The rate of rotation of a solid disk with a radius of $8 m$ and mass of $5 kg$ constantly changes from $5 Hz$ to $17 Hz$ . If the change in rotational frequency occurs over $6 s$ , what torque was applied to the disk?

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Answer 1 · 2017-03-02T07:46:36

The torque was $=2010.6Nm$

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 solid disc, rotating about the center is

$I=1/2*mr^2$

$=1/2*5*8^2= 160 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(17-5)/6*2pi$

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

So the torque is $tau=160*(4pi) Nm=2010.6Nm$

The rate of rotation of a solid disk with a radius of 8 m8 m and mass of 5 kg5 kg constantly changes from 5 Hz5 Hz to 17 Hz17 Hz. If the change in rotational frequency occurs over 6 s6 s, what torque was applied to the disk?