What torque would have to be applied to a rod with a length of $1 m$ and a mass of $2 kg$ to change its horizontal spin by a frequency of $5 Hz$ over $4 s$ ?

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Answer 1 · 2017-01-18T14:19:35

The torque is $=1.31Nm$

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 $I=1/12*mL^2$

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

The rate of change of angular velocity is

$(domega)/dt=(5)/4*2pi$

$=((5pi)/2) rads^(-2)$

So the torque is $tau=1/6*(5pi)/2 Nm=5/12piNm=1.31Nm$

What torque would have to be applied to a rod with a length of 1 m1 m and a mass of 2 kg2 kg to change its horizontal spin by a frequency of 5 Hz5 Hz over 4 s4 s?