A disk-shaped pulley of mass M and radius R is rotating at angular velocity w. T

Question

A disk-shaped pulley of mass M and radius R is rotating at angular velocity w.  Th friction in the bearing is so small to be ignored.  A brake shoe is being pressed against the pulley to stop it.  The brake shoe is pushed against with the same force and the coefficient of friction is the same.

A m-10kg R-0.8 m w-12

B m-10 kg R-0.4 m w-24

C m-20 kg R-0.4 m w-6

D m-40 kg R-0.2 m w-3

E m-20kg  R-0.2 m w-3

F m-30 kg R-0.6 m W-8

Rank on the angle the pulley rotates through before the pulley stops (explain with math)

Explanation / Answer

KA = 0.5*IA*WA^2 = 0.5*0.5*M*R^2*wA^2 = 0.25*10*0.8^2*12^2 = 230.4 J

KB = 0.5*IB*WB^2 = 0.5*0.5*M*R^2*wB^2 = 0.25*10*0.4^2*24^2 = 230.4 J

KC = 0.5*IC*WC^2 = 0.5*0.5*M*R^2*wC^2 = 0.25*20*0.4^2*6^2 = 28.8 J

KD = 0.5*ID*WD^2 = 0.5*0.5*M*R^2*wD^2 = 0.25*40*0.2^2*3^2 = 3.6 J

KE = 0.5*IE*WE^2 = 0.5*0.5*M*R^2*wE^2 = 0.25*20*0.2^2*3^2 = 1.8 J

KF = 0.5*IF*WF^2 = 0.5*0.5*M*R^2*wF^2 = 0.25*30*0.6^2*8^2 = 172.8 J


work done = change in kinetic enrgy

T*theta = chnage in kinetic enrgy

here T(torque) is constant

so. theta is proportional to chnage in kinetic energy

KA = KB < KF < KC < KD < KE

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