What torque would have to be applied to a rod with a length of $3 m$ and a mass of $4 kg$ to change its horizontal spin by a frequency $5 Hz$ over $1 s$ ?

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Answer 1 · 2017-02-20T06:31:54

The torque is $=94.2Nm$

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 is $I=1/12*mL^2$

$=1/12*4*3^2= 3 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(5)/1*2pi$

$=(10pi) rads^(-2)$

So the torque is $tau=3*(10pi) Nm=30piNm=94.2Nm$

What torque would have to be applied to a rod with a length of 3 m3 m and a mass of 4 kg4 kg to change its horizontal spin by a frequency 5 Hz5 Hz over 1 s1 s?