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

1 Answer
Nov 14, 2017

Answer:

94.25 Nm

Explanation:

Torque is rate of change of angular momentum.

#T = (I2piDeltaf)/t#

#= ((mr^2)/3 2piDeltaf)/t# ……[∵ Moment of inertia of rod about an axis through its one end and perpendicular to plane is #(mr^2)/3#]

#= ((4 kg × (3 m)^2)/3 2pi × 5Hz)/(4s)#

#= 94.25 Nm#