What torque would have to be applied to a rod with a length of $12 m$ and a mass of $7 kg$ to change its horizontal spin by a frequency $5 Hz$ over $8 s$ ?

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Answer 1 · 2018-01-11T16:30:10

The torque for the rod rotating about the center is $=329.9Nm$
The torque for the rod rotating about one end is $=1319.5Nm$

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 $m=7kg$

The length of the rod is $L=12m$

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

$I=1/12*mL^2$

$=1/12*7*12^2= 84 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(5)/8*2pi$

$=(5/4pi) rads^(-2)$

So the torque is $tau=84*(5/4pi) Nm=329.9Nm$

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

$I=1/3*mL^2$

$=1/3*7*12^2=336kgm^2$

So,

The torque is $tau=336*(5/4pi)=1319.5Nm$

What torque would have to be applied to a rod with a length of 12 m12 m and a mass of 7 kg7 kg to change its horizontal spin by a frequency 5 Hz5 Hz over 8 s8 s?