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 #9 Hz# over #6 s#?

1 Answer
Feb 21, 2016

Answer:

#L=(25*pi)/2#

Explanation:

#L=alpha*I#
#"(L:torque ;alpha :angular acceleration ;I:the moment of inertia)"#
#alpha=(Delta omega)/(Delta t)#
#alpha=(2*pi*Delta f)/(Delta t)#
#alpha=(2*pi*9)/6#
#alpha=3*pi#
#I=1/12*m*l^2 " for a rod spinning around its center"#
#I=1/12*2*5^2=25/6#
#L=3*pi*25/6#
#L=(25*pi)/2#