# Challenge problem, week of 11/29/10

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Here's a problem I encountered sometime in the first few weeks of mechanics last semester. Post an answer if you get one, or ideas if you have them. I can provide hints as necessary (diagrams, possibly useful tools, etc.) maybe starting later in the week. At this point I am considering turning these around on a weekly basis, and so I'll probably post a solution Monday, 12/6. If Mr. Fullerton wants to give extra credit for such things, I imagine one would want to have it done before then for extra points. And show all work! Whether by beautiful $\LaTeX$ here or by mailing be the back of a napkin, documenting your thinking is imperative.

Suppose a hockey puck sits atop a spherical, frictionless surface of radius R. The puck is then given a very small nudge, just so that it begins moving slightly (this nudge imparts a negligible amount of momentum onto the puck). Through what vertical distance does the puck descend before leaving the surface?

Edited by Anthony Frachioni
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it seems too easy but wouldn't the answer just be R? because once it travels down a vertical distance of R it would be at the widest part of the sphere, even with the center, and would thus proceed to fall straight down rather than maintain contact with the sphere's surface.

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Hi Elliot,

It does seem too easy! And I understand why you think the answer must be R, because I did too when I first read the problem. Unfortunately, it's not that easy. As the puck slides down the surface of the sphere, it accelerates both out and down. At some point, the out component of the puck's velocity becomes so great that gravity cannot keep it on the sphere. I encourage you to convince yourself of this by rolling a marble off a basketball or some other similar surface. (Rolling adds extra complexity to the problem, but the same principle applies.)

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So either few people have looked at this, or it's so hard that nobody wants to begin it. In any case, Fullerton has asked me to give a hint. I'll start with a small hint; I'll add to it as necessary. Draw a free body diagram. You never outgrow those.

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Maybe I'm forgetting something critical, but what force is acting on the puck such that it accelerates outward as well? The Normal force on the basketball to the puck couldn't be unbalanced I figure, so I'm left wondering what is unaccounted for.

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The Normal force on the basketball to the puck couldn't be unbalanced

Why?

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I have limited knowledge of the normal force, as I learned in physics B that it is simply a "reactionary force" always with equal magnitude to the initiating force unless deformation occurred. In physics C we spent perhaps 5 minutes learning that the normal force is a summation of resistance forces from everything that the initial force is "pushing on." With that in mind, the normal force is the only force in play that I imagine would be able to cause the puck to leave the basketball early. BUT I assume because there is no deformation to the basketball or the puck that the normal force can be no more than equal to the weight of the puck at any time, and so there is no acceleration outward. Please disprove my statement.

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I was attempting to recreate the situation on a real basketball but using a longer object on top (I didn't have a hockey puck), and it does indeed look as though the long object leaves the basketball before getting halfway down. I am beginning to think that this is because the fulcrum location on the puck changes as it moves down the basketball until the force of gravity on the puck causes it to rotate until it leaves the basketball (because it is orientated more vertically at this point).

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Wait just a minute, this outward acceleration doesn't have anything to do with the "centrifugal force" does it?

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Sorry for taking awhile on this one, the semester's just started up here. Lots of physics to be done! And maybe it's a good thing, willorn seems to have been thinking about this a fair amount. I'll respond to these posts in the order in which they were posted.

7) On the normal force: The most important thing about the normal force is that it is normal, or perpendicular, to the surface of contact. However, it's not always equal in magnitude to the "initiating force" that was mentioned. You asked for proof, so suppose I dropped a marble down a crack in the sidewalk. The crack has very steep walls, and the marble gets jammed between them because the crack gets narrower with increasing depth (see picture).

Now try to draw a free body diagram for the marble. Only gravity and normal forces exist (and normal forces are perpendicular to the nearly vertical surfaces). Also, the marble is in equilibrium with it's environment and isn't accelerating, so the vector sum of all forces must be zero. When you're done, you'll have long normal force vectors to satisfy the static condition in the vertical direction (which, presuming I made the walls steep enough, are longer than the gravity vector). Recall Newton does not require that force be conserved. If a force like gravity is applied to some body or system, physics does not forbid the system from responding with forces of greater magnitude.

