The rate of rotation of a solid disk with a radius of $6 m$ and mass of $5 kg$ constantly changes from $32 Hz$ to $18 Hz$ . If the change in rotational frequency occurs over $8 s$ , what torque was applied to the disk?

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Answer 1 · 2017-06-29T12:43:28

The torque was $=989.6Nm$

The torque is the rate of change of angular momentum

$tau=(dL)/dt=(d(Iomega))/dt=I(domega)$

For a solid disc, $I=(mr^2)/2$

So, $I=5*(6)^2/2=90kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(32-18)/8*2pi$

$=(7/2pi) rads^(-2)$

So the torque is $tau=90*(7/2pi) Nm=315piNm=989.6Nm$

The rate of rotation of a solid disk with a radius of 6 m6 m and mass of 5 kg5 kg constantly changes from 32 Hz32 Hz to 18 Hz18 Hz. If the change in rotational frequency occurs over 8 s8 s, what torque was applied to the disk?