What torque would have to be applied to a rod with a length of $6 m$ and a mass of $3 kg$ to change its horizontal spin by a frequency $15 Hz$ over $6 s$ ?

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Answer 1 · 2017-04-14T07:17:20

The torque, for the rod rotating about the center, is $=141.4Nm$
The torque, for the rod rotating about one end, is $=565.5Nm$

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, rotating about the center is

$I=1/12*mL^2$

$=1/12*3*6^2= 9 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(15)/6*2pi$

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

So the torque is $tau=9*(5pi) Nm=45piNm=141.4Nm$

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

$I=1/3*mL^2$

$=1/3*3*6^2=36$

So,

The torque is $tau=36*(5pi)=180pi=565.5Nm$

What torque would have to be applied to a rod with a length of 6 m6 m and a mass of 3 kg3 kg to change its horizontal spin by a frequency 15 Hz15 Hz over 6 s6 s?