What torque would have to be applied to a rod with a length of #5 m# and a mass of #2 kg# to change its horizontal spin by a frequency of #2 Hz# over #2 s#?

1 Answer
Mar 27, 2017

Answer:

The torque, for the rod rotating about the center is #=26.2Nm#
The torque, for the rod rotating about one end is #=104.7Nm#

Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of a rod, rotating about the center is

#I=1/12*mL^2#

#=1/12*2*5^2= 25/6 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(2)/2*2pi#

#=(2pi) rads^(-2)#

So the torque is #tau=25/6*(2pi) Nm=25/3piNm=26.2Nm#

The moment of inertia of a rod, rotating about one end is

#I=1/3*mL^2#

#=1/3*2*5^2=50/3kgm^2#

So,

The torque is #tau=50/3*(2pi)=100/3pi=104.7Nm#