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

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Answer 1 · 2017-04-26T15:35:51

The torque $=15.1Nm$

The torque is the rate of change of angular momentum

$tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

For the object, $I=(mr^2)$

So, $I=3*(2)^2s=12kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(3-2)/5*2pi$

$=(2/5pi) rads^(-2)$

So the torque is $tau=12*(2/5pi) Nm=15.1Nm$

An object with a mass of 3 kg 3 kg is traveling in a circular path of a radius of 2 m2 m. If the object's angular velocity changes from 2 Hz 2 Hz to 3 Hz 3 Hz in 5 s 5 s, what torque was applied to the object?