Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.5 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answer #1

(a) Use the law of conservation of mechanical energy
incorporating both translational and rotational kinetic energy. The
moment of inertia, I, for the basketball is close to 2mr²/3:

mgh = mv²/2 + I?²/2

mgh = mv²/2 + (2mr²/3)(v/r)²/2

gh = 5v²/6

h = 5v²/6g

= 5(6.5m/s)²/6(9.8m/s²)

**= 3.6m**

(b) The value of I for a frozen orange juice can (which, I suppose,
would be modeled as a solid cylinder) would be close to mr²/2, so
it’s speed would be found similarly to part (a):

mgh = mv²/2 + (1/2)(mr²/2)(v/r)²

v = ?[4gh/3]

= ?[4(9.8m/s²)(3.6m)/3]

= 6.86m/s

Hope this helps.

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