What torque would have to be applied to a rod with a length of 4m and a mass of 8kg to change its horizontal spin by a frequency 9Hz over 2s?

1 Answer
Jan 11, 2018

The torque for the rod rotating about the center is =301.6Nm
The torque for the rod rotating about one end is =1206.4Nm

Explanation:

The torque is the rate of change of angular momentum

τ=dLdt=d(Iω)dt=Idωdt

The mass of the rod is m=8kg

The length of the rod is L=4m

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

I=112mL2

=112842=10.67kgm2

The rate of change of angular velocity is

dωdt=922π

=(9π)rads2

So the torque is τ=10.67(9π)Nm=301.6Nm

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

I=13mL2

=13842=42.67kgm2

So,

The torque is τ=42.67(9π)=5045π=1206.4Nm