What torque would have to be applied to a rod with a length of #8 m# and a mass of #8 kg# to change its horizontal spin by a frequency #18 Hz# over #4 s#?

1 Answer
Apr 1, 2017

Answer:

The torque, for the rod rotating about the center is #=1206.4Nm#
The torque, for the rod rotating about one end is #=4825.5Nm#

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*8*8^2= 42.67 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(18)/4*2pi#

#=(9pi) rads^(-2)#

So the torque is #tau=42.67*(9pi) Nm=1206.4Nm#

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

#I=1/3*mL^2#

#=1/3*8*8^2=170.67kgm^2#

So,

The torque is #tau=170.67*(9pi)=4825.5Nm#