What torque would have to be applied to a rod with a length of $8 m$ and a mass of $8 kg$ to change its horizontal spin by a frequency $6 Hz$ over $9 s$ ?

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Answer 1 · 2017-02-18T09:32:41

The torque is $=178.7Nm$

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

$=1/12*8*8^2= 42.67 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(6)/9*2pi$

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

So the torque is $tau=42.67*(4pi)/3 Nm=178.7Nm$

What torque would have to be applied to a rod with a length of 8 m8 m and a mass of 8 kg8 kg to change its horizontal spin by a frequency 6 Hz6 Hz over 9 s9 s?