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 #15 Hz# over #6 s#?

1 Answer
Dec 28, 2016

Answer:

#tau≈65Nm#

Explanation:

The net torque, #tau#, can be expressed in terms of the moment of inertia and angular acceleration, where #tau=Ialpha#.

Assuming that this is a thin, rigid rod, with the axis of rotation through its center, the moment of inertia, #I#, of the rod is given by #1/12ML^2#, where #M# is the mass of the rod and #L# is its length. If the rod is rotated about its end, #I=1/3ML^2#. I will show the calculation for the axis of rotation through the center. We are given both #L# and #M#. Thus,

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

#I=50/12kgm^2#

The angular acceleration of the rod as it rotates, #alpha#, can be calculated from the given values of time and frequency, where #alpha=(Δomega)/( Δt)#. We can find #omega# from the given change in frequency of #15Hz#, as #omega=2pif#.

#omega=2pi(15s^-1)#

#omega=30pi(rad)/s#

Because the problem states that the frequency changed by this amount, this is # Δomega#. We are given that this took place over #6s#, which is our # Δt# value. Thus,

#alpha=(30pi(rad)/s)/(6s)#

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

Now that we have values for #I# and #alpha#, we can calculate the torque.

#tau=Ialpha#

#tau=50/12kgm^2*5pi (rad)/s^2#

#tau=(250pi)/12Nm#

#tau≈65Nm#