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 $24 H z$ $24 Hz$ to $15 H z$ $15 Hz$ . 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 · 2017-04-21T14:30:15

The torque was $= 2290.2 N m$ $=2290.2Nm$

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 is $= I$ $=I$

For the solid disc, $I = \frac{m r^{2}}{2}$ $I=(mr^2)/2$

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{24 - 15}{5} \cdot 2 π$ $(domega)/dt=(24-15)/5*2pi$

$= (\frac{18}{5} π) r a d s^{- 2}$ $=(18/5pi) rads^(-2)$

So the torque is $τ = 202.5 \cdot (\frac{18}{5} π) N m = 2290.2 N m$ $tau=202.5*(18/5pi) Nm=2290.2Nm$

The rate of rotation of a solid disk with a radius of 9m9 m and mass of 5kg5 kg constantly changes from 24Hz24 Hz to 15Hz15 Hz. If the change in rotational frequency occurs over 5s5 s, what torque was applied to the disk?