What torque would have to be applied to a rod with a length of $7 m$ and a mass of $9 kg$ to change its horizontal spin by a frequency $12 Hz$ over $9 s$ ?

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Answer 1 · 2018-01-24T18:23:21

The torque for the rod rotating about the center is $=307.9Nm$
The torque for the rod rotating about one end is $=1231.5Nm$

The torque is the rate of change of angular momentum

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

The mass of the rod is $m=9kg$

The length of the rod is $L=7m$

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

$I=1/12*mL^2$

$=1/12*9*7^2= 36.75 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(12)/9*2pi$

$=(8/3pi) rads^(-2)$

So the torque is $tau=36.75*(8/3pi) Nm=307.9Nm$

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

$I=1/3*mL^2$

$=1/3*9*7^2=147kgm^2$

So,

The torque is $tau=147*(8/3pi)=1231.5Nm$

What torque would have to be applied to a rod with a length of 7 m7 m and a mass of 9 kg9 kg to change its horizontal spin by a frequency 12 Hz12 Hz over 9 s9 s?