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 14Hz over 6s?

1 Answer
Feb 24, 2017

The torque (for the rod rotating about the center) is =61.6Nm
The torque (for the rod rotating about one end) is =244.3Nm

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=1462π

=(143π)rads2

So the torque is τ=256(143π)Nm=1759πNm=61.1Nm

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

I=13mL2

=13252=503kgm2

So,

The torque is τ=503(143π)=7009π=244.3Nm