The rate of rotation of a solid disk with a radius of $5 m$ $5 m$ and mass of $5 k g$ $5 kg$ constantly changes from $11 H z$ $11 Hz$ to $24 H z$ $24 Hz$ . If the change in rotational frequency occurs over $1 s$ $1 s$ , what torque was applied to the disk?

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Answer 1 · 2017-01-29T07:50:34

The torque was $= 5105.1 N m$ $=5105.1Nm$

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

$= \frac{1}{2} \cdot 5 \cdot 5^{2} = \frac{125}{2} k g m^{2}$ $=1/2*5*5^2= 125/2 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{24 - 11}{1} \cdot 2 π$ $(domega)/dt=(24-11)/1*2pi$

$= (26 π) r a d s^{- 2}$ $=(26pi) rads^(-2)$

So the torque is $τ = \frac{125}{2} \cdot (26 π) N m = 125 \cdot 13 π N m = 5105.1 N m$ $tau=125/2*(26pi) Nm=125*13piNm=5105.1Nm$

The rate of rotation of a solid disk with a radius of 5m5 m and mass of 5kg5 kg constantly changes from 11Hz11 Hz to 24Hz24 Hz. If the change in rotational frequency occurs over 1s1 s, what torque was applied to the disk?