What torque would have to be applied to a rod with a length of #6 m# and a mass of #3 kg# to change its horizontal spin by a frequency #5 Hz# over #4 s#?

1 Answer
Sep 3, 2016

Answer:

#tau=657.75# #N-m#

Explanation:

The torque is given by:
#tau=I*dot omega=I*omega/t#
where #I# is the moment of Inertia of the horizontal spinning rod which is given by: #I=(m*l^2)/12#
and #omega#, the angular velocity is given by #omega=2pi*f#

So, from the given data
#I=(3*6^2)/12=9# #kg-m^2#
#omega=2pi * 5=31.4# #(rad)/s#
#t=4# #s#

Thus we can now evaluate the torque
#tau=I*omega/t#
#tau=9*31.4/4=657.75# #N-m#