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 $3 Hz$ over $8 s$ ?

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Answer 1 · 2017-01-16T15:14:35

The torque is $21.2Nm$

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 is $I=1/12*mL^2$

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

The rate of change of angular velocity is

$(domega)/dt=(3)/8*2pi$

$=((3pi)/4) rads^(-2)$

So the torque is $tau=9*(3pi)/4 Nm=27/4piNm=21.2Nm$

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 3 Hz3 Hz over 8 s8 s?