## Slip, sliding away

Posted by Chris on January 12, 2011 – 2:33 pm

A smooth sphere of radius r is glued to a table top. A small mass is resting at the exact top dead centre. It is given a very small disturbance and starts to slide over the surface of the sphere. How far below the top of the sphere does the object get before separating from the sphere?

Ignore the effects of rotation (the moment of inertia of the mass is negligible).

Neglect friction, buoyancy, air-resistance and all other such complications. This is not intended to be a trick (i.e. joke) problem.

January 12th, 2011 at 3:04 pm

The side, relative to the top, Assuming that your sphere or object is not dense enough to cling to the other.

January 12th, 2011 at 3:57 pm

Hi Spencer, your’e right to ignore the mutual graviational attraction.

But your main answer is wrong. The object doesn’t get as far as the sphere’s “equator” before losing contact with the sphere.

January 12th, 2011 at 4:41 pm

this is just an intuitive guess from hazily remembered school physics and maths, but would it be where the parabola that would describe the path of the object if there were no sphere there intersects the circle that is the cross section of the sphere along with the object is falling?

January 12th, 2011 at 5:06 pm

Hi misterfahrenheit. I think that is correct, if so, we both have good intuition.

I haven’t thought about how to solve the problem that way. But even as I’m typing this response, I think I can see how to do it that way. But I suspect that method will require calculus. There is a more direct approach that only require quite simple mechanics knowledge.

I’m might even award bonus points to someone who can determine that parabola, or equivalently say where the ball lands.

January 12th, 2011 at 6:57 pm

(1/3 * radius) below the top.

January 12th, 2011 at 8:05 pm

The ball lands (5^(1/2) * radius ) away from the center.

January 12th, 2011 at 8:26 pm

Hi Madhura. I can confirm that your first answer is correct. I’m embarassed to say that I’ve no idea about the second answer, sorry But I can say that your answer looks plausible.

I’ve been busy on the girls, girls, girls problem. Because that page has got a little heated, I’m going to spend a bit more time on that page. Then I’ll check your second result.

If you think you can explain how you go your results, you could save me a bit of sweat. I have a write up for the first result. I appreciate that descibing geometry can be difficult on this site.

January 12th, 2011 at 9:36 pm

My answer to the second part (where the ball lands) seems to be incorrect. I think it’s

1.462 * radius

@Chris: Well, its really hard to write mathematical notations here. I’ll give it a try as soon as I get a little break.

January 12th, 2011 at 9:56 pm

Hi Madhura. Thank you. I’m a scumbag for asking you to do the work. The real reason is that I’m not really interested in the actual answer (usually) but how it was determined. I also am under the delusion that that’s what most people want as well.

I think I have finished my efforts on the other blog. But, guess what, I’ve got to duck out. It’s 5 am here in the UK, and I reckon that I’d be about half an hour doing part 2 on paper, and another half hour (or more) typing it up; so I’m going to bed instead

But don’t feel bad if you don’t get round to it. I like working these problems out and posting the solution(s).

January 13th, 2011 at 2:50 pm

Hi Madhur. I’m not surprised that you haven’t posted and that you revised your answer to part two. The parabola stuff is horribly tedious; I’ve spent at least an hour on it, and still haven’t finished. Admittedly, I’m trying a parabola only approach and not using the centripetal force idea (which is what I assumed you used initially).

January 13th, 2011 at 7:24 pm

Here is how I derived the answer for the seconds part using the centripetal force idea. (Hope Chris would post how it’s done using parabola only approach)

From the first part you get

1. The velocity of the ball when it separates from the sphere. i.e. v = ((2/3)gr)^0.5 making an angle of 48.19 degree with the horizontal plane.

r : radius of the sphere

g : gravitational acceleration

2. The exact position of the ball when it separates from the sphere.

i.e. r/3 below the top (vertically) and (5^0.5)r/3 = 0.745r from the center (horizontally).

Then to get the answer for the second part you apply the ’suvat’ equations in the vertical direction to find the time from the separations till the ball lands.

