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

1 Answer
May 13, 2017

The torque for the rod rotating about the center is =14.14Nm
The torque for the rod rotating about one end is =56.55Nm

Explanation:

The torque is the rate of change of angular momentum

tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt

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

I=1/12*mL^2

=1/12*2*3^2= 1.5 kgm^2

The rate of change of angular velocity is

(domega)/dt=(3)/2*2pi

=(3pi) rads^(-2)

So the torque is tau=1.5*(3pi) Nm=4.5piNm=14.14Nm

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

I=1/3*mL^2

=1/3*2*3^2=6kgm^2

So,

The torque is tau=6*(3pi)=18pi=56.55Nm