What torque would have to be applied to a rod with a length of #3 m# and a mass of #4 kg# to change its horizontal spin by a frequency #7 Hz# over #1 s#?

Question

1056 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-26T08:36:40

The torque (for the rod rotating about the center) is #=131.95Nm#
The torque (for the rod rotating about one end) is #=527.8Nm#

The torque is the rate of change of angular momentum

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

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

#I=1/12*mL^2#

#=1/12*4*3^2= 3 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(7)/1*2pi#

#=(14pi) rads^(-2)#

So the torque is #tau=3*(14pi) Nm=42piNm=131.95Nm#

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

#I=1/3*mL^2#

#=1/3*4*3^2=12kgm^2#

So,

The torque is #tau=12*(14pi)=527.8Nm#