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Imagine a frictionless pulley (a solid cylinder) of unknown mass M and radius r = 0.200 m which is used to draw water from a well. A bucket of mass m = 1.50 kg is attached to a massless cord wrapped around the pulley. The bucket starts from rest at the top of the well and falls for t = 3.00 s before hitting the water h = 7.65 m below the top of the well.

(a) What is the linear acceleration of the falling bucket?

(b) What is the angular acceleration of the pulley?

(c) What is the tension in the cord?

(d) What is the torque that is applied to the pulley due to the cord? (use the tension from the previous question)

(e) Using the torque and the angular acceleration, find the moment of inertia of the pulley.

(f) Using the moment of inertia, find the mass of the pulley.

(g) What is the change in the potential energy of the bucket?

(h) What is the velocity of the bucket when it hits the water? (use the previously calculated linear acceleration)

(i) What is the angular velocity of the pulley when the bucket hits the water? (use the relation between the angular velocity and the linear velocity)

(j) What is the angular momentum of the pulley when the bucket hits the water? (use the moment of inertia and the angular velocity of the pulley)

(k) What is the kinetic energy of the bucket when it hits the water?

(l) What is the kinetic energy of the pulley when the bucket hits the water?

(m) What is the total kinetic energy of the system when the bucket hits the water?

(n) Is the total kinetic energy smaller, equal to (within +/- 1%), or larger than the potential energy change? ...smaller, larger, or equal to (within 1%)

(o) If the pulley was not frictionless, the kinetic energy would be smaller, equal to, or larger than the potential energy change? ...smaller, larger, eqaul to

Added Mon, 27 Jul '15

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Solutions

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Part A)

Apply d = vot + .5at2

7.65 = 0 + .5(a)(3)2

Part B)

alpha = a/r

alpha = 1.7/.2

Part C)

From the free body diagram of the falling bucket

mg - T = ma

(1.5)(9.8) - T = 1.5a

T = 14.7 - 1.5a

T = 14.7 - 1.5(1.7)

Part D)

Torque = Tr

Torque = 12.2(.2)

Part E)

Torque = I(alpha)

2.43 = I(8.5)

Part F)

I for a disk is .5mr^2

.286 = .5(M)(.2)^2

Part G)

PE = mgh

PE = (1.5)(9.8)(7.65)

Part H)

vf = vo + at

vf = 0 + (1.7)(3)

vf = 5.1 m/s

Part I)

w = v/r

w = 5.1/.2

Part J

L = Iw

L = (.286)(25.5)

L = 7.29 kg m^2/s

Part K)

KE = .5mv^2

KE = .5(1.5)(5.1)^2

Part M)

KE = .5Iw2

KE = .5(.286)(25.5)2

KE = 93.0 J

Part N)

Total KE = 93 + 19.5

Part O)

Smaller by conservation of Energy. Some energy would go into overcoming friction, which is Non-Conservative Energy

Added Mon, 27 Jul '15

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