S = ut + 1/2at^2

S : the distance the ball has to travel till it lands = 5r/3

u : vertical component of the velocity = v Sin(48.19)

a : g

From that you get t = 1.3159*(r/g)^0.5

Then we can apply S = ut in the horizontal direction

S : the distance traveled by ball in horizontal direction

u : horizontal component of the velocity = v Cos(48.19)

t : the time found from the above part

So from this you get S = 0.716r

To get the distance from the center you have to add the horizontal distance from the center to the point where the ball left the sphere.

i.e. The final answer is 0.716r + 0.745r = 1.461r

January 13th, 2011 at 8:15 pm

Hi Madhura. Thank you for posting that. Did you like the problem? You’ll notice that the physics problems aren’t popular. THe only ones that seem to get attention are the buoyancy (rock in a boat on a lake) and a few air resistance ones (does it take longer going up or coming down when you throw something up). They’re usually good for a laugh

I will post the parabola only method (if it works, but I can’t see why it shouldn’t).

But I’m afraid that I got myself back onto the Girls, girls, girls blog. I’m loving that problem to bits. Simple things for simple minds.

January 14th, 2011 at 7:59 am

Howdy. This one is interesting to me. I would really like to get more of an explanation from the experts here why that is how we solve this one.

I’m not mathematically challenged but dusty as far as being conversant. So don’t be afraid to use terms that I might need to go look up.

Logically, I see that the object doesn’t separate at the sphere’s equator. The horizontal motion gained from it’s path from the top will carry it away from the sphere earlier.

But how do we determine the point at which the object’s ‘falling parabola’ is away from the path on the sphere?

Thank you.

January 14th, 2011 at 3:39 pm

Hi cazayoux. When the mass starts to move, it is doing so on a curved path. So there must be a centripetal force mv²/r keeping it in that path. The component of the gravitational field perpendicular to the surface of the sphere is able to provide that. But as the mass moves ever faster, the required centripetal force becomes greater. Eventually the isn’t enough g left to do the job. That’s the point where the mass departs from the sphere.

As a minor note, as the required centripetal force is increasng, the component of g normal to the sphere is decreasing (it gets to 0 on the equator).

I’m trying (but have spent 99% of my available time on the GGG blog) to do it using misterfahrenheit suggestion.

Using only KE = mg(2r-h) allows defining a family of parabolas. But we need to solve for the unique one that is a tangent to the sphere. The maths is surpringly tedious. Just defining the trajectory is bad enough, but we also need the angle at each point on the tajectory, and then solve, wht I suspect is going to be, an unpleasant simultaneous equation involving the sphere and the parabola to find the unique tangent.

There is a trivial solution for the trajectory, i.e. when then trajectory is purely vertical and touches the sphere at the equator. Clearly that one is in not appropriate.

January 16th, 2011 at 11:46 am

Centripetal force based solution: When the mass, m, has dropped through a vertical distance d, it will have gained KE = mgd = ½ m v² => v² = 2gd

Let A be the angle that the beetle is at (measured from the centre of the sphere. Angle = 0 at top. The component of g normal to the sphere’s surface is g cosA. You’ll need to make a sketch to follow all this:

But cosA = (r-d)/r, adjacent side = r-d, hypoteneuse = r

At the point of departure, the centripetal force required to hold the beetle on is exactly the same as the normal component of g holding the beetle on the sphere, and so mv²/r = 2gdm/r = mg cos(A) = mg (r-d)/r

=> 2gdm/r = mg(r-d)/r => 2d = (r-d) => d = r/3 is the answer.

Note if let h be the vertical height at the point of departure, measured from the ground, that d = 2r-h, so using d = r/3 => h = 5r/3. Also note that the angle of the tangent to the sphere at the point of departure is A. I mention that because of the upcoming pure parabola solution. But that looks really messy, but I have the equations for the bulk of it.

January 17th, 2011 at 6:58 am

After a week away, here’s my question on this one…

Size of sphere? If it is the size of a marble – centripetal force (kinetic energy) will have a limited effect, versus a sphere the size of my ego (huge!!!) where speed can come into play and centripetal force will have a greater effect?

January 17th, 2011 at 11:54 am

Hi Karl, it’s 1/3 of the radius, no matter what that is. Although a g-field is required, the strength is unimportant.

January 18th, 2011 at 7:55 am

I’ve finally realised why I’ve been failing to crack this. There is a whole family of parabolas that are tangent to the sphere. Those parabolic tangents stay outside the sphere. The one I want is at the end of the family and is the first one to just fail to enter the sphere. One “end” of that family is the pure vertical parabola that touches the sphere at it’s equator. There is a continuous family from there, up to the “critical” one that I’m seeking. It will be the case that the radius of curvature of the parabola will be r at the point of contact.

Armed with that realisation, I’ll see if I can finally do it. I kept on having one unknown too many. It’s still fiddly though, the equations are all horribly tedious to write out.