# Big sphere, small sphere

#### ssj5harsh

I'm not sure if this is a brain teaser.
There is a large sphere and a small spherical particle starts from rest from the top of the large sphere. At what point on the sphere will it fall off? The experiment is performed on earth under the influence of gravity. (Btw, I do not know the answer to this question myself).

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#### Doc Al

Mentor
ssj5harsh said:
I'm not sure if this is a brain teaser.
Sounds like homework to me. :grumpy:

#### ssj5harsh

Well, my teacher introduced it to us saying that it's a brain teaser. But he said it would need calculus(which we haven't studied yet), so I shouldn't bother myself . I was just wondering if there was a simpler method. (If this isn't in the spirit of this section of the forum, feel free to delete it.)

(Btw, the other questions he gave were:
1. There are 'n' ants on a rod of length 'l', randomly placed, moving randomly. What is on average the least distance each ant covers before it strikes into another ant.
2. In a new method of storing information, supercomputers assign letters from 00 to 99 to each symbol (according to a set code). Then all the text is converted into nos. and a decimal point is put before it. The fraction is then etched on a 6-inch rod and kept safely(free from dust, moisture, etc).This way the encyclopedia britannica(or even something bigger) is on a 6-inch rod. Other than being practically impossible, find the theoretical limitation of this method.
3. There are 1023 bottles. One is a poison bottle. You have ten persons(to sacrifice for the purpose). Find a method to identify the poison bottle.)

Just thought you should know Last edited:

#### Doc Al

Mentor
Calculus isn't needed to solve this problem. (I'll move it to the homework help section, in case you wish to try it.)

#### ssj5harsh

Right then, Here's what I've got. The forces on the small particle are weight downward and the reaction force radially outward. Putting in equations,( 'x' is the angle the reaction force makes withlthe vertical)
mg-Fcos(x)= mg
which gives:
Fcos(x)=0
Thus force is zero or angle is 90 degrees. Reaction force is zero when it loses contact. This doesn't tell us anything about the angle. And now we dont know what else to do. So we're stuck.

#### NateTG

Homework Helper
Here's a hint:

What would happen if the marble started out with an initial velocity of $\sqrt{gR}$ (Where $g$ is the acceleration due to gravity, and $R$ is the radius of the sphere.)?

#### ssj5harsh

I'm sorry, I don't quite get it. The intial velocity would not make any difference in the equations, would it?Btw, the problem states that it is left from the top of the sphere, so it probably starts from rest. All I can understand is that sqrt(rg) is the escape velocity from the earth. Is this somehow related to circular motion? I, for one, don't see the connection.

#### NateTG

Homework Helper
$\sqrt{Rg}$ is very unlikely to be the Earth's escape velocity.

Do you know anything about centripetal acceleration?

#### OlderDan

Homework Helper
ssj5harsh said:
I'm not sure if this is a brain teaser.
There is a large sphere and a small spherical particle starts from rest from the top of the large sphere. At what point on the sphere will it fall off? The experiment is performed on earth under the influence of gravity. (Btw, I do not know the answer to this question myself).
The fact that the shape of the particle was specified suggests that maybe you are to take that into consideration in the problem. Were you told anything about rolling vs sliding, and if the small sphere is solid or hollow? If not, then you can probably assume sliding.

#### ssj5harsh

Well, sorry, the escape velocity of the earth is sqrt(2Rg) if R is the radius of the earth.

As for the centripetal acceleration, it is the acceleration towards the centre in circular motion. It can be due to Gravitational, Electromagnetic or other such force. But, it shouldn't apply to this problem since the forces(both electromagnetic and gravitational) are insignificant between the particle and the sphere, wrt to the earth's gravity.We can even consider the earth to be flat for this question, right?

My doubt exists due to the fact that my common sense tells me that the particle should attain a velocity high enough to "jump" off the sphere before it reaches the equator of the sphere. I am not able to prove this using physics.

Older Dan, the teacher stated this problem loosely because it isn't part of our course. It was just a way to catch our attention and set us thinking right from the first lecture. I believe it worked .

#### OlderDan

Homework Helper
ssj5harsh said:
Well, sorry, the escape velocity of the earth is sqrt(2Rg) if R is the radius of the earth.

