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InterviewSolution
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A solid flywheel of 20 kg mass and 120 mm radius revolves at `600 "revmin"^(-1)`. With what force must a brake lining be pressed against it for the flywheel to stop in 3s, if the coefficient of friction is 0.1 ? |
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Answer» Given, revolutions per minute, `n_(0)=600 "rev min"^(-1)` `=(600)/(60)"rev s"^(-1)` Revolutions per second, `n_(0)=10 "rev s"^(-1)` So, initial angular velocity, `omega_(0)=2pin_(0)=(2pi)(10)=20pi "rad s"^(-1)` Let `alpha` be the constant angular retardation, then applying `omega = omega_(0)=alpha t` or `0=(20 pi)-3(alpha)` or `alpha=(20)/(3)pi "rad s"^(-2)` Further, `alpha=(tau)/(I)` Here, torque `tau=muNR" "("R=radius")` or `tau=muFR" "("F=applied force")` `=N=F and I=(1)/(2)mR^(2)` From the above equations, `(20)/(3)pi=(muFR)/((1)/(2)mR^(2))=(2muF)/(mR)` or Force, `F=(10pimR)/(3mu)` Substituting the values, we have `F=(10xx22xx20xx0.12)/(3xx7xx0.1)` or `F=2.51.43 N`. |
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