What torque would have to be applied to a rod with a length of $5 m$ and a mass of $2 kg$ to change its horizontal spin by a frequency $8 Hz$ over $8 s$ ?

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Answer 1 · 2017-02-05T07:33:21

The torque is $=26.2Nm$

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 $I=1/12*mL^2$

$=1/12*2*5^2= 25/6 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $tau=25/6*(2pi) Nm=25/3piNm=26.2Nm$

What torque would have to be applied to a rod with a length of 5 m5 m and a mass of 2 kg2 kg to change its horizontal spin by a frequency 8 Hz8 Hz over 8 s8 s?