What torque would have to be applied to a rod with a length of $8 m$ and a mass of $8 kg$ to change its horizontal spin by a frequency $7 Hz$ over $4 s$ ?

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Answer 1 · 2018-01-17T15:02:41

The torque for the rod rotating about the center is $=469.1Nm$
The torque for the rod rotating about one end is $=1876.6Nm$

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 $L=8m$

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

$I=1/12*mL^2$

$=1/12*8*8^2= 42.67 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(7)/4*2pi$

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

So the torque is $tau=42.67*(7/2pi) Nm=469.1Nm$

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

$I=1/3*mL^2$

$=1/3*8*8^2=170.67kgm^2$

So,

The torque is $tau=170.67*(7/2pi)=1876.6Nm$

What torque would have to be applied to a rod with a length of 8 m8 m and a mass of 8 kg8 kg to change its horizontal spin by a frequency 7 Hz7 Hz over 4 s4 s?