What torque would have to be applied to a rod with a length of $6 m$ and a mass of $2 kg$ to change its horizontal spin by a frequency $2 Hz$ over $5 s$ ?

Question

1275 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-11T16:20:31

The torque is $=15.1Nm$

The torque is the rate of change of angular momentum

$tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

The [moment of inertia of a rod is $I=1/12*mL^2$

$=1/12*2*6^2= 6 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(2)/5*2pi$

$=((4pi)/5) rads^(-2)$

So the torque is $tau=6*(4pi)/5 Nm=24/5piNm=15.1Nm$

What torque would have to be applied to a rod with a length of 6 m6 m and a mass of 2 kg2 kg to change its horizontal spin by a frequency 2 Hz2 Hz over 5 s5 s?