The moment of interia of ring is 0.40 kg $m^{2}$ . if it is rotating at a rate of 2100 revolutions per minute, calculate the torque required to stop it in 2.0 seconds. What will be the work done?

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Solution

Given,

$I = 0.04 k g . m^{2}$

$w_{1} = 2100 r e v / m i n = 70 π r a d / s$

$w_{2} = 0$

work done = change in kinetic energy

$= \frac{1}{2} I w_{1}^{2} - \frac{1}{2} I w_{2}^{2}$

$= \frac{1}{2} I (w_{1}^{2} - w_{2}^{2})$

$= \frac{1}{2} \times 0.4 ((70 π)^{2} - (0)^{2})$

$= 0.2 \times 4900 π^{2}$

$= 980 π^{2}$

$\approx 9800 J$

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The moment of interia of ring is 0.40 kg m2. if it is rotating at a rate of 2100 revolutions per minute, calculate the torque required to stop it in 2.0 seconds. What will be the work done?

Given,I=0.04kg.m2w1=2100rev/min=70πrad/sw2=0work done = change in kinetic energy=12Iw21−12Iw22=12I(w21−w22)=12×0.4((70π)2−(0)2)=0.2×4900π2=980π2≈9800J

The moment of interia of ring is 0.40 kg $m^{2}$ . if it is rotating at a rate of 2100 revolutions per minute, calculate the torque required to stop it in 2.0 seconds. What will be the work done?