}\right)}{-\left(m r^{2}\right)\left(\frac{a_{i}}{t_{2}}\right)} $$

\(1=\left(\frac{1}{2}\right)\left(\frac{t_{2}}{t_{1}}\right)\)

\(t_{2}=2 t_{1}\)

Hence, we conclude that the ring \((\mathrm{B})\) takes

longer time to come to rest.