An object with a mass of # 7 kg# is traveling in a circular path of a radius of #9 m#. If the object's angular velocity changes from # 5 Hz# to # 6 Hz# in # 8 s#, what torque was applied to the object?

2 Answers
Jan 13, 2018

Answer:

442.26 Nm

Explanation:

Torque acting on a particle is given as, #tau# = I#prop#
(Where, I is the moment if inertia and #prop# is angular acceleration)
Now,# I = m(r^2)#
Or, I= 567 SI unit
#prop# = #(domega) / dt #
= (#omega# final -#omega# initial)/#t# where #omega# is angular velocity
= #(2pi/t)# (n final - n initial) (as omega=2#pi#n) where n is frequency
= 0.78 #(rad)/(s^2)#
Hence, #tau# = #(567*0.78)#N.m
Or,442.26 N.m

Jan 13, 2018

Answer:

The torque is #=445.3Nm#

Explanation:

The torque is the rate of change of angular momentum

#tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt#

where #I# is the moment of inertia

The mass of the object is #m=7kg#

The radius of the path is #r=9m#

For the object, the moment of inertia is #I=mr^2#

So, #I=7*(9)^2=567kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(f_2-f_1)/t*2pi=(6-5)/8*2pi#

#=(1/4pi)rads^(-2)#

So,

The torque is #tau= 567*1/4pi=445.3Nm#