What torque would have to be applied to a rod with a length of $3 m$ and a mass of $1 kg$ to change its horizontal spin by a frequency of $5 Hz$ over $2 s$ ?

Question

1353 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-30T06:25:27

The torque for the rod rotating about the center is $=11.78Nm$
The torque for the rod rotating one end is $=47.12Nm$

The torque is the rate of change of angular momentum

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

The mass of the rod is $m=1kg$

The length of the rod is $L=3m$

The moment of inertia of a rod, rotating about the center is

$I=1/12*mL^2$

$=1/12*1*3^2= 0.75 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $tau=0.75*(5pi) Nm=11.78Nm$

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

$I=1/3*mL^2$

$=1/3*1*3^2=3kgm^2$

So,

The torque is

$=3*(5pi)=47.12Nm$

What torque would have to be applied to a rod with a length of 3 m3 m and a mass of 1 kg1 kg to change its horizontal spin by a frequency of 5 Hz5 Hz over 2 s2 s?