# What torque would have to be applied to a rod with a length of 4 m and a mass of 3 kg to change its horizontal spin by a frequency of 3 Hz over 2 s?

Nov 18, 2016

The torque applied to the rod is 37.68 Newton-metre.

#### Explanation:

Torque is defined as the product of moment of inertia and angular acceleration
$\implies \tau = I \alpha$
Moment of inertia of rod is ${I}_{\text{rod}} = \left(\frac{M {L}^{2}}{12}\right)$ and angular acceleration is change in angular velocity with respect to time
$\implies \alpha = \frac{\omega}{t}$
angular velocity $\omega = 2 \pi \nu$ as frequency is given in above case.
Now the torque becomes $\tau = \left[\frac{M {L}^{2}}{12}\right] \left[\frac{2 \pi \nu}{t}\right]$
Substituting the given values we get,
$\tau = \left[\frac{\left(3 k g\right) {\left(4 m\right)}^{2}}{12}\right] \left[\frac{2 \times 3.14 \times 3 H z}{2 s}\right]$
$\therefore$torque$\tau = 37.68$N-m.