A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from a speed of 180 revolutions per minute in 3 s. Assuming that the retardation is uniform, then the value of retarding torque `tau` in Nm isA. `10 pi`B. `20 pi`C. `30 pi`D. `40 pi`

A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from

1.	A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from a speed of 180 revolutions per minute in 3 s. Assuming that the retardation is uniform, then the value of retarding torque `tau` in Nm isA. `10 pi`B. `20 pi`C. `30 pi`D. `40 pi`
Answer» Correct Answer - B `I=m K^(2)=40xx0.25=10 kg m^(2)` `alpha = (omega_(2)-omega_(1))/(t)=(0-2pi xx 3)/(3)= - 2pi` `tau = I alpha =-2pi xx 10=-20 pi`

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A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from a speed of 180 revolutions per minute in 3 s. Assuming that the retardation is uniform, then the value of retarding torque `tau` in Nm isA. `10 pi`B. `20 pi`C. `30 pi`D. `40 pi`

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