# What torque would have to be applied to a rod with a length of 4"m" and a mass of 2"kg" to change its horizontal spin by a frequency 9"Hz" over 8"s"?

Torque ($\tau$) = moment of inertia (I)X angular acceleration ($\alpha$ )
I = $\frac{1}{12} M {L}^{2}$ , where mass M =2kg, Length L= 4m
$\alpha = \frac{\mathrm{do} m e g a}{\mathrm{dt}}$ = $2 \pi$xrate of change of frequency