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 $7 Hz$ over $5 s$ ?

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Answer 1 · 2017-04-27T06:34:23

The torque for the rod rotating about the center is $=1.47Nm$
The torque for the rod rotating about one end is $=5.86Nm$

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

$I=1/12*mL^2$

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

The rate of change of angular velocity is

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

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

So the torque is $tau=1/6*(14/5pi) Nm=7/15piNm=1.47Nm$

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

$I=1/3*mL^2$

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

So,

The torque is $tau=2/3*(14/5pi)=28/15pi=5.86Nm$

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 7 Hz7 Hz over 5 s5 s?