What torque would have to be applied to a rod with a length of $5 m$ and a mass of $2 kg$ to change its horizontal spin by a frequency of $18 Hz$ over $6 s$ ?

Question

1244 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-27T09:51:37

The torque for the rod rotating about the center is $=78.54Nm$
The torque for the rod rotating about one end is $=314.16Nm$

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

$I=1/12*mL^2$

$=1/12*2*5^2= 4.17 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(18)/6*2pi$

$=(6pi) rads^(-2)$

So the torque is $tau=4.17*(6pi) Nm=78.54Nm$

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

$I=1/3*mL^2$

$=1/3*2*5^2=50/3kgm^2$

So,

The torque is $tau=50/3*(6pi)=100pi=314.16Nm$

What torque would have to be applied to a rod with a length of 5 m5 m and a mass of 2 kg2 kg to change its horizontal spin by a frequency of 18 Hz18 Hz over 6 s6 s?