Question

A hollowed sphere of mass M is rolling without slipping on the horizontal floor with a linear velocity of its center of mass. It coasts an incline having an angle θ. What is the maximum height h reached before rolling back down the incline.

Answer 1

Best way is to use conservation of mechanical energy.
Are you familiar with this method?

If not, you should study the motion of the sphere up the inclined plane, find its acceleration and work out maximum height attained. But this method takes more time.

Answer 2

On thing you will need in either case is the moment of inertia of the hollow sphere.
You should find it in our textbook or online.

Answer 3

Conservation of energy? I guess there will be torque due to the friction acting on the sphere.

Answer 4

Q: "Conservation of energy? I guess there will be torque due to the friction acting on the sphere."

A: Yes there will. But since the sphere is rolling and the incline is fixed, the friction force does no work at all. So the system is conservative.

Answer 5

Power of a force \(\vec F\) acting at a point A on a system is dot-product of force and velocity of point of application: \(P=\vec F.\vec v(A)\)

In the case of pure rolling on a fixed base, point of contact \(I_{wheel}\) has instantaneous zero velocity. So instantaneous power is zero, so work done \(W=\int P(t)\,dt\) is zero as well.

This is the miracle of the rolling wheel: there is a force due to friction accelerating the wheel, but this force does not work at all.

Note that this is valid if the base is fixed. For a moving base, the friction force will do work because point of contact \(I_{wheel}\) will be moving too.

Answer 6

Hmm..yes. True!

Answer 7

Have you found the solution?

Answer 8

No
I have to solve with motion of the sphere and the moment of inertia
but I really can't understand my answer
i calculate the kinetic energy (linear and rotational ) and equal them to P.E ( mgh )
0.5 m v^2 + 0.2 I w^2 = mgh
w = v/R
and then ?
how the final answer will look like ?

Answer 9

What is this 0.2 in your rotational KE?

Actually,
KErot = ½ I ω²

For a hollow shere, I = ⅔ MR²

The rest is ok. Solve for h.

Answer 10

oh I sorry meant 0.5
0.5 m v^2 + 0.5 * (2/3 m R^2) (v^2/R^2) = mgh
0.5 v^2 + 1/3 v^2 = gh
h= (5/6 v^2 )/g
am I right ?

Answer 11

Correct. Well done!