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 $5 H z$ $5 Hz$ to $17 H z$ $17 Hz$ . If the change in rotational frequency occurs over $6 s$ $6 s$ , what torque was applied to the disk?

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Answer 1 · 2017-06-10T14:42:51

The torque was $= 31.42 N m$ $=31.42Nm$

The torque is the rate of change of angular momentum

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

where $I$ $I$ is the moment of inertia

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

So, $I = 5 \cdot \frac{{(1)}^{2}}{2} = \frac{5}{2} k g m^{2}$ $I=5*(1)^2/2=5/2kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{17 - 5}{6} \cdot 2 π$ $(domega)/dt=(17-5)/6*2pi$

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

So,

The torque is $τ = \frac{5}{2} \cdot (4 π) = 10 π = 31.42 N m$ $tau=5/2*(4pi)=10pi=31.42Nm$

The rate of rotation of a solid disk with a radius of 1m1 m and mass of 5kg5 kg constantly changes from 5Hz5 Hz to 17Hz17 Hz. If the change in rotational frequency occurs over 6s6 s, what torque was applied to the disk?