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 $3 Hz$ over $4 s$ ?

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Answer 1 · 2017-03-09T14:25:46

The torque (rod rotating about the center) is $=0.785Nm$
The torque (rod rotating about one end) is $=3.14Nm$

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

$(domega)/dt=(3)/4*2pi$

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

So the torque is $tau=1/6*(3/2pi) Nm=1/4piNm=0.785Nm$

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 $tau=2/3*(3/2pi)=pi=3.14Nm$

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 3 Hz3 Hz over 4 s4 s?