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 # 9 Hz# in # 3 s#, what torque was applied to the object?

1 Answer

Answer:

The torque applied to the object is 53.33 Newton-metre.

Explanation:

As we know torque #tau = Ialpha#
So angular acceleration #alpha #= (change in angular velocity #omega)//(#change in time t)
#=> alpha = (omega_2-omega_1)//dt#
#=>alpha=5//3 m/s^2#
And torque now becomes #tau= mr^2alpha[I=mr^2]#
By substituting all values we get,
#tau=2kgxx(4m)^2xx(5/3m/s^2)=53.33#Newton-metre.