The rate of rotation of a solid disk with a radius of $1 m$ and mass of $5 kg$ constantly changes from $6 Hz$ to $24 Hz$ . If the change in rotational frequency occurs over $6 s$ , what torque was applied to the disk?

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Answer 1 · 2017-10-04T05:58:43

The torque was $=47.1Nm$

The torque is the rate of change of angular momentum

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

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

The mass of the dics is $m=5kg$

The radius of the disc is $r=1m$

So, the moment of inertia is $I=5*(1)^2/2=2.5kgm^2$

The rate of change of the angular velocity is

$(domega)/dt=(24-6)/6*2pi=6pirads^-2$

So,

The torque is

$tau=2.5*6pi=47.1Nm$

The rate of rotation of a solid disk with a radius of 1 m1 m and mass of 5 kg5 kg constantly changes from 6 Hz6 Hz to 24 Hz24 Hz. If the change in rotational frequency occurs over 6 s6 s, what torque was applied to the disk?