The rate of rotation of a solid disk with a radius of $2 m$ and mass of $5 kg$ constantly changes from $27 Hz$ 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 · 2017-05-21T20:21:48

The torque was $=70.7Nm$

The torque is the rate of change of angular momentum

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

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

$I=5*2^2/2=10kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(27-18)/8*2pi$

$=(9/4pi) rads^(-2)$

So the torque is $tau=10*(9/4pi) Nm=45/2piNm=70.7Nm$

The rate of rotation of a solid disk with a radius of 2 m2 m and mass of 5 kg5 kg constantly changes from 27 Hz27 Hz to 18 Hz18 Hz. If the change in rotational frequency occurs over 8 s8 s, what torque was applied to the disk?