Conservation of energy of object rolling down incline

[Raziel2701]

Homework Statement

A marble of mass M and radius R rolls without slipping down the track on the left from a height h1, as shown below. The marble then goes up the frictionless track on the right to a height h2. Find h2. (Use the following as necessary: M, R, and h1.)

http://imgur.com/sZIyQ

The Attempt at a Solution

[tex]Mgh_1=\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2[/tex]

[tex]\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2=Mgh_2 +\frac{1}{2}mv^2[/tex]

I don't know why but these equations for conservation of energy are not giving me the right answer which is supposed to be:

[tex]\frac{5}{7}h_1 +\frac{2}{7}R[/tex]

So I know that my conservation of energy equations are incorrect but I don't know where nor why.

I'm setting potential at the top to be equal to kinetic at the bottom, and the kinetic at the bottom to be equal to potential at h2 plus linear kinetic since there is no rotational kinetic because the track is frictionless. Yet I'm not getting anywhere with those because that velocity at h2 is different than the velocity at the bottom, so I have three unknowns and two equations.

I'm using the moment of inertia to be 2/5 MR^2 and also the relationship that w=v/r to simplify things. Bottom line is, are my conservation of energy equations correct?

 

[da_nang]

since there is no rotational kinetic because the track is frictionless.

Just becuase there's no friction doesn't mean it can't rotate.

 

[chaoseverlasting]

You've also neglected friction while the marble is rolling down the incline.

 

[Doc Al]

[tex]Mgh_1=\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2[/tex]

This first equation is fine. Note that in order to roll without slipping, we must assume that the downward segment of the track has friction; only the upward segment is frictionless.

[tex]\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2=Mgh_2 +\frac{1}{2}mv^2[/tex]

This one needs correction. As was pointed out already, you neglect the rotational KE of the marble as it reaches the top. (Since there's no friction on the upward segment, and thus no torque on the marble, its rotational motion cannot change on its upward climb.)

Hint: Use your first equation to solve for the translational KE at the bottom in terms of h1.

You've also neglected friction while the marble is rolling down the incline.

Actually, he didn't. His first equation incorporates that.

 

[Raziel2701]

[tex]\frac{1}{2}Mgh_1 - \frac{1}{2}I\omega^2=\frac{1}{2}Mv^2[/tex]

On that one, should I put in w=v/r? My fear is that I'm canceling the R, and getting v back to put into this one equation:

[tex]Mgh_1 - \frac{1}{2}I\omega^2 +\frac{1}{2}I\omega^2=Mgh_2 +\frac{1}{2}Mv^2 +\frac{1}{2}I\omega^2[/tex]

Well, on the LHS of the equation the rotational terms cancel and I guess that's ok. On the RHS I have a 1/2Mv^2 term where I would suppose the velocity is different than at the bottom right? So do you see how I don't know how to go forwards from here?

[tex]Mgh_1=Mgh_2 +\frac{1}{2}Mv^2 +\frac{1}{2}I\omega^2[/tex]

Should I solve for h_2 in terms of v, or omega, or is there something else, another equation I should consider?

 

[Doc Al]

[tex]\frac{1}{2}Mgh_1 - \frac{1}{2}I\omega^2=\frac{1}{2}Mv^2[/tex]

On that one, should I put in w=v/r?

Yes, express ω in terms of v. Then solve for the translational KE. Then you can move on to fix the second equation.

My fear is that I'm canceling the R, and getting v back to put into this one equation:

[tex]Mgh_1 - \frac{1}{2}I\omega^2 +\frac{1}{2}I\omega^2=Mgh_2 +\frac{1}{2}Mv^2 +\frac{1}{2}I\omega^2[/tex]

Well, on the LHS of the equation the rotational terms cancel and I guess that's ok.

No, this is just going in circles.

When the marble slides up the ramp, what changes? What doesn't change? What's conserved?

 

[chaoseverlasting]

Actually, he didn't. His first equation incorporates that.

Because of the rotational energy term, right?

 

[Doc Al]

Because of the rotational energy term, right?

Yep.

 

[Raziel2701]

When the marble slides up the ramp, the rotation should stop because there's nothing for the marble to rotate against isn't it? There's no torque on the marble.

So only the translational kinetic energy is transformed into potential correct?

If so, I think I got this one.

 

[Doc Al]

When the marble slides up the ramp, the rotation should stop because there's nothing for the marble to rotate against isn't it?

No.

There's no torque on the marble.

If there's no torque on the marble (which is true) then what's to stop its rotation? Any change in rotational speed requires a torque.

So only the translational kinetic energy is transformed into potential correct?

That happens to be true, but not because the rotational KE goes away. Whatever rotational KE it has at the bottom of the ramp will equal the rotational KE at the top. So it drops out of your conservation of energy equation.

 

[mistasong]

So solving the first equation gives 1/2 mv^2 = 5/7 mgh1.
Plugging into the second equation then gives me h2 = 5/7h1 + (2v^2)/(10g).
however that does not match the answer given above.

 

[Doc Al]

So solving the first equation gives 1/2 mv^2 = 5/7 mgh1.

Good.

Plugging into the second equation then gives me h2 = 5/7h1 + (2v^2)/(10g).

What second equation?

however that does not match the answer given above.

Your answer is incorrect, but I think the answer given in post #1 is also incorrect.

 

[mistasong]

Well after I solved for 1/2mv^2 I don't know what to do.

 

[mistasong]

Actually I got it now
Thank you so much Doc Al