number 9 9. You came for the physics, you stayed for the problem sets^2. On Prob

Question

number 9

9. You came for the physics, you stayed for the problem sets^2. On Problem Set #3, you were disappointed at the lines for Pirates of the Caribbean and ended up at the Stan Family Robinson treehouse. You now realize that approximating the water, bucket system as an Atwood's Machine with a massless, frictionless pulley was a gross oversimplification. Instead, the pulley must have mass and it moment of inertia. They seem to be using n. barrel as the pulley, so treat it as a thin ring with total mass of 18.0 kg and n radius of 0.800 m. A filled bucket having a mass of 2.65 kg (and attached to a massless string wrapped around the barrel) is released from rest. What is the angular speed of the barrel after the bucket has fallen a distance of 12.0 m? Assume that the falling bucket rotates the barrel with no slipping, and use conservation of energy to solve this problem.

Explanation / Answer

Suppose after falling 12m bucket will have velocity v .

so KE of bucket = m v^2 /2

then angular speed of barrel = v / r

KE of barrel will be = Iw^2 /2 = ( M r^2 ) ( v/r)^2 /2 = Mv^2 /2

using energy conservation,

work done by gravity = change in KE

mgh = mv^2 /2 + M v^2 /2

2.65 x 9.81 x 12 = 2.65 v^2 /2   +   18v^2 / 2

v = 5.50 m/s

angular speed w = v/r = 5.50 / 0.800 = 6.87 m/s

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