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 $16 Hz$ over $4 s$ ?

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Answer 1 · 2017-01-24T13:57:06

The answer is $226.2Nm$

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

$(domega)/dt=(16)/4*2pi$

$=(8pi) rads^(-2)$

So the torque is $tau=9*(8pi) Nm=72piNm=226.2Nm$

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 16 Hz16 Hz over 4 s4 s?