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 $6 Hz$ in $5 s$ , what torque was applied to the object?

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Answer 1 · 2017-07-19T06:31:12

The torque was $=60.3Nm$

The torque is the rate of change of angular momentum

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

The moment of inertia of the object is

$I=mr^2$

$=3*2^2= 12 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(6-2)/5*2pi$

$=(8/5pi) rads^(-2)$

So the torque is $tau=12*(8/5pi) Nm=96/5piNm=60.3Nm$

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 6 Hz 6 Hz in 5 s 5 s, what torque was applied to the object?