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?

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Answer 1 · 2016-03-09T10:46:15

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

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

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

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

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

An object with a mass of 2 kg 2 kg is traveling in a circular path of a radius of 4 m4 m. If the object's angular velocity changes from 2 Hz 2 Hz to 5 Hz 5 Hz in 3 s 3 s, what torque was applied to the object?