Calculating Height of Rolling Marble

[cristina]

A marble of mass M and radius R rolls without slipping down the track on the left from a height h1. The marble then goes up the frictionless track on the right to a height h2 where h2<h1 . Find h2.
I don't know how to think of this one, any hints please?

 

[ShawnD]

If there was friction on h1 and no friction on h2, there is no way h2 is higher than h1.

On h1, the friction makes it roll. So you need to know the moment of inertia.

$$I = \frac{2}{5}mR^2 + mR^2$$

$$I = \frac{7}{5}mR^2$$


At the bottom of the first track, the rotational energy will equal the initial gravitational energy

$$\frac{1}{2}I \omega ^2 = mgh$$

$$\frac{7}{10}mR^2 \omega ^2 = mgh_1$$

$$\omega ^2 = \frac{10gh_1}{7R^2}$$


Now write the energy from rolling down h1 a different way, this time rotating around the centre with the centre of rotation moving down the slope

$$\frac{1}{2}I \omega ^2 + \frac{1}{2}mv^2 = mgh_1$$

$$\frac{1}{2}mv^2 = mgh_1 - \frac{1}{2}I \omega ^2$$

$$\frac{1}{2}mv^2 = mgh_1 - \frac{1}{5}mR^2 (\frac{10gh_1}{7R^2})$$

$$\frac{1}{2}mv^2 = mgh_1 - \frac{10}{35}mgh_1$$

Now that kinetic energy (the left side of that equation) is the one that makes the marble go up to h2, so make an equation for that.

$$mgh_2 = mgh_1 - \frac{10}{35}mgh_1$$

$$mgh_2 = \frac{25}{35}mgh_1$$

cancel the m and the g

$$h_2 = \frac{25}{35}h_1$$



Can somebody confirm this answer?

 

[cristina]

May you please give some explanation before confirming the answer?
Thank you.

 

[Doc Al]

May you please give some explanation before confirming the answer?

ShawnD's analysis is correct, but can be simplified. Here's how to understand what's going on:

When the ball rolls down the track there is friction causing it to roll without slipping. Remember that the condition for rolling without slipping is V = ωR.

The ball starts with an initial gravitational PE = mgh₁. As the ball rolls down, that PE is converted to translational KE (1/2mV²) and rotational KE about the center of mass (1/2Iω²). Remember that the two KEs are tied by V = ωR.

When the ball rolls up the track, there is no friction so the rotational KE remains fixed. Only the translational KE is converted to gravitational PE = mgh₂. So, solve for the translational KE at the bottom of track using:
mgh₁ = 1/2mV² + 1/2Iω², along with: V = ωR.
Remember you are solving for 1/2mV², so don't do any unnecessary work. Your answer will be some multiple of mgh₁. Now set that equal to mgh₂, and solve for h₂.