Which sphere reaches the bottom of the inclined plane first

  • #1

Homework Statement



Two spheres are placed side by side on an inclined plane and released at the same time. Both spheres roll down the inclined plane without slipping.

(a) Using FBD, explain what force provides the torque allowing the sphere to roll down the inclined plane.
(b) Which sphere reaches the bottom of the inclined plane first and why?
(c) How do the kinetic energies of the two spheres compare at the bottom of the inclined plane?

Homework Equations



Tau = Fr sin theta

The Attempt at a Solution



(a) Friction is the only force providing the torque. Fg is applied directly from the centre of mass of the spheres so it has r=0 and so cannot provide any torque. F_N is directed towards the centre of mass so it has sin theta = 0.

(b) Here's where I'm getting a little confused. I used energy to solve this. For the hollow sphere, most of the mass needs to be rotated, so it has more KE_rot than KE_trans. But for the solid sphere, most of the mass will effectively be sliding across, so it has more KE_trans than KE_rot. So with greater KE_trans the solid sphere will have a greater speed and so will reach the bottom first.

However, I'm pretty sure this isn't entirely right bc I'm ignoring the energy released by friction, but I'm not sure how to determine how much energy was lost to friction for each sphere.

(c) I think this is the same issue with what I found in (b). Without knowing how much loss of energy friction caused, how can I determine how much KE is left —I'm assuming they're talking about the total KE here.
 

Answers and Replies

  • #2
PeroK
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Homework Statement



Two spheres are placed side by side on an inclined plane and released at the same time. Both spheres roll down the inclined plane without slipping.

(a) Using FBD, explain what force provides the torque allowing the sphere to roll down the inclined plane.
(b) Which sphere reaches the bottom of the inclined plane first and why?
(c) How do the kinetic energies of the two spheres compare at the bottom of the inclined plane?

Homework Equations



Tau = Fr sin theta

The Attempt at a Solution



(a) Friction is the only force providing the torque. Fg is applied directly from the centre of mass of the spheres so it has r=0 and so cannot provide any torque. F_N is directed towards the centre of mass so it has sin theta = 0.

(b) Here's where I'm getting a little confused. I used energy to solve this. For the hollow sphere, most of the mass needs to be rotated, so it has more KE_rot than KE_trans. But for the solid sphere, most of the mass will effectively be sliding across, so it has more KE_trans than KE_rot. So with greater KE_trans the solid sphere will have a greater speed and so will reach the bottom first.

However, I'm pretty sure this isn't entirely right bc I'm ignoring the energy released by friction, but I'm not sure how to determine how much energy was lost to friction for each sphere.

(c) I think this is the same issue with what I found in (b). Without knowing how much loss of energy friction caused, how can I determine how much KE is left —I'm assuming they're talking about the total KE here.
That looks good so far. Think a bit more about how friction is acting in the case of a rolling sphere. Hint: note that the sphere is rolling without slipping.
 
  • #3
kuruman
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Be careful here. Some points to consider.
(a) If the spheres roll without slipping and you ignore rolling resistance, air resistance and such (as is usually the case in such problems), there is no loss of energy to friction.
(b) What is the fraction KErot/mgh? Yes, the more massive sphere starts with more PE, but if the same fraction of its potential energy is converted to rotational, would it have more speed at the bottom? You can also compare the linear acceleration of the two spheres using the FBD as the problem suggests.
 
  • #4
@kuruman — Oh... I thought that the reason why they explicitly asked part (a) was to let people know that there was friction at play? So am I safe to assume no friction in cases like these (at all times)?

As for (b) in your post, so I'm guessing your point is I should not be using energy but rather linear acceleration? However, I'm wondering if it is rotational acceleration instead because from what I've got, if one of the spheres has mass m and the ramp has angle theta, mg sin theta - F_f = ma. Since F_f is negligible, mg sin theta = ma so m cancels out and g sin theta = a. However, since both sphere are on the same ramp they have same linear acceleration?
 
  • #5
PeroK
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@kuruman — Oh... I thought that the reason why they explicitly asked part (a) was to let people know that there was friction at play? So am I safe to assume no friction in cases like these (at all times)?

As for (b) in your post, so I'm guessing your point is I should not be using energy but rather linear acceleration? However, I'm wondering if it is rotational acceleration instead because from what I've got, if one of the spheres has mass m and the ramp has angle theta, mg sin theta - F_f = ma. Since F_f is negligible, mg sin theta = ma so m cancels out and g sin theta = a. However, since both sphere are on the same ramp they have same linear acceleration?
Without friction both spheres will slip without rotation and move simply under the influence of the linear gravitational acceleration. There must be friction to initiate rolling.

However, your assumption that energy is lost to rolling friction is not correct. Can you see why? It's not an easy question.
 
  • #6
kuruman
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@kuruman — Oh... I thought that the reason why they explicitly asked part (a) was to let people know that there was friction at play? So am I safe to assume no friction in cases like these (at all times)?

As for (b) in your post, so I'm guessing your point is I should not be using energy but rather linear acceleration? However, I'm wondering if it is rotational acceleration instead because from what I've got, if one of the spheres has mass m and the ramp has angle theta, mg sin theta - F_f = ma. Since F_f is negligible, mg sin theta = ma so m cancels out and g sin theta = a. However, since both sphere are on the same ramp they have same linear acceleration?
I wrote that there is no loss of energy to friction. This does not mean that there is no friction as @PeroK pointed out. You can use either energy or linear acceleration to answer this problem. Energy considerations will get you to the answer faster, but figuring out the angular and linear accelerations will be more instructive. Consider this: If you dropped the two spheres at the same time in free fall (no incline), they would hit the ground at the same time even though the masses are different. What is the same and what is different between that case and the one you have here?
 
