Nonuniform ball. A ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.45 m. The initial height of the ball is h = 0.39 m. At the loop bottom, the magnitude of the normal force on the ball is 2.01Mg. 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 = βMR^2, but β is not 0.4 as it is for a ball of uniform density. Determine β.
Here is a hint: In order for the ball to have the given normal force, what must be its speed at the bottom of the ramp? Conserve mechanical energy for the ball rolling down the ramp, taking into account that it will have both translational and rotational kinetic energy.

Question

anonymous · Answer

At the loop bottom, the magnitude of the normal force on the ball is 2.01Mg

IDK maybe I am reading this wrong but shouldn't force be given in newtons?

anonymous · Answer

I believe it is 2.01*Mass*gravity.
At rest, the normal force would be 1*M*g.
the additional acceleration is 1.01*M*g, which is related to the rotational motion around the .045m radius circle.

anonymous · Answer

aaa

anonymous · Answer

The answer is .716