Rotational Motion-Up an Incline

[abspeers]

Homework Statement

A solid sphere (S), a thin hoop (H), and a solid disk (D),
all with the same radius, are allowed to roll down an
inclined plane without slipping. At the bottom, they run into another inclined plane and begin rolling upwards. Which will travel the furthest up the inclined plane, neglecting all frictional forces?

Thin Hoop KE=1/2mv^2 + 1/2(MR^2)w^2

Solid Sphere KE=1/2mv^2 + 1/2(2/5MR^2)w^2

Solid Disk KE=1/2mv^2 + 1/2(1/2MR^2)w^2

Homework Equations

1/2mv^2 + 1/2Iw^2 = mgh

The Attempt at a Solution

A previous question had asked which reaches the bottom first and I had found that to be the sphere, followed by the disk, and finally the thin hoop. I suspect that the sphere would travel the furthest up the incline because it has the greatest KE at the bottom of the first incline, but I am unsure if this assumption is correct.

Thank you for the help!

 

[tiny-tim]

Welcome to PF!

Hi abspeers! Welcome to PF!

… I suspect that the sphere would travel the furthest up the incline because it has the greatest KE at the bottom of the first incline, but I am unsure if this assumption is correct.

You're right, it's not correct …

never mind the KE at the bottom, its the KE and PE at the start that matters, isn't it?

 

[kerimek]

Your assumption is correct. The sphere will travel the furthest up the incline due to its greater kinetic energy at the bottom of the first incline. This is because the sphere has both translational and rotational kinetic energy, while the hoop and disk only have rotational kinetic energy. Therefore, the sphere has a greater total kinetic energy and will be able to overcome the gravitational potential energy to travel further up the incline. Additionally, the moment of inertia for a solid sphere is smaller than that of a disk or hoop, meaning it requires less energy to rotate and therefore has more energy available for translational motion.