What torque would have to be applied to a rod with a length of $3 m$ and a mass of $8 kg$ to change its horizontal spin by a frequency $8 Hz$ over $9 s$ ?

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Answer 1 · 2017-02-11T17:46:49

The torque is $=33.5Nm$

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 $I=1/12*mL^2$

$=1/12*8*3^2=6 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(8)/9*2pi$

$=((16pi)/9) rads^(-2)$

So the torque is $tau=6*(16pi)/9 Nm=33.5Nm$

What torque would have to be applied to a rod with a length of 3 m3 m and a mass of 8 kg8 kg to change its horizontal spin by a frequency 8 Hz8 Hz over 9 s9 s?