The rate of rotation of a solid disk with a radius of $9 m$ and mass of $5 kg$ constantly changes from $32 Hz$ to $18 Hz$ . If the change in rotational frequency occurs over $6 s$ , what torque was applied to the disk?

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Answer 1 · 2017-02-20T14:18:09

The torque was $=2968.8Nm$

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 is $I=1/2*mr^2$

$=1/2*5*9^2= 405/2 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(32-18)/6*2pi$

$=((14pi)/3) rads^(-2)$

So the torque is $tau=405/2*(14pi)/3 Nm=2968.8Nm$

The rate of rotation of a solid disk with a radius of 9 m9 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 6 s6 s, what torque was applied to the disk?