What torque would have to be applied to a rod with a length of $6 m$ and a mass of $8 kg$ to change its horizontal spin by a frequency $4 Hz$ over $2 s$ ?

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Answer 1 · 2017-02-09T07:48:17

The torque is $=301.6Nm$

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 $I=1/12*mL^2$

$=1/12*8*6^2= 24 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $tau=24*(4pi) Nm=96piNm=301.6Nm$

What torque would have to be applied to a rod with a length of 6 m6 m and a mass of 8 kg8 kg to change its horizontal spin by a frequency 4 Hz4 Hz over 2 s2 s?