What torque would have to be applied to a rod with a length of $4 m$ and a mass of $8 kg$ to change its horizontal spin by a frequency $9 Hz$ over $2 s$ ?

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Answer 1 · 2018-01-11T14:30:03

The torque for the rod rotating about the center is $=301.6Nm$
The torque for the rod rotating about one end is $=1206.4Nm$

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 $m=8kg$

The length of the rod is $L=4m$

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

$I=1/12*mL^2$

$=1/12*8*4^2= 10.67 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $tau=10.67*(9pi) Nm=301.6Nm$

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

$I=1/3*mL^2$

$=1/3*8*4^2=42.67kgm^2$

So,

The torque is $tau=42.67*(9pi)=504/5pi=1206.4Nm$

What torque would have to be applied to a rod with a length of 4 m4 m and a mass of 8 kg8 kg to change its horizontal spin by a frequency 9 Hz9 Hz over 2 s2 s?