The rate of rotation of a solid disk with a radius of $6 m$ and mass of $5 kg$ constantly changes from $11 Hz$ to $24 Hz$ . If the change in rotational frequency occurs over $3 s$ , what torque was applied to the disk?

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Answer 1 · 2017-06-23T05:14:07

The torque was $=2540.4Nm$

The torque is the rate of change of angular momentum

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

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

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

The rate of change of angular velocity is

$(domega)/dt=(24-11)/3*2pi$

$=(26/3pi) rads^(-2)$

So the torque is $tau=90*(26/3pi) Nm=2450.4Nm$

The rate of rotation of a solid disk with a radius of 6 m6 m and mass of 5 kg5 kg constantly changes from 11 Hz11 Hz to 24 Hz24 Hz. If the change in rotational frequency occurs over 3 s3 s, what torque was applied to the disk?