The rate of rotation of a solid disk with a radius of $5 m$ $5 m$ and mass of $5 k g$ $5 kg$ constantly changes from $12 H z$ $12 Hz$ to $10 H z$ $10 Hz$ . If the change in rotational frequency occurs over $9 s$ $9 s$ , what torque was applied to the disk?

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Answer 1 · 2018-04-25T05:41:35

The torque was $= 87.27 N m$ $=87.27Nm$

The torque is the rate of change of angular momentum

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

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

The mass of the disc is $m = 5 k g$ $m=5kg$

The radius of the disc is $r = 5 m$ $r=5m$

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

So, $I = 5 \cdot \frac{{(5)}^{2}}{2} = 62.5 k g m^{2}$ $I=5*(5)^2/2=62.5kgm^2$

The initial angular velocity is

$ω_{0} = 2 π f_{0} = 2 \cdot 12 \cdot π = 24 π r a d s^{- 1}$ $omega_0=2pif_0=2*12*pi=24pirads^-1$

The final angular velocity is

$ω_{1} = 2 π f_{1} = 2 \cdot 10 \cdot π = 20 π r a d s^{- 1}$ $omega_1=2pif_1=2*10*pi=20pirads^-1$

The time is $t = 9 s$ $t=9s$

The angular acceleration is

$α = \frac{ω_{1} - ω_{0}}{t} = \frac{24 π - 20 π}{9} = \frac{4}{9} π r a d s^{- 2}$ $alpha=(omega_1-omega_0)/t=(24pi-20pi)/9=4/9pirads^-2$

So,

The torque is $τ = 62.5 \cdot \frac{4}{9} π = 87.27 N m$ $tau= 62.5*4/9pi=87.27Nm$

The rate of rotation of a solid disk with a radius of 5 m5m5 m and mass of 5 kg5kg5 kg constantly changes from 12 Hz12Hz12 Hz to 10 Hz10Hz10 Hz. If the change in rotational frequency occurs over 9 s9s9 s, what torque was applied to the disk?