# Sliding body

1. Apr 25, 2010

### benf.stokes

1. The problem statement, all variables and given/known data

A point particle of mass m placed on top of a semi-sphere of radius R and mass M, which is based on a frictionless surface. At one instant, the particles begin to slide over the semi-sphere.
Obtain an equation for the angle at the point where the particle loses contact with the semi-sphere.
How do M and m get into the picture?

Thanks

2. Relevant equations

$$m\cdot g\cdot \cos(\theta) - N = m\cdot \frac{v^2}{R}$$
$$m\cdot g\cdot R+0= \frac{1}{2}\cdot m\cdot v^2+m\cdot g\cdot R\cdot \cos(\varphi)$$

3. The attempt at a solution

$$\mbox{Using conservation of mechanical energy comes:}$$

$$m\cdot g\cdot R+0= \frac{1}{2}\cdot m\cdot v^2+m\cdot g\cdot R\cdot \cos(\varphi)$$

$$\mbox{Solving for v^2 gives:}$$

$$v^2 = 2\cdot g\cdot R\cdot (1-\cos(\varphi))$$

$$\mbox{Using the centripetal force condition yields:}$$

$$m\cdot g\cdot \cos(\theta) - N = m\cdot \frac{v^2}{R}$$

$$\mbox{When contact is lost N = 0:}$$

$$v^2 = g\cdot R\cdot \cos(\theta) \hspace{1 pt} \mbox{being \theta \hspace{0.5 pc} \mbox{the angle at which contact is lost and so:}}$$

$$2\cdot g\cdot R\cdot (1-\cos(\theta)) = g\cdot R\cdot \cos(\theta) \hspace{0.5 pc} \mbox{which upon solving gives:}$$

$$\theta=\cos^{-1}(\frac{2}{3})$$

$$\mbox{But I don't know where do M and m figure in this solution.}$$

Last edited: Apr 25, 2010
2. Apr 25, 2010

### thebigstar25

the question is, do they have to be there ? .. have a look at your solution and tell us what do you conclude from it?

3. Apr 25, 2010

.Sometimes questions contain red herrings

4. Apr 25, 2010

### benf.stokes

In my solution M and m don't appear, so they would be irrelevant to the question. But why would my instructor introduce M and m then int he question?

5. Apr 25, 2010

I suppose to see if you're on the ball.M never appears in the solution and m cancels out.As I said they are red herrings.

6. Apr 25, 2010

### Staff: Mentor

Hint: The hemisphere is on a frictionless surface. (You solved a different problem.)

7. Apr 25, 2010

The way I read the question is that the surface on which M slides is frictionless.I got the same answer as benf.

8. Apr 25, 2010

### Staff: Mentor

Right! (I thought I just said that.)
Did you take into account the motion of the hemisphere?

9. Apr 25, 2010

Sorry I meant the surface on which m slides.Looking at the question again it seems that benf and myself both misinterpreted it(but it does have a trace of ambiguity,that's my excuse anyway).Because of the way I read the question I did not consider any movement of M.

10. Apr 25, 2010

### benf.stokes

How do I take in account the motion of the hemisphere??

11. Apr 26, 2010

### Staff: Mentor

How are the motions of the hemisphere and the sliding mass related? What does Newton's 3rd law tell you? How can you revise the conservation of energy expression?

Analyze the forces from the accelerating frame of the hemisphere.

12. Apr 27, 2010

### benf.stokes

I did so but m and M continue to not appear

13. Apr 27, 2010

### Staff: Mentor

Show what you did. (What you wrote in post #1 assumes the hemisphere to be stationary.)

14. Apr 29, 2010

### benf.stokes

$$\mbox{Let the aceleration of the hemisphere be equal to A and the radius of the hemisphere R. Then the transversal force is equal to:}$$

$$m\cdot g\cdot \sin\theta+mA \left( \cos \right) \,\theta=ma$$

$${\it Ft}=m{\frac {d}{dt}}v \left( t \right) ={\frac {mv \left( t \right) {\it dv}}{{\it ds}}}={\frac {mv \left( t \right) {\it dv}}{Rd \theta }}$$

Or

$$\left( g \left( \sin \right) \,\theta+A \left( \cos \right) \,\theta \right) Rd\theta =v{\it dv}$$

$$\mbox{Integrating both sides yields:}$$

$$\int_0^\theta \! g\cdot R\cdot \sin(\theta) \, d\theta + \int_0^\theta \! A\cdot R\cdot \cos(\theta) \, d\theta = \int_0^v \! v \, dv$$

Or

$${v}^{2}=2\,gR \left( 1- \left( \cos \right) \,\theta \right) +AR \left( \sin \right) \,\theta$$

$$\mbox{Using the centripetal aceleration condition:}$$

$$mg \left( \cos \right) \,\theta+mA \left( \sin \right) \,\theta-N={ \frac {m{v}^{2}}{R}}$$

$$\mbox{When the body loses contact with the surface N = 0 and so the previous equation becomes:}$$

$${v}^{2}=R \left( g \left( \cos \right) \,\theta+A \left( \sin \right) \,\theta \right)$$

$$\mbox{Solving both equations in order to v^2 gives:}$$

$$R \left( g\cos \left( \theta \right) +A\sin \left( \theta \right) \right) =2\,gR \left( 1-\cos \left( \theta \right) \right) +AR\sin \left( \theta \right)$$

$$\mbox{Having reduced this to a problem of math I still don't have m or M appearing here}$$

Thanks for the help

Last edited: Apr 29, 2010
15. May 2, 2010

### Staff: Mentor

OK. But realize that A and the horizontal acceleration of the sliding body are not independent.

A is not a constant. It varies as the body slides down.

I would approach the problem by first figuring out the sliding body's velocity (in the accelerating frame of the hemisphere) as a function of theta using conservation of energy. Don't neglect the relationship between the speeds of hemisphere and sliding body. (That's where m and M come in.)

It seems like a nasty bit of algebra.

16. May 3, 2010

I have come to a full stop on this question.Intuition sort of tells me that the hemisphere should accelerate in the horizontal plane but when I sketch force diagrams which ignore friction I can see no resultant horizontal force that will cause this acceleration.The weight of the small mass acts vertically on the hemisphere and has no horizontal component.What,if anything,am I overlooking?

17. May 3, 2010

### Staff: Mentor

The hemisphere and mass exert a normal force on each other. That normal force has a horizontal component which accelerates the hemisphere. (Just to nitpick, the weight of the small mass acts vertically on the small mass, not the hemisphere.)

18. May 4, 2010