# 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 7 Hz over 3 s?

Jan 22, 2018

$244.4 N . m$

#### Explanation:

Torque is defined as the rate of change in angular momentum,

i.e $\tau = \frac{\mathrm{dL}}{\mathrm{dt}} = \frac{d \left(I \omega\right)}{\mathrm{dt}} = I \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

Now,moment of inertia($I$) of a rod w.r.t its one end is $\frac{M {l}^{2}}{3}$,(as here its length is the radius of the circle in which it moves)

so,its $I = 16.67 K g . {m}^{2}$ (given $l$=5 m and $M$=$2 K g$)

Now,its rate of change in angular velocity is $\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{7}{3} \cdot 2 \pi$ (as, $\omega = 2 \pi \cdot \nu$ where $\nu$ is frequency)

So,torque acting is $16.67 \cdot \left(\frac{7}{3}\right) \cdot 2 \pi$ 0r, $244.4 N . m$