# 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 #5 Hz# over #4 s#?

##### 1 Answer

#### Answer:

#### Explanation:

The net torque,

Assuming that this is a thin, rigid rod, with the axis of rotation through its center, the moment of inertia,

#I=1/12*2kg*(5m)^2#

#=25/6kgm^2#

The angular acceleration of the rod as it rotates,

#alpha=(Δomega)/( Δt)# .

We can find

#omega=2pi(5s^-1)#

#=10pi(rad)/s#

Because the problem states that the frequency *changed* by this amount, this is

#alpha=(10pi(rad)/s)/(4s)#

#=(5pi)/2(rad)/s^2#

Now that we have values for

#tau=Ialpha#

#=25/6kgm^2*(5pi)/2(rad)/s^2#

#=(125pi)/12Nm#

#≈33Nm#