The rate of rotation of a solid disk with a radius of $9 m$ $9 m$ and mass of $5 k g$ $5 kg$ constantly changes from $21 H z$ $21 Hz$ to $12 H z$ $12 Hz$ . If the change in rotational frequency occurs over $5 s$ $5 s$ , what torque was applied to the disk?

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Answer 1 · 2017-11-10T06:16:00

The torque was $= 2290 N m$ $=2290Nm$

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

where $I$ $I$ is the moment of inertia

The mass of the disk is $m = 5 k g$ $m=5kg$

The radius is $r = 9 m$ $r=9m$

For the solid disk, $I = \frac{1}{2} m r^{2}$ $I=1/2mr^2$

So, $I = \frac{1}{2} \cdot 5 \cdot {(9)}^{2} = 202.5 k g m^{2}$ $I=1/2*5*(9)^2=202.5kgm^2$

The angular velocity is

$ω = 2 \times π \times f$ $omega=2xxpixxf$ where the frequency is $= f$ $=f$

And the rate of change of angular velocity is

$alpha=(d omega)/dt=(Deltaomega)/t=(42pi-24pi)/5=(18/5pi)rads^-2$

The torque is

$tau=202.5*18/5pi=2290Nm$

The rate of rotation of a solid disk with a radius of 9 m9m9 m and mass of 5 kg5kg5 kg constantly changes from 21 Hz21Hz21 Hz to 12 Hz12Hz12 Hz. If the change in rotational frequency occurs over 5 s5s5 s, what torque was applied to the disk?