# What torque would have to be applied to a rod with a length of 8 m and a mass of 8 kg to change its horizontal spin by a frequency 2 Hz over 5 s?

Jul 19, 2017

$T = 107.233 N m$

#### Explanation:

I will suppose that the rode turns around one of its ends.

So let's calculate its moment of inertia :

$I = \frac{1}{12} m {l}^{2} = \frac{1}{12} \cdot 8 \cdot 64 = \frac{128}{3} k g {m}^{2}$

(If the rod does not turn around one of its ends the only thing changing is its moment of intertia. For example if it turns around its center the moment of inetria is then $I = \frac{1}{3} m {l}^{2}$)

The change is the angular velocity will be :

Δω=2piΔf=2pi2=4pi(rads)/s

The angular acceleration is :

α=(Δω)/(Δt)=(4pi)/5(rads)/s^2

So for the torque we have :

T=Iα=128/3*(4π)/5=(512pi)/15=107.233Nm