As for the centripetal acceleration, it is the acceleration towards the centre in circular motion. It can be due to Gravitational, Electromagnetic or other such force. But, it shouldn't apply to this problem since the forces(both electromagnetic and gravitational) are insignificant between the particle and the sphere, wrt to the earth's gravity.We can even consider the earth to be flat for this question, right?

My doubt exists due to the fact that my common sense tells me that the particle should attain a velocity high enough to "jump" off the sphere before it reaches the equator of the sphere. I am not able to prove this using physics.

Older Dan, the teacher stated this problem loosely because it isn't part of our course. It was just a way to catch our attention and set us thinking right from the first lecture. I believe it worked .
"Jump off" is not a very accurate description of the motion, but "going off on a tangent" (initially) would work pretty well. Do you know how to use energy conservation to find the speed change of a particle experiencing a constant gravitational force?

You are right about the equator, if you mean a horizontal equator. The particle will be off the sphere before it gets there.

#### ssj5harsh

Oh, thanks OlderDan. Using energy conservation didn't strike me at all!!:surprised
So, I'll begin with an equation of motion(since acceleration is constant).
u=0
v=ysinx (downwards)
(let 'x' be the angle of the reaction force when it leaves the sphere.'y' is the velocity)
a=g (downwards)

Taking downward positive,
v^2 = u^2 + 2as

y^2 = 2gs/sin^2(x)

By Law of conservation Of energy,
KE1 + PE1 = KE2 + PE2
0 + 2mgs= (my^2)/2 + mg(R-s)
This gives,
sin^2(x)=1 (if m,s != 0)
Thus x is 90 degrees or 270 degrees. Both are equally valid since the ball can fall either way. Hence the ball does indeed fall off at the equator if it slides in the absence of friction.
Thanks a lot OlderDan and NateTG.

#### Doc Al

Mentor
ssj5harsh said:
So, I'll begin with an equation of motion(since acceleration is constant).
u=0
v=ysinx (downwards)
(let 'x' be the angle of the reaction force when it leaves the sphere.'y' is the velocity)
a=g (downwards)
The acceleration is not constant. And "a = g" only if the ball is freely falling, which it is not (it's sliding along the surface of the sphere).

As you realize, the ball will break contact with the sphere at some point. Measure the position of the ball in terms of its angle from the vertical. Apply conservation of energy to find the ball's speed at any angle. Then consider what conditions cause the ball to leave the surface: To maintain contact, the centripetal force must be sufficient to keep the ball pressed against the surface. (What force provides the centripetal force?) But as the ball's speed increases, at some point that force will not be enough to provide the needed centripetal acceleration. Express that mathematically to solve for the angle that the ball loses contact with the sphere.

#### ssj5harsh

I can't believe I made such a silly mistake. Another mistake I made was to not consider the radius of the small sphere.
Right so, By conservation of energy, If it falls by h,
0 + mg(2R + r)= (mv^2)/2 + mg(2R+r-h)
gh = (v^2)/2
v^2 = 2gh

Also, (R+r)cosx = R+r-h
Hence h= (R+r)(1-cosx)

v^2 = 2g(R+r)(1-cosx)

Here, the centripetal force is the gravitational force of the earth. It is mgcosx.
Hence, for motion, mv^2/(R+r) = mgcosx (Is this right?)
Hence, v^2 = (R+r)gcosx

At height h,
2g(R+r)(1-cosx) = (R+r)gcosx
2-2cosx = cosx
Hence cosx = 2/3
x=48.19 degrees

I am confused whether the equation of centripetal force should apply here, please clarify. I think it should, since the velocity v is indeed in a tangent.
Otherwise, I think it is correct.

#### Doc Al

Mentor
Looks good to me.

For the ball to remain in contact with the sphere, it must have a centripetal acceleration (since it follows a circular path). Since the amount of centripetal force that gravity can provide is limited, at some point it will be insufficient to keep the ball on its circular path--that's when the ball is moving too fast to maintain contact.

#### kurushio95

wait, if it was rest at the exact top of the sphere, wouldn't that mean it was balancing?

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