An object with a mass of # 2 kg# is traveling in a circular path of a radius of #4 m#. If the object's angular velocity changes from # 4 Hz# to # 5 Hz# in # 3 s#, what torque was applied to the object?

1 Answer
Sep 14, 2017

We will need to know the angular acceleration that existed during those 3 s. If this was a linear motion problem we could use the suvat formula
#v = u + a*t#
The equivalent formula in angular motion is
#omega_f = omega_i + alpha*t#

Using that last formula,
#5*(2*pi " radians")/s = 4*(2*pi " radians")/s + alpha*3 s#

Solving for #alpha#,
#alpha = ((2*pi " radians")/s)/(3 s) = 2.09 " radians"/s^2#

Next we need the angular motion equivalent of F=m*a (Newton's 2nd Law). The equivalent is

#tau = I*alpha#

I will assume the object can be considered a point mass. This requires that the dimensions of the object be much smaller than the 4 m radius. With that assumption, the rotational inertia, #I#, of the rotating mass is
#I = m*r^2 = 2 kg*(4 m)^2 = 32 kg*m^2#

So now we can calculate the torque, #tau#.

#tau = 32 kg*m^2 * 2.09 " radians"/s^2 = 67 kg*m^2/s^2#
#tau = 67 N*m#

You might ask how the combination of units #kg*m^2/s^2# simplified into #N*m#. Note that you can group #kg*m/s^2# apart from the other m. And #kg*m/s^2# is the definition of the Newton.

Torque is the product of force and the length of the lever arm. So, we have our answer.

I hope this helps,
Steve