What torque would have to be applied to a rod with a length of $1 m$ and a mass of $1 kg$ to change its horizontal spin by a frequency of $7 Hz$ over $3 s$ ?

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Answer 1 · 2017-06-08T05:23:09

The torque for the rod rotating about the center is $=1.22Nm$
The torque for the rod rotating about one end is $=4.89Nm$

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

$I=1/12*mL^2$

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

The rate of change of angular velocity is

$(domega)/dt=(7)/3*2pi$

$=(14/3pi) rads^(-2)$

So the torque is $tau=1/12*(14/3pi) Nm=7/18piNm=1.22Nm$

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

$I=1/3*mL^2$

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

So,

The torque is $tau=1/3*(14/3pi)=14/9pi=4.89Nm$

What torque would have to be applied to a rod with a length of 1 m1 m and a mass of 1 kg1 kg to change its horizontal spin by a frequency of 7 Hz7 Hz over 3 s3 s?