What torque would have to be applied to a rod with a length of $6 m$ and a mass of $3 kg$ to change its horizontal spin by a frequency $6 Hz$ over $5 s$ ?

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Answer 1 · 2017-06-11T11:45:01

The torque for the rod rotating about the center is $=67.86Nm$
The torque for the rod rotating about one end is $=271.43Nm$

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

$I=1/12*mL^2$

$=1/12*3*6^2= 9 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(6)/5*2pi$

$=(12/5pi) rads^(-2)$

So the torque is $tau=9*(12/5pi) Nm=108/5piNm=67.86Nm$

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

$I=1/3*mL^2$

$=1/3*3*6^2=36kgm^2$

So,

The torque is $tau=36*(12/5pi)=432/5pi=271.43Nm$

What torque would have to be applied to a rod with a length of 6 m6 m and a mass of 3 kg3 kg to change its horizontal spin by a frequency 6 Hz6 Hz over 5 s5 s?