What torque would have to be applied to a rod with a length of $5$ and a mass of $5 kg$ to change its horizontal spin by a frequency of $3 Hz$ over $4 s$ ?

Question

1585 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-25T11:36:29

The torque is $=294.5 Nm$

The torque is the rate of change of angular momentum

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

The moment of inertia (spinning around the middle of the rod) is $I=1/12ml^2$

$=1/2*5*5^2= 125/2 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(3)/4*2pi$

$=3/2pi rads^(-2)$

So the torque is $tau=125/2*3/2pi Nm=375/4piNm=294.5Nm$

What torque would have to be applied to a rod with a length of 5 5 and a mass of 5 kg5 kg to change its horizontal spin by a frequency of 3 Hz3 Hz over 4 s4 s?