# 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?

Nov 14, 2017

94.25 Nm

#### Explanation:

Torque is rate of change of angular momentum.

$T = \frac{I 2 \pi \Delta f}{t}$

$= \frac{\frac{m {r}^{2}}{3} 2 \pi \Delta f}{t}$ ……[∵ Moment of inertia of rod about an axis through its one end and perpendicular to plane is $\frac{m {r}^{2}}{3}$]

= ((4 kg × (3 m)^2)/3 2pi × 5Hz)/(4s)

$= 94.25 N m$