The rate of rotation of a solid disk with a radius of $6 m$ $6 m$ and mass of $5 k g$ $5 kg$ constantly changes from $12 H z$ $12 Hz$ to $15 H z$ $15 Hz$ . If the change in rotational frequency occurs over $12 s$ $12 s$ , what torque was applied to the disk?

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Answer 1 · 2017-02-16T15:02:28

The torque was $= 141.4 N$ $=141.4N$

The torque is the rate of change of angular momentum

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

The moment of inertiaof a solid disc is $I = \frac{1}{2} \cdot m r^{2}$ $I=1/2*mr^2$

$= \frac{1}{2} \cdot 5 \cdot 6^{2} = 90 k g m^{2}$ $=1/2*5*6^2= 90 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{15 - 12}{12} \cdot 2 π$ $(domega)/dt=(15-12)/12*2pi$

$= (\frac{π}{2}) r a d s^{- 2}$ $=(pi/2) rads^(-2)$

So the torque is $τ = 90 \cdot (\frac{π}{2}) N m = 45 π N m = 141.4 N m$ $tau=90*(pi/2) Nm=45piNm=141.4Nm$

The rate of rotation of a solid disk with a radius of 6m6 m and mass of 5kg5 kg constantly changes from 12Hz12 Hz to 15Hz15 Hz. If the change in rotational frequency occurs over 12s12 s, what torque was applied to the disk?