# Interesting mechanics problem

## Homework Statement

This week’s physics puzzle is a classic often used in mechanics. A point-like marble of mass m is situated on top of an immobile hemisphere of radius R. It is given a tiny push and starts rolling down. At what angle θ will it detach from the hemisphere’s surface and fly off? Assume there is no friction between the marble and the hemisphere, and treat the marble as a point object.

## Homework Equations

v^2/r=centripetal acceleration=omega^2*r
omega*r=circumferential speed

## The Attempt at a Solution

This is just an interesting problem i saw at a blog. A solution had been given but the link is broken now. I haven't been able to find the answer and i would really like your thoughts on it.Here's what i've come up with, i should however tell you that i am not very skilled at calculus, never having had any formal education in it.

I think the angle the ball makes with the hemisphere is 90°-theta
So the gravitational acceleration can be taken apart as a
tangential acceleration:g*cos(90°-theta) =g*sin(theta)
and a
normal acceleration:g*sin(90°-theta) =g*cos(theta)
(which i think serves as a centripetal acceleration here)

if integrate the tangential acceleration to get the speed i get: -g*cos(theta)
which surprises me because it's negative (have i already made a mistake?)

Now i think that a at some angle theta the speed will be so large that the centripetal acceleration can't hold it anymore and it flies off. So all that is left for me to do is to find a way to compare the centripetal acceleration i would i expect with a given speed to the centripetal acceleration i actually have. The moment that the expected centripetal acceleration is larger than the given, i now that is the angle theta the ball will fly off.
I'm guessing i'll have to use the expressions omega*r and omega^2*r to compare them but i don't immediately see how.
Thanks ( and if i'm on the right track just a hint will do :) )

#### Attachments

• hemisphere.gif
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## Answers and Replies

Hmmm, I think you've got a couple of mistakes in there...

First, when you integrate an equation for acceleration so you can get velocity, you need to integrate with respect to time, not position (or angular position, as you've done here). A good check for this is to look at the units. If you start with acceleration (e.g. ft/sec^2) and want velocity (ft/sec) you need to multiply by something with time units.

Second, you should define a positive direction to be used consistently across your analysis. You said you were surprised to see a negative velocity, but you don't define a positive direction. If you choose up to be positive, then the velocity of the ball rolling down the hemisphere will be negative...

That being said - I don't think you need to do any integration to solve this problem, but you're not too far off. What condition is true the instant that the marble is no longer on the surface of the hemisphere? Think in terms of forces...

Is this enough of a hint?

-Kerry

Doc Al
Mentor

## The Attempt at a Solution

This is just an interesting problem i saw at a blog. A solution had been given but the link is broken now. I haven't been able to find the answer and i would really like your thoughts on it.Here's what i've come up with, i should however tell you that i am not very skilled at calculus, never having had any formal education in it.
You won't need any calculus.

I think the angle the ball makes with the hemisphere is 90°-theta
So the gravitational acceleration can be taken apart as a
tangential acceleration:g*cos(90°-theta) =g*sin(theta)
and a
normal acceleration:g*sin(90°-theta) =g*cos(theta)
(which i think serves as a centripetal acceleration here)
OK. (This is the centripetal acceleration that gravity alone could provide.)

if integrate the tangential acceleration to get the speed i get: -g*cos(theta)
which surprises me because it's negative (have i already made a mistake?)
v = ∫a dt, not dθ, but these are not independent variables. There's a better way to find the speed: Consider conservation of energy.

Now i think that a at some angle theta the speed will be so large that the centripetal acceleration can't hold it anymore and it flies off. So all that is left for me to do is to find a way to compare the centripetal acceleration i would i expect with a given speed to the centripetal acceleration i actually have. The moment that the expected centripetal acceleration is larger than the given, i now that is the angle theta the ball will fly off.
I'm guessing i'll have to use the expressions omega*r and omega^2*r to compare them but i don't immediately see how.
Hint: ac = v²/r, so find v² for a given theta. (See previous hint above.)

hmm thanks for the hint.
I think that the moment the ball would fly off the surface the tangential component of the force should be greater than the normal component, is that right?
If that's the case the angle would be 45° which does seem rather logical.
I'm not sure if that's what you meant though :\

i think i can find the height for a given theta and assuming R is equal to 1: is it 1-cos(theta)? then the v would be m*g - m*g*cos(theta)
and thus i find an expected centripetal acceleration which is (v^2)/1 and then i just have to consider the equality of the expected acceleration and the actual acceleration?
it does seem to be a really messy equation...

