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 $6 H z$ $6 Hz$ to $27 H z$ $27 Hz$ . 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 · 2016-12-26T14:46:22

THe torque is $= 206.2 N m$ $=206.2Nm$

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 of a solid disc is $I = \frac{1}{2} m r^{2}$ $I=1/2mr^2$

$= \frac{1}{2} \cdot 1 \cdot 5^{2} = \frac{25}{2} = 12.5 k g m^{2}$ $=1/2*1*5^2= 25/2=12.5 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{27 - 6}{8} \cdot 2 π$ $(domega)/dt=(27-6)/8*2pi$

$= (\frac{21 π}{4}) π r a d s^{- 2}$ $=((21pi)/4)pi rads^(-2)$

So the torque is $τ = 12.5 \cdot \frac{21 π}{4} N m = 65.625 π N m = 206.2 N m$ $tau=12.5*(21pi)/4 Nm=65.625piNm=206.2Nm$

The rate of rotation of a solid disk with a radius of 1m1 m and mass of 5kg5 kg constantly changes from 6Hz6 Hz to 27Hz27 Hz. If the change in rotational frequency occurs over 8s8 s, what torque was applied to the disk?