  • #7
@PeroK — So just to clarify, the force of friction I drew on the FBD would be a rolling friction?

I'm probably off here, but I was thinking maybe since F_f is applied at the same point of each sphere so there isn't really any work done...?

@kuruman — your different masses reaching the ground at the same time example almost makes me want to think that rotational inertia doesn't matter, but that doesn't seem to make sense to me...

so there's rolling motion and friction and normal forces involved in this situation that we don't have in the example you gave me. The masses aren't exactly given so I'm guessing that they are different (same as the example you gave).
 
  • #8
PeroK
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@PeroK — So just to clarify, the force of friction I drew on the FBD would be a rolling friction?

I'm probably off here, but I was thinking maybe since F_f is applied at the same point of each sphere so there isn't really any work done...?
The simplest way to think about it is that the point of contact of the sphere is instantaneously at rest, hence you effectively have static friction, which does no work.

You can, however, calculate the magnitude of the friction force, the resultant rotational accleration and linear opposition to gravity and what comes out is an equation for total kinetic energy (linear plus rotational) that equals the gravitational PE lost. That confirms that if the friction force is just right to maintain rolling without slipping then no energy is lost. Note that static friction has this capacity to act variably up to a maximum value.
 
  • #9
kuruman
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@PeroK — So just to clarify, the force of friction I drew on the FBD would be a rolling friction?

I'm probably off here, but I was thinking maybe since F_f is applied at the same point of each sphere so there isn't really any work done...?

@kuruman — your different masses reaching the ground at the same time example almost makes me want to think that rotational inertia doesn't matter, but that doesn't seem to make sense to me...

so there's rolling motion and friction and normal forces involved in this situation that we don't have in the example you gave me. The masses aren't exactly given so I'm guessing that they are different (same as the example you gave).
I will answer for @PeroK (unless he answers first) and say yes, friction does no work here in the sense that it does not dissipate energy to heat. All it does is convert some of the potential energy to rotational kinetic energy about the center of mass while the rest of the potential energy goes into translational KE of the center of mass. If friction were not there, all of the potential energy would be converted to translational KE of the center of mass.

If the problem does not specify that the masses or the radii of the spheres are equal, you may not assume that they are. The question may be confusing now but, if you think it through, it will make sense.
 
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  • #10
Okay. So going back to the problem, for part (b), I don't have to worry that I left friction out of consideration etc. Both spheres are rolling w/o slipping so friction does convert some GPE to KE_rot in both cases, but I think my original reasoning still works there.

For (c), assuming they're talking about the total KE that is KE_rot + KE_trans, since the rotational friction causes no loss of ME, the KE should be the same at the bottom for both spheres by the conservation of energy?
 
  • #11
kuruman
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Okay. So going back to the problem, for part (b), I don't have to worry that I left friction out of consideration etc. Both spheres are rolling w/o slipping so friction does convert some GPE to KE_rot in both cases, but I think my original reasoning still works there.

For (c), assuming they're talking about the total KE that is KE_rot + KE_trans, since the rotational friction causes no loss of ME, the KE should be the same at the bottom for both spheres by the conservation of energy?
The KE will not be the same if the masses are different because the total mechanical energy at the start is mgh.
 
  • #12
PeroK
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Okay. So going back to the problem, for part (b), I don't have to worry that I left friction out of consideration etc. Both spheres are rolling w/o slipping so friction does convert some GPE to KE_rot in both cases, but I think my original reasoning still works there.

For (c), assuming they're talking about the total KE that is KE_rot + KE_trans, since the rotational friction causes no loss of ME, the KE should be the same at the bottom for both spheres by the conservation of energy?
I think that's correct. Although, for part b) you might like to think about whether the frictional force is the same in both cases.
 
  • #13
PeroK
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The KE will not be the same if the masses are different because the total mechanical energy at the start is mgh.
My guess is that the problem is for two spheres of the same mass and radius: one hollow, one solid.

But, that's just a guess!
 
  • #14
kuruman
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My guess is that the problem is for two spheres of the same mass and radius: one hollow, one solid.

But, that's just a guess!
That will make a difference. There is no mention of one solid and one hollow in the statement of the problem. @JessicaHelena please clarify.
 
  • #15
@kuruman, PeroK — I'm sorry for the confusion, turns out I had omitted the thing in the brackets after the question statement (one hollow, one solid), as I've been saying (I think). There's no mention of whether the masses are the same, but the solution (which I don't think is too helpful frankly) assumes the masses to be equal.
 
  • #16
@PeroK — so if the frictional forces are the same then KE_rot would be the same?


If I didn't know the masses were the same then I don't think I'd be able to figure that out, but since the solution assumes the masses to be equal, then the spheres would have the same normal force and since they're on the same ramp, they'd have same mu and so same frictional force. Please point out anything wrong with this if there is.
 
  • #17
PeroK
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@PeroK — so if the frictional forces are the same then KE_rot would be the same?


If I didn't know the masses were the same then I don't think I'd be able to figure that out, but since the solution assumes the masses to be equal, then the spheres would have the same normal force and since they're on the same ramp, they'd have same mu and so same frictional force. Please point out anything wrong with this if there is.
They have the same maximum static frictional force. But, static friction does not always act to the maximum.

For example, on a slight incline, the max friction would be high, but rolling would be slow. So, a smaller frictional force is acting. And, the opposite on a steeper incline.
 
  • #18
@PeroK — oh okay... but now I'm not quite sure how to finish up part (b)...
 
  • #19
PeroK
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@PeroK — oh okay... but now I'm not quite sure how to finish up part (b)...
I think the answers you gave are right.
 
  • #20
kuruman
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I think the answers you gave are right.
Specifically, your argument to (b) applies to a hollow vs. solid sphere race regardless of mass or radius as long as there is rolling without slipping.
 
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