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
94.25 Nm
Explanation:
Torque is rate of change of angular momentum.
#= ((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#
