# 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 9 Hz over 3 s?

Torque ($\tau$) is defined as change in rate of angular momentum ($L$)i.e $\frac{\mathrm{dL}}{\mathrm{dt}}$ i.e $\frac{d \left(I \omega\right)}{\mathrm{dt}}$ i.e $I \frac{\mathrm{do} m e g a}{\mathrm{dt}}$ (where,$I$nis the moment of inertia)
Now,moment of inertia of a rod w.r.t its one end along an axis perpendicular to its plane is $\frac{M {L}^{2}}{3}$ i.e $2 \cdot {\left(5\right)}^{2} / 3$ or $\frac{50}{3} K g . {m}^{2}$
Now, as the radius of the circle isn't mentioned,so the rod can move about its mid point as well,in that case $I = \frac{M {L}^{2}}{12}$ i.e $\frac{25}{6} K g . {m}^{2}$
Now, rate of change in angular velocity = $\frac{\mathrm{do} m e g a}{\mathrm{dt}}$ = $2 \pi \frac{\nu 1 - \nu 2}{t}$ i.e $2 \pi \cdot \frac{9}{3}$ or $6 \pi$
So, $\tau = \frac{50}{3} \cdot 6 \pi$ or, $314.16 N . m$ or, $\frac{25}{6} \cdot 6 \pi$ i.e $78.54 N . m$ for the $2 n d$ case