The rate of rotation of a solid disk with a radius of $4 m$ $4 m$ 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 $8 s$ $8 s$ , what torque was applied to the disk?

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Answer 1 · 2017-03-23T06:48:03

THe torque was $= 282.7 N m$ $=282.7Nm$

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 a solid disc, is

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

$= \frac{1}{2} \cdot 5 \cdot 4^{2} = 40 k g m^{2}$ $=1/2*5*4^2= 40 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{21 - 12}{8} \cdot 2 π$ $(domega)/dt=(21-12)/8*2pi$

$= (\frac{9}{4} π) r a d s^{- 2}$ $=(9/4pi) rads^(-2)$

So the torque is $τ = 40 \cdot (\frac{9}{4} π) N m = (90 π) N m = 282.7 N m$ $tau=40*(9/4pi) Nm=(90pi)Nm=282.7Nm$

The rate of rotation of a solid disk with a radius of 4m4 m and mass of 5kg5 kg constantly changes from 21Hz21 Hz to 12Hz12 Hz. If the change in rotational frequency occurs over 8s8 s, what torque was applied to the disk?