The rate of rotation of a solid disk with a radius of $1 m$ $1 m$ and mass of $5 k g$ $5 kg$ constantly changes from $16 H z$ $16 Hz$ to $27 H z$ $27 Hz$ . 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 · 2017-03-27T12:08:00

The torque was $= 14.4 N m$ $=14.4Nm$

The torque is the rate of change of angular momentum

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

mass, $m = 5 k g$ $m=5kg$

radius, $r = 1 m$ $r=1m$

The moment of inertia of a solid disc is

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

$= \frac{1}{2} \cdot 5 \cdot 1^{2} = 2.5 k g m^{2}$ $=1/2*5*1^2= 2.5 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{(27 - 16)}{12} \cdot 2 π$ $(domega)/dt=((27-16))/12*2pi$

$= (\frac{11}{6} π) r a d s^{- 2}$ $=(11/6pi) rads^(-2)$

So the torque is $τ = 2.5 \cdot (\frac{11}{6} π) N m = 14.4 N m$ $tau=2.5*(11/6pi) Nm=14.4Nm$

The rate of rotation of a solid disk with a radius of 1 m1m1 m and mass of 5 kg5kg5 kg constantly changes from 16 Hz16Hz16 Hz to 27 Hz27Hz27 Hz. If the change in rotational frequency occurs over 12 s12s12 s, what torque was applied to the disk?