Rotational Motion-Up an Incline

AI Thread Summary
A solid sphere, thin hoop, and solid disk are analyzed for their motion down and up an incline without slipping. The kinetic energy (KE) equations for each shape indicate that the sphere has the greatest KE at the bottom of the incline. However, the discussion highlights that the potential energy (PE) at the start of the incline is crucial for determining how far each object will travel upwards. The conclusion suggests that the initial conditions, rather than just the KE at the bottom, dictate the distance traveled up the incline. Understanding the balance of KE and PE is essential for solving this problem accurately.
abspeers
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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!
 
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Welcome to PF!

Hi abspeers! Welcome to PF! :smile:
abspeers said:
… 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? :wink:
 

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