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 $6 Hz$ over $2 s$ ?

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Answer 1 · 2017-03-09T07:00:42

The torque (rod rotating about the center) is $=3.14Nm$
The torque (rod rotating about one end) is $=12.57Nm$

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

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

$=(6pi) rads^(-2)$

So the torque is $tau=1/6*(6pi) Nm=piNm=3.14Nm$

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

$I=1/3*mL^2$

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

So,

The torque is $tau=2/3*(6pi)=4pi=12.57Nm$

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 6 Hz6 Hz over 2 s2 s?