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 $2 Hz$ over $3 s$ ?

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Answer 1 · 2017-04-14T07:22:12

The torque, for the rod rotating about the center, is $=0.35Nm$
The torque, for the rod rotating about one end, is $=1.40Nm$

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

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

$=(4/3pi) rads^(-2)$

So the torque is $tau=1/12*(4/3pi) Nm=1/9piNm=0.35Nm$

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 $tau=1/3*(4/3pi)=4/9pi=1.40Nm$

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 2 Hz2 Hz over 3 s3 s?