An object with a mass of $3 kg$ is traveling in a circular path of a radius of $3 m$ . If the object's angular velocity changes from $4 Hz$ to $5 Hz$ in $8 s$ , what torque was applied to the object?

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Answer 1 · 2017-01-16T15:09:53

The torque is $=21.2Nm$

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 $I=mr^2$

$=3*3^2= 27 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(5-4)/8*2pi$

$=((pi)/4) rads^(-2)$

So the torque is $tau=27*(pi)/4 Nm=27/4piNm=21.2Nm$

An object with a mass of 3 kg 3 kg is traveling in a circular path of a radius of 3 m3 m. If the object's angular velocity changes from 4 Hz 4 Hz to 5 Hz 5 Hz in 8 s 8 s, what torque was applied to the object?