I may be drifting a little from the question at hand, but hopefully I've helped clear up some confusion.

8) Assume the bottom surface of the puck is tangent to the sphere whenever the puck and sphere are in contact. A puck is used in this problem just so that it's clear that the object doesn't roll down the sphere. Without friction, even a ball won't roll, but this helps be more explicit. Also, a basketball and hockey puck may not be the best props for trying this at home (rubber on rubber is kinda sticky). Try rolling a tennis ball down an exercise ball if you've got one, pretending the tennis ball doesn't roll. You won't get the same answer because the tennis ball is rolling, but you'll at least see it fly off into space (and preview part two of the problem...) and get a general sense of what's happening.

9) The dreaded centrifugal force... Intro physics teachers everywhere deny its very existence, which is one of a series of lies told to their students (it's for your own good). It does exist, but this problem can be solved without thinking about it. If you like, we can talk more about this often confusing detail, but maybe another thread would be a more appropriate place for that conversation. You're on the right track! Just quantify the phenomenon in terms of centripetal, not centrifugal, forces.

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After the conclusion of my AP tests, I think I've finally found a nice solution to this problem! I will employ the wonderful Latex to derive my answer!

I began with an FBD diagram, just like my momma (Cermak) taught me.

[ATTACH=CONFIG]131[/ATTACH]

I began by examining my FBD and drawing trigonometric conclusions. (After receiving word from Frach, I realized I neglected to mention the Normal; It is important to note that the Normal would necessarily be zero for the puck to fly off the basketball, and that the normal force is a function of the centripetal force and therefore:$cos\theta = (Fc-Fn)/mg \Rightarrow Fn = 0 \Rightarrow cos\theta = Fc/mg$

[ATTACH=CONFIG]132[/ATTACH]

Then I used my given knowledge that the basketball is frictionless in order to determine that the total energy is conserved in the basketball puck system.

[ATTACH=CONFIG]133[/ATTACH]

Later of course, I realized I had unnecessarily taken the square root of velocity, oh well. I substitute the centripetal acceleration from the first eqution with its representative expression and substitute my new velocity! At this time I am very hopeful.

[ATTACH=CONFIG]135[/ATTACH]

And now I am stuck. But here comes the part that took some more observation, if we look at the FBD again, it can be seen that an equivalent theta exists accidentally in the very diagram I just drew! This comes as a result of the basic geometric concept of Alternate Interior Angles. The length that was previously ignored can be expressed as the difference between R, (the total radius) and "k" (the height at which the puck leaves the basketball).

[ATTACH=CONFIG]136[/ATTACH]

Using a new expression of the same angle theta, I can now set up and solve for what seems to be an answer that is either perfect or too good to be true. Note that I was able to substitute the "h" for "k" in the expression because k represents the vertical height the puck has fallen.

$\inline cos\theta = 2h/r = \left ( r-k \right )/r \Rightarrow 2k/r = \left ( r-k \right )/r \Rightarrow 2k = r-k \Rightarrow k = r/3$

Edited by willorn

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Great analysis willorn... can you elaborate on the centripetal force is caused by in your problem?

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Willorn:

Yo momma raised you right.

As we discussed on Thursday, you took a big step near the top and didn't explain yourself. I first thought you found the answer by mistake. Like FizziksGuy pointed out, a discussion of centripetal and normal forces would probably be a good idea in order to completely describe your solution.

Another idea: keep track of which symbols are variable and which are constant. For example, a lowercase v might represent velocity. Throw a subscript or something on it when you want to talk about a particular velocity, like the velocity at which the puck flies away.

Good job! I'm happy somebody sat down and worked through this. Make sure you get a star or a point or whatever it is you get.

As willorn has demonstrated, the puck descends through a distance of one third the sphere's radius. This result is independent of gravity and the mass of the puck (on the moon, on the sun, with big puck, with little puck, the takeoff position doesn't change). Cool!

This problem has been solved. If you're up for more fun with this sphere, see part II. Otherwise, I have started another thread with a new challenge.

Edited by Anthony Frachioni

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