Solving Block & Sphere Angle Problem w/ Gravity

[mrdoe]

Homework Statement

A 5kg block starts at the top of a fixed frictionless sphere of radius 10m. The ball is given a slight nudge, enough to get it going by the force of gravity (this happens on Earth) but of negligible magnitude. What is the angle, relative to the vertical line from the bottom to the top of the sphere, at which the block leaves the surface of the sphere?

Homework Equations

(I don't know)

The Attempt at a Solution

I don't even know where to begin with this. Perhaps consider normal force at angle theta, and integrate or something like that?

 

[rock.freak667]

Perhaps you should consider the block at a general angle θ and then see what should happen for the block to stay on the sphere.

 

[Doc Al]

Yes, consider all the forces acting on the block and apply Newton's 2nd law. (No need for integration.) Make use of the fact that it's a sphere. Come up with some criterion for when the block loses contact.

 

[mrdoe]

The problem is, I don't know what the criterion is (is my brain working today?)

It has something to do with the block's velocity and theta, but I don't know what.

 

[mrdoe]

bump
I really don't know how to do this =(

 

[Fightfish]

Hint: Consider the motion of the block; how would you describe it?
By 'ball', do you mean the sphere or ... the block?

 

[mrdoe]

block (sry)

 

[mrdoe]

Well, the motion of the block is at first on the surface of the sphere, then off at a tangent, i guess. I don't know what the significance of this is though...

 

[Fightfish]

Well, the motion of the block is at first on the surface of the sphere, then off at a tangent, i guess. I don't know what the significance of this is though...

Ok, we are not interested in the part after it slides off the sphere, so the part we are interested in considering is the motion along the surface of the sphere. The path along surface of the sphere is curved...circular in shape when considered two-dimensionally. Does that ring a bell?

 

[mrdoe]

Yes it's circular in shape. So you mean I should consider the component of the forces acting on the block perpendicular to the tangent?

 

[mrdoe]

The problem is that the pseudo-"centripetal" force is entirely canceled out by the normal force. Am I right? So it has something to do with a circle/centripetal pseudoforce but I'm not getting it.

 

[mrdoe]

Is it just that at some angle theta, the velocity is no longer in balance with the pseudocentripetal force and the ball therefore leaves the surface?

 

[Fightfish]

Is it just that at some angle theta, the velocity is no longer in balance with the pseudocentripetal force and the ball therefore leaves the surface?

"Pseudocentripetal force"? It's a real net force, not a pseudo-force. And no, the normal force does not cancel out with the normal component of gravitational force, otherwise there is no net radial force and no centripetal force. The idea is close; what happens when the block is just about to leave the surface of the sphere?

 

[mrdoe]

"Pseudocentripetal force"? It's a real net force, not a pseudo-force. The idea is close; what happens when the block is just about to leave the surface of the sphere?

The centripetal net force is not large enough to keep the block in "orbit"?

 

[Fightfish]

The centripetal net force is not large enough to keep the block in "orbit"?

Yup! (radial force would be more accurate though =)) And the net radial force on the block at the point of sliding off the sphere is given by...? Centripetal force required to keep block in circular motion given by...?

 

[mrdoe]

The centripetal acceleration has to be v^2/r and thus force must be v^2/20...

 

[mrdoe]

Wait: there is no centripetal force.

For angle of theta with respect to vertical, there's only a force of $$mg\sin\theta$$, which is parallel to the inclined plane (or the sphere's tangent).
Therefore, the block should fly off the sphere at 0 degrees. What am I getting wrong here..

 

[mrdoe]

Bump, I really need help on this one.

 

[Doc Al]

Wait: there is no centripetal force.

Sure there is.

For angle of theta with respect to vertical, there's only a force of $$mg\sin\theta$$, which is parallel to the inclined plane (or the sphere's tangent).

What happened to the radial component of gravity?

Answer these questions. In general, what forces act on the block? (Hint: I'm thinking of two forces acting.) At the moment the block is about to lose contact, what can you say about one of those forces?

 

[mrdoe]

There are two forces acting on a block on an inclined plane (the plane tangent to the sphere at the tangent point of the block to the sphere): gravity and normal. The net force is down the inclined plane, the magnitude of this is mg sin(theta) where theta is the inclination of the plane.

