What torque would have to be applied to a rod with a length of 5m and a mass of 2kg to change its horizontal spin by a frequency of 2Hz over 3s?

1 Answer
Sep 29, 2017

The torque for the rod rotating about the center is =17.45Nm
The torque for the rod rotating about one end is =69.81Nm

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

=112252=256kgm2

The rate of change of angular velocity is

dωdt=232π

=(43π)rads2

So the torque is τ=256(43π)Nm=509πNm=17.45Nm

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

I=13mL2

=13252=503kgm2

So,

The torque is τ=503(43π)=2009π=69.81Nm