Question

A ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.47 m. The initial height of the ball is h = 0.35 m. At the loop bottom, the magnitude of the normal force on the ball is 2.0 Mg. The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central sphere (of a different uniform density). The rotational inertia of the ball can be expressed in the general form I = βMR2, but β is not 0.4 as it is for a ball of uniform density. Determine β.

Answer #1

Conservation of energy

Potential energy lost = KE gained

M . g . h = 1/2 . M . v^2 + 1/2 . I . w^2

I = ß M R^2

w = v / R

Mgh = 0.5 Mv^2 + 0.5 ß MR^2 (v/R)^2 = 0.5 Mv^2 + 0.5ßMv^2

The normal force supplies the centripetal force and the gravity
support force

2 Mg = Mg + M v^2 / r

v = root (rg) = sqrt(0.47*9.8) = 2.147 m/s

Back to the energy equation - cancel the "M's"

9.81*0.35 = 0.5* 2.147^2 + 0.5 ß *2.147^2

3.4335 = 2.3048 +2.3048ß

1.1287 = 2.3048ß

ß = 0.4897 = 0.5(appr)

ß = 0.5

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