What torque would have to be applied to a rod with a length of $8 m$ and a mass of $3 kg$ to change its horizontal spin by a frequency $18 Hz$ over $9 s$ ?

Question

1297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T07:20:51

The torque (for the rod rotating about the center) is $=201.1Nm$
The torque (for the rod rotating about one end) is $=804.2Nm$

The torque is the rate of change of angular momentum

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

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

$I=1/12*mL^2$

$=1/12*3*8^2= 16 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(18)/9*2pi$

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

So the torque is $tau=16*(4pi) Nm=64piNm=201.1Nm$

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

$I=1/3*mL^2$

$=1/3*3*8^2=64kgm^2$

So,

The torque is $tau=64*(4pi)=804.2Nm$

What torque would have to be applied to a rod with a length of 8 m8 m and a mass of 3 kg3 kg to change its horizontal spin by a frequency 18 Hz18 Hz over 9 s9 s?