Last edited:
Doc Al
Mentor
I think that the moment the ball would fly off the surface the tangential component of the force should be greater than the normal component, is that right?
Why would you think that?

What are all the forces acting on the ball at any time?

only the tangential and normal component of gravity

Doc Al
Mentor
only the tangential and normal component of gravity
That would be true if the hemisphere wasn't there. (But then it would just be a falling body!) What other force acts?

does the speed of the point-object also create a centrifugal force?

Doc Al
Mentor
does the speed of the point-object also create a centrifugal force?
"Centrifugal force" is only used when describing things from an accelerated reference frame; forget about it.

What I'm looking for is much simpler than that. If the hemisphere wasn't there, the ball would just fall straight down. So what must be exerting a force on the ball?

the hemisphere would, is it just an action-reaction kind of force?

Doc Al
Mentor
the hemisphere would,
Yes, the hemisphere exerts a normal force on the ball.
is it just an action-reaction kind of force?
You could call it a "reaction" force.

So there are two forces acting on the ball: gravity and the normal force. When the ball loses contact with the surface, what happens to the normal force?

it's gone, only gravity is left pointing downwards

Doc Al
Mentor
it's gone, only gravity is left pointing downwards
Right. If the ball moves too fast, gravity cannot provide enough centripetal force to keep the ball on the hemisphere. Find the angle at which that happens.

i understand that intuitively but i don't now how to translate the part: moving too fast.
in a formula. that's thing i've been stuck on the most i think. i know the centripetal force is m*g*cos(theta) , meaning it gets lower and lower but i don't know how i relate the speed to that expression

Doc Al
Mentor
i understand that intuitively but i don't now how to translate the part: moving too fast.
in a formula. that's thing i've been stuck on the most i think. i know the centripetal force is m*g*cos(theta) , meaning it gets lower and lower but i don't know how i relate the speed to that expression
Think this way: m*g*cos(theta) is the radial component of gravity, which provides the centripetal force. How does that relate to speed? (Use F = ma.) How does speed relate to theta? (Use energy conservation.)

i think i finally got it, i had used a wrong formula to find the speed yesterday, saying it was m*g*h. obviously that's only the potential energy and the speed is the root of (2*g*h)
so v^2 (which is essentially the centripetal acceleration, assuming R=1) is 2*g*h
when the two forces are in equilibrum, the ball is about to fly off.
So 2*g*h*m=m*g*cos(theta)
simplifying:2h=cos(theta)
(h=1-cos(theta) )
2-2*cos(theta)=cos(theta)
cos(theta)=2/3
theta=48°11'23"
is that right? it's not unfeasible,but i expected a 'nicer' angle...

I'm quite sure i'm right now though. The height at which the ball loses contact with the hemisphere is 1/3

Last edited:
Doc Al
Mentor
Good!

Just for the record, here's how I think of it:

The speed of the ball at any point satisfies v^2 = 2gΔh = 2gr(1 - cosθ). Thus to maintain contact with the hemisphere, the ball must be centripetally accelerated: a_c = v^2/r = 2g(1 - cosθ). The only centripetal force available is the radial component of gravity, mgcosθ. We can find the spot where gravity is just enough to provide the required acceleration via Newton's 2nd law:
F = ma_c
mgcosθ = m2g(1 - cosθ)
cosθ = 2(1 - cosθ)
cosθ = 2/3

If the original height of the ball is y = r, it falls off at height y = 2r/3.

thanks again!
this thread can be locked or forgotten now :)

fluidistic
Gold Member
thanks again!
this thread can be locked or forgotten now :)

Not at all! This is a common problem so this thread can help many students.