What torque would have to be applied to a rod with a length of 12m and a mass of 6kg to change its horizontal spin by a frequency 5Hz over 9s?

1 Answer
Jun 7, 2016

The torque needed is 80πNm

Explanation:

Assuming rotation around the center of mass and rotation only ever happens in one axis. rev is full revolution (=360=2πrad)

Moment of inertia of a rod is
I=ml212=612212kgm2=72kgm2
The angular velocity has to change by 5Hz=5revs during 9s.
Thus angular acceleration is ε=59revs2

Moment of inertia is rotational analogue of mass and torque is angular brother of force.

We could then use our intuition from Newtons second law to rotational motion. As the force F=ma, the torque M=Iε=7259kgm2revs2=40kgm2s22πrad
=80πNm

As you see rad=mm is dimensionless so it could vanish in presence of other units. Also torque has dimension identical with energy (even Nm=J), but it shouldn't be associated, because it's entirely different concept. Torque is measured in newtonmeters.

It is funny thou, that almost every mechanical quantity (lenght, velocity, acceleration, mass, momentum, force) has its own angular sibling.