The rate of rotation of a solid disk with a radius of $6 m$ and mass of $5 kg$ constantly changes from $11 Hz$ to $15 Hz$ . If the change in rotational frequency occurs over $3 s$ , what torque was applied to the disk?

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Answer 1 · 2016-12-10T12:04:10

The torque is $=754kgm^2s^(-1)$

The torque is $tau=Ialpha$

$tau=I(domega)/dt$

The moment of inertia of a solid disc $I=(mr^2)/2$

$I=5*6^2/2=5*18=90 kg m^2$

$(domega)/dt=(15-11)/3*2pi=(8/3pi )rads^(-1)$

So, the torque is

$tau=90*8/3pi=240pi=754kgm^2s^(-1)$

The rate of rotation of a solid disk with a radius of 6 m6 m and mass of 5 kg5 kg constantly changes from 11 Hz11 Hz to 15 Hz15 Hz. If the change in rotational frequency occurs over 3 s3 s, what torque was applied to the disk?