What torque would have to be applied to a rod with a length of 4m and a mass of 5kg to change its horizontal spin by a frequency of 3Hz over 7s?

1 Answer
Aug 24, 2017

The torque for the rod rotating about the center is =17.95Nm
The torque for the rod rotating about one end is =71.81Nm

Explanation:

The torque is the rate of change of angular momentum

τ=dLdt=d(Iω)dt=Idωdt

The moment of inertia of a rod, rotating about the center is

I=112mL2

=112542=203kgm2

The rate of change of angular velocity is

dωdt=372π

=(67π)rads2

So the torque is τ=203(67π)Nm=407πNm=17.95Nm

The moment of inertia of a rod, rotating about one end is

I=13mL2

=13542=803kgm2

So,

The torque is τ=803(67π)=1607π=71.81Nm