What torque would have to be applied to a rod with a length of $3 m$ and a mass of $1 kg$ to change its horizontal spin by a frequency of $15 Hz$ over $2 s$ ?

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Answer 1 · 2017-05-12T06:45:11

The torque for the rod rotating about the center is $=35.3N$
The torque for the rod rotating about one end is $=141.2N$

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, rotating about the center is

$I=1/12*mL^2$

$=1/12*1*3^2= 0.75 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(15)/2*2pi$

$=(15pi) rads^(-2)$

So the torque is $tau=0.75*(15pi) Nm=35.3Nm$

The moment of inertia of a rod, rotating about one end is

$I=1/3*mL^2$

$=1/3*1*3^2=3kgm^2$

So,

The torque is $tau=3*(15pi)=141.2Nm$

What torque would have to be applied to a rod with a length of 3 m3 m and a mass of 1 kg1 kg to change its horizontal spin by a frequency of 15 Hz15 Hz over 2 s2 s?