What torque would have to be applied to a rod with a length of $5 m$ and a mass of $2 kg$ to change its horizontal spin by a frequency of $2 Hz$ over $2 s$ ?

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Answer 1 · 2017-03-27T09:55:08

The torque, for the rod rotating about the center is $=26.2Nm$
The torque, for the rod rotating about one end is $=104.7Nm$

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

$I=1/12*mL^2$

$=1/12*2*5^2= 25/6 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $tau=25/6*(2pi) Nm=25/3piNm=26.2Nm$

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

$I=1/3*mL^2$

$=1/3*2*5^2=50/3kgm^2$

So,

The torque is $tau=50/3*(2pi)=100/3pi=104.7Nm$

What torque would have to be applied to a rod with a length of 5 m5 m and a mass of 2 kg2 kg to change its horizontal spin by a frequency of 2 Hz2 Hz over 2 s2 s?