Question

A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is

• A. $0.1\,kgm^2s^{-1}$
• B. $0.04\,kgm^2s^{-1}$
• C. $0.025\,kgm^2s^{-1}$
• D. $0.01\,kgm^2s^{-1}$

Answer 1

The Correct Option is A

Rate of change of angular momentum is equal to the torque:
\(\tau=\frac {dL}{dt}\)

\(\tau= \frac {(L_f-L_i)}{△ t}\)
Where,
\(L_f\)= Finial angular momentum
\(L_i\)= Initial angular momentum.
As external force is mentioned, therefore we can say the torque is retarding which will be in negative.
\(-0.02\ Nm=\frac {0-Li}{5}\)

\(0.02\ Nm=\frac {Li-0}{5}\)
\(Li=0.10\ kgm^2s^{-1}\)

So, the correct option is (A): \(0.10\ kgm^2s^{-1}\).

Answer 2

\(C = 0.02\) N/m
\(t = 5\) sec
Torque is applied for \(5 \ sec\) and torque is rate of change of angular momentum
Therefore, angular momentum lost due to decelaration \(= \bar C\times \Delta t = (0.02) \times 5 = 0.1\)
change in angular momentum = final angular momentum - initial angular momentum
\(⇒\) 0.1 = 0 - Initial angular momentum
\(⇒\) Initial angular momentum = \(0.1 \ kg m^2s^{-1}\)

So, the correct option is (A): \(0.1 \ kg m^2s^{-1}\)