What torque would have to be applied to a rod with a length of $4 m$ and a mass of $9 kg$ to change its horizontal spin by a frequency $7 Hz$ over $18 s$ ?

Question

1261 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-01T06:38:17

The torque for the rod rotating about the center is $=29.3Nm$
The torque for the rod rotating about one end is $=117.3Nm$

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, rotating about the center is

$I=1/12*mL^2$

$=1/12*9*4^2= 12 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(7)/18*2pi$

$=(7/9pi) rads^(-2)$

So the torque is $tau=12*(7/9pi) Nm=28/3piNm=29.3Nm$

The moment of inertia of a rod, rotating about one end is

$I=1/3*mL^2$

$=1/3*9*4^2=48kgm^2$

So,

The torque is $tau=48*(7/9pi)=112/3pi=117.3Nm$

What torque would have to be applied to a rod with a length of 4 m4 m and a mass of 9 kg9 kg to change its horizontal spin by a frequency 7 Hz7 Hz over 18 s18 s?