What torque would have to be applied to a rod with a length of $4 m$ and a mass of $5 kg$ to change its horizontal spin by a frequency of $3 Hz$ over $7 s$ ?

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Answer 1 · 2017-05-30T06:30:04

The torque for the rod rotating about the center is $=17.95Nm$
The torque for the rod rotating about one end is $=71.81Nm$

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

$I=1/12*mL^2$

$=1/12*5*4^2= 6.67 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(3)/7*2pi$

$=(6/7pi) rads^(-2)$

So the torque is $tau=6.67*(6/7pi) Nm=17.95Nm$

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

$I=1/3*mL^2$

$=1/3*5*4^2=26.67kgm^2$

So,

The torque is $tau=26.67*(6/7pi)=71.81Nm$

What torque would have to be applied to a rod with a length of 4 m4 m and a mass of 5 kg5 kg to change its horizontal spin by a frequency of 3 Hz3 Hz over 7 s7 s?