What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #3 Hz# over #2 s#?

1 Answer
Nov 18, 2016

Answer:

The torque applied to the rod is 37.68 Newton-metre.

Explanation:

Torque is defined as the product of moment of inertia and angular acceleration
#=>tau=Ialpha#
Moment of inertia of rod is #I_"rod"=((ML^2)/12)# and angular acceleration is change in angular velocity with respect to time
#=>alpha=omega/t#
angular velocity #omega=2pinu# as frequency is given in above case.
Now the torque becomes #tau=[(ML^2)/12][(2pinu)/t]#
Substituting the given values we get,
#tau=[[(3kg)(4m)^2]/12][(2xx3.14xx3Hz)/(2s)]#
#:.#torque# tau=37.68#N-m.