Conservation of energy of object rolling down incline

AI Thread Summary
The discussion focuses on the conservation of energy for a marble rolling down an incline and then up a frictionless track. The initial energy equation, Mgh1 = (1/2)Mv^2 + (1/2)Iω^2, is correctly set up, but confusion arises regarding the rotational kinetic energy as the marble moves up the ramp. Participants clarify that while there is no torque on the marble during its ascent, the rotational kinetic energy remains constant. The final height h2 is derived from the conservation equations, but discrepancies in the expected answer indicate potential errors in the calculations. Ultimately, the conversation emphasizes the importance of correctly applying conservation principles in rotational motion.
Raziel2701
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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


Mgh_1=\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2

\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2=Mgh_2 +\frac{1}{2}mv^2

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

\frac{5}{7}h_1 +\frac{2}{7}R

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?
 
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Raziel2701 said:
since there is no rotational kinetic because the track is frictionless.
Just becuase there's no friction doesn't mean it can't rotate.
 
You've also neglected friction while the marble is rolling down the incline.
 
Raziel2701 said:
Mgh_1=\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2
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.

\frac{1}{2}*Mv^2 +\frac{1}{2}*I*\omega^2=Mgh_2 +\frac{1}{2}mv^2
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.

chaoseverlasting said:
You've also neglected friction while the marble is rolling down the incline.
Actually, he didn't. His first equation incorporates that.
 
\frac{1}{2}Mgh_1 - \frac{1}{2}I\omega^2=\frac{1}{2}Mv^2

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:

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

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?

Mgh_1=Mgh_2 +\frac{1}{2}Mv^2 +\frac{1}{2}I\omega^2

Should I solve for h_2 in terms of v, or omega, or is there something else, another equation I should consider?
 
Raziel2701 said:
\frac{1}{2}Mgh_1 - \frac{1}{2}I\omega^2=\frac{1}{2}Mv^2

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:

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

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?
 
Doc Al said:
Actually, he didn't. His first equation incorporates that.

Because of the rotational energy term, right? :redface:
 
chaoseverlasting said:
Because of the rotational energy term, right?
Yep.
 
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.
 
  • #10
Raziel2701 said:
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.
 
  • #11
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.
 
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  • #12
mistasong said:
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.
 
  • #13
Well after I solved for 1/2mv^2 I don't know what to do.
 
  • #14
Actually I got it now
Thank you so much Doc Al
 
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