Somewhere I'm getting this wrong.

 

[mrdoe]

Are you talking about the gravity between the sphere and the block? I honestly don't think that's a force to consider in this problem.

 

[mrdoe]

The real problem is that the normal force counteracts any centripetal force.

 

[Doc Al]

There are two forces acting on a block on an inclined plane (the plane tangent to the sphere at the tangent point of the block to the sphere): gravity and normal.

Right! And when the block is just about to lose contact with the surface, what happens to the normal force?

 

[Doc Al]

The real problem is that the normal force counteracts any centripetal force.

They aren't equal! They don't cancel. The centripetal force is the net radial force acting on the block--that's definitely not zero, else the block can't move in a circle along the sphere.

 

[mrdoe]

The normal force goes away when the block loses contact. By that time, the only acceleration is due to gravity. Am I correct? However, assume T is the time at which the block loses contact. Then at T-e where e is infinitely small, the block still has the normal force. So it wouldn't have lost contact in the first place? ...
The attached image will show why I think the normal force counters the centripetal. Please correct me.

http://img34.imageshack.us/img34/4008/questionv.png

 

[Doc Al]

The normal force goes away when the block loses contact. By that time, the only acceleration is due to gravity. Am I correct?

Right.

However, assume T is the time at which the block loses contact. Then at T-e where e is infinitely small, the block still has the normal force. So it wouldn't have lost contact in the first place? ...

No. At T-e the normal force will just be very small. Just find the point where it's zero.

The attached image will show why I think the normal force counters the centripetal. Please correct me.

The diagram doesn't speak for itself. I suspect you are assuming that the normal force equals mgcosθ. You're probably confusing this with the case of an inclined plane, where there is zero acceleration perpendicular to the surface--but that's not the case in this problem. The fact that the surface is curved makes a big difference.

 

[mrdoe]

OK I'll do it the way I think it should be done. Correct me if there are mistakes.

We have $$mg\cos\theta$$ as the centripetal force. At some velocity $$v$$, $$\frac{v^2}{20} > 5g\cos\theta$$. Then $$\frac{v^2}{100} > g\cos\theta$$ and using KE/PE, we know that setting the base of the ball to be PE = 0, we have mgh = 49.05(20) = 981J and at height $$10+10\cos\theta$$, we have $$490.5(1+\cos\theta)=\frac{mv^2}{2}$$ and $$\frac{v^2}{20}=\frac{490.5(1+\cos\theta)}{50}$$. Then $$\frac{490.5(1+\cos\theta)}{50}>49.5\cos\theta$$ and $$1+\cos\theta>5\cos\theta$$ so $$\cos\theta<0.25$$ and therefore $$\theta=75.52.$$

Am I right?

 

[Doc Al]

We have $$mg\cos\theta$$ as the centripetal force.

Right.

At some velocity $$v$$, $$\frac{v^2}{20} > 5g\cos\theta$$.

No. Do this step over.

Rather than plug in numbers right away, I suggest you solve it algebraically. Only plug in numbers at the last step to get a numerical answer.

 

[mrdoe]

Well $$\frac{v^2}{r} = \frac{v^2}{10} > 5g\cos\theta$$. Sorry

Then later on we have $$\frac{v^2}{10} = \frac{490.5(1+\cos\theta)}{25}$$ and $$1+\cos\theta>2.5\cos\theta$$ so $$1.5\cos\theta < 1$$ so $$\cos\theta < 2/3$$ and we have $$\theta = 48.19$$ degrees.

 

[Doc Al]

Well $$\frac{v^2}{r} = \frac{v^2}{10} > 5g\cos\theta$$. Sorry

Then later on we have $$\frac{v^2}{10} = \frac{490.5(1+\cos\theta)}{25}$$ and $$1+\cos\theta>2.5\cos\theta$$ so $$1.5\cos\theta < 1$$ so $$\cos\theta < 2/3$$ and we have $$\theta = 48.19$$ degrees.

Good!

But just for the exercise, I strongly recommend that you solve it algebraically. Then you'll see that the answer doesn't depend on the radius or the mass. (And it actually requires less work.)