What torque would have to be applied to a rod with a length of #1 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #6 Hz# over #4 s#?

1 Answer
Feb 10, 2016

Answer:

The angular acceleration is #(3pi) text{radians} * s^-2# and the moment of inertia is 2 kg #m^2#. The required torque is the product of these, which is 6#pi# Nm, or approximately 18.8 Nm

Explanation:

First find the angular acceleration, #alpha#.

#alpha = text{change in angular velocity} / text{time}#

The angular velocity is 6 Hz, and 1 Hz means that a circle of 2#pi# radians is traversed in 1 second. Thus the angular velocity is 6 (2 #pi# radians) #s^-1#, which is 12 #pi# radians #s^-1#

#alpha# = 6 (2 #pi# *radians) #s^-1#) / 4 s = (3 #pi# * radians) #s^-2#

Torque = moment of inertia * #alpha#

moment of inertia depends on the shape of the spinning object and can be complicated. Fortunately for a mass on a rod it's simple:
moment of inertia = mass * #text{rod length}^2#
so:
moment of inertia = 2kg * #1m^2# = 2 kg #m^2#

Now we have:

Torque = (2 kg #m^2#) * ((#3 pi) text{radians} (s^-2)#)

Torque = 6#pi# Nm,
or approximately :
Torque #~# 18.8 Nm