The rate of rotation of a solid disk with a radius of $8 m$ $8 m$ and mass of $5 k g$ $5 kg$ constantly changes from $2 H z$ $2 Hz$ 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 · 2017-01-18T06:50:24

The torque was $= 2010.6 N m$ $=2010.6Nm$

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 inertia of the disc is $I = \frac{1}{2} \cdot m r^{2}$ $I=1/2*mr^2$

$= \frac{1}{2} \cdot 5 \cdot 8^{2} = 160 k g m^{2}$ $=1/2*5*8^2= 160 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{12 - 2}{5} \cdot 2 π$ $(domega)/dt=(12-2)/5*2pi$

$= (4 π) r a d s^{- 2}$ $=(4pi) rads^(-2)$

So the torque is $τ = 160 \cdot (4 π) N m = 640 π N m = 2010.6 N m$ $tau=160*(4pi) Nm=640piNm=2010.6Nm$

The rate of rotation of a solid disk with a radius of 8m8 m and mass of 5kg5 kg constantly changes from 2Hz2 Hz to 12Hz12 Hz. If the change in rotational frequency occurs over 5s5 s, what torque was applied to the disk?