What torque would have to be applied to a rod with a length of $12 m$ and a mass of $6 kg$ to change its horizontal spin by a frequency $7 Hz$ over $9 s$ ?

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Answer 1 · 2017-07-11T06:34:58

The torque for the rod rotating about the center is $=351.9Nm$
The torque for the rod rotating about one end is $=1407.4Nm$

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

$I=1/12*mL^2$

$=1/12*6*12^2= 72 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(7)/9*2pi$

$=(14/9pi) rads^(-2)$

So the torque is $tau=72*(14/9pi) Nm=112piNm=351.9Nm$

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

$I=1/3*mL^2$

$=1/3*6*12^2=288kgm^2$

So,

The torque is $tau=288*(14/9pi)=1407.4Nm$

What torque would have to be applied to a rod with a length of 12 m12 m and a mass of 6 kg6 kg to change its horizontal spin by a frequency 7 Hz7 Hz over 9 s9 s?