Question

Some coconuts arrange ten monkeys like bowling pens. The coconuts then climb a curved hill until they are 100 meters above the ground, from which point they release an object from rest, hoping to score a strike. (A) (i) Find the final speed of the object if it is a uniform solid sphere that rolls without slipping. (ii) How much faster would the sphere reach the bottom if it were to slide down the hill in the absence of friction? (B) Find the final speed of the object if it is a uniform cylindrical shell with inner radius equal to one-half the outer radius, which rolls without slipping.

Answer 1

Part Ai)

For a solid sphere, I = 2/5mr²

At the bottom, it will have KE rotational and KE translational

KEr = .5Iw² and w = v/r, so w² = v²/r²

Thus KEr = .5(2/5)(mr²)(v²/r²) which simplifies to .2mv²

So, by conservation of Energy

PE = KEr + KEt

mgh = .2mv² + .5mv² (mass cancels)

(9.8)(100) = (.7v²)

v = 37.4 m/s

Part Aii)

Without friction, there would be no rotational KE, so...

PE = KE

mgh = .5mv²

(9.8)(100) + .5v²

v = 44.3 m/s

That is 44.3 - 37.4 = 6.87 m/s faster

Part B)

I for the shell with the two radii is .5(M)(b² - a²)

In this case, I = .5m(r² - (.5r)²) = .375mr²

KE = .5(.375mr²)(v²/r²) = .1875mv²

Thus mgh = .1875mv² + .5mv²

(9.8)(100) = (.6875)v²

v = 37.8 m/s