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

1 Answer
Jun 7, 2016

Answer:

The torque needed is #80pi N*m#

Explanation:

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

Moment of inertia of a rod is
#I=(ml^2)/12=(6*12^2)/12kg*m^2=72kg*m^2#
The angular velocity has to change by #5Hz=5(rev)/s# during #9s#.
Thus angular acceleration is #epsilon=5/9 (rev)/s^2#

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=I epsilon=72*5/9kg*m^2*(rev)/s^2=40(kg*m^2)/s^2*2pi rad#
#=80pi N*m#

As you see #rad=m/m# is dimensionless so it could vanish in presence of other units. Also torque has dimension identical with energy (even #N*m=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.