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 $22 H z$ $22 Hz$ . 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 · 2017-06-02T08:16:13

The torque is $= 15.7 N m$ $=15.7Nm$

The torque is the rate of change of angular momentum

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

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

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

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

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{22 - 16}{12} \cdot 2 π$ $(domega)/dt=(22-16)/12*2pi$

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

So the torque is $τ = 2.5 \cdot (π) N m = 5 π N m = 15.7 N m$ $tau=2.5*(pi) Nm=5piNm=15.7Nm$

The rate of rotation of a solid disk with a radius of 1m1 m and mass of 5kg5 kg constantly changes from 16Hz16 Hz to 22Hz22 Hz. If the change in rotational frequency occurs over 12s12 s, what torque was applied to the disk?