An object with a mass of # 2 kg# is traveling in a circular path of a radius of #4 m#. If the object's angular velocity changes from # 2 Hz# to # 5 Hz# in # 3 s#, what torque was applied to the object?

1 Answer
Mar 9, 2016

Answer:

Torque: #\tau = \frac{dL}{dt}=\frac{d}{dt}(I\omega)= I\dot(\omega)=2\pi.mr^2\dot{f} = 16\pi \quad N.m#

Explanation:

#\omega = 2\pif; \qquad \dot{\omega} = 2\pi\dot{f}; \qquad I = mr^2#

Torque: #\tau = \frac{dL}{dt}=\frac{d}{dt}(I\omega)= I\dot(\omega)=2\pi.mr^2\dot{f}#

#m=2\quad kg; \qquad r = 4\quad m; \dot{f}=\frac{\Delta f}{\Delta t}=(5Hz-3Hz)/(3\quad s)=1s^{-2}#

#\tau = 2\pi.(2\quad kg)(4\quad m)^2(1s^{-2})=16\pi \quad N.m#