An object with a mass of # 3 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 1 Hz# to # 3 Hz# in # 5 s#, what torque was applied to the object?

1 Answer
Jan 4, 2016

Answer:

#=60.75N.m#.

Explanation:

From rotational dynamics we get that #sumvectau=Ivecalpha#, where #tau# is the torque, #I# is the moment of inertia, and #alpha# is the angular acceleration.

But since angular acceleration is the rate of change in angular velocity, we get that #alpha=(domega)/(dt)#.

We then first need to convert the linear frequency in Hertz into angular frequency in rad/s.

#alpha=(domega)/(dt)=((3xx2pir)-(1xx2pir))/5=5.0265rad//s^2#.

Moment of inertia #I=sum_jm_jr_(perpj)^2=int_mr^2dm#.

So in this case for a point mass, #I=mr^2=3xx2^2=12kg.m^2#.

Therefore resultant torque is #tau=I alpha#

#=12xx5.0265#

#=60.75N.m#.