#013 Nonuniform ball . In the figure, a ball of mass M and radius R rolls smooth

Question

#013

Nonuniform ball. In the figure, a ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.51 m. The initial height of the ball is h = 0.37 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 = ?MR2, but ? is not 0.4 as it is for a ball of uniform density. Determine ?.

Explanation / Answer

Here ,

for B (beta)

let the speed of ball at the bottom is v

as Normal force , N = 2 * Mg

2 * M* g = M * g + M * v^2/r

9.8 = v^2/0.51

v = 2.235 m/s

Now , angular speed , w = 2.235/R rad/s

for the speed of ball at the bottom

Using conservation of energy

0.5 * M * V^2 + 0.5 * I * w^2 = M* g * h

0.5 * M * V^2 + 0.5 * * M *R^2* w^2 = M* g * h

0.5 * 2.235^2 + 0.5 * *R^2* (2.235/R)^2 = 9.8 * 0.37

0.5 * 2.235^2 + 0.5 * * (2.235)^2 = 9.8 * 0.37

solving for

= 0.452

the value of is 0.452

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