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 $7 Hz$ over $5 s$ ?

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Answer 1 · 2017-02-22T10:16:52

The torque (rod rotating about the center ) is $=79.2Nm$
The torque (rod rotating about one end ) is $=316.7Nm$

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

$(domega)/dt=(7)/5*2pi$

$=(14/5pi) rads^(-2)$

So the torque is $tau=9*(14/5pi) Nm=126/5piNm=79.2Nm$

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 $tau=36*(14/5pi)=504/5pi=316.7Nm$

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 7 Hz7 Hz over 5 s5 s?