# The Circle (sorry for my slacking)

1. Jan 17, 2012

### Plutonium88

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

A ball on top of a circle is acted on by a FG. The ball rolls to a certain point on the circle, and detaches. Ffr Is negligable.

Given info:
Diameter Of the circle, Mass of ball no ffr
x = the distance from the balls posistion to the top of the big circle.
x is what is needed to be found (Teal in diagram)

Okay... So i think i might have it....

http://s13.postimage.org/fpzmhstnb/New_Bitmap_Image_3.png [Broken]

So what i did, was assumed was that the distance from the top to the center (1/2D) and the distance from the object to the center (1/2) were the adjacent and opposite sides of an isoceles right angle triangle.... ( i don't know if this is okay..)

Now I sovled the angle with TanA = 1/2D/1/2D = 1(TanInverse) = 45 degrees

Now with this angle...

I assumed that angle it traveled would be the angle it left at... (45 Degrees) and any corresponding angle woudl be 45 degrees anyway..

So i am left with the forces of FN which = 0 at the point it detaches

and i am left with the Force of gravity...

I have come to conclude that gravity is what allows centripetal accellearation...

So because the angle of FN = 45 degrees, when i make my Frame of reference so that the X direction is paralell to FN, i end up with force of gravity with the angle and 2 components.

So..

Fnety = mgsin45
may = mgsin45
ay = gsin45

Fnetx = mgcos45 - FN
Fnetx = mgcos45 - 0
max = mgcos45
ax=gcos45

okay.. so now since i know that gravitys X component is what provides the centripetal force...
R = D/2

Fc = Fnetx

Mv^2/R = mgcos45
V =√[Dgcos45/2]

Now.. Comparing total mechanical energy..

ET1 = EG1 = mgD
Et2 = EK2 + EG2 = 1/2mv^2 + mg(d-x)

Et1 = Et2

mgD = 1/2mv^2 + mg(d-x)

gx = 1/2v^2
x = dgcos45/2/2g
x = D (gcos45/2g)
x = D (cos45/2)
x = D([√2]/8)

and this is my answer... x = D([√2]/8)

so yea.. i'm hoping this strategy is right.. everything seems to add up mathematically.. i'm still not sure about this angle though.. i'm not sure if that is correct way to solv eit.

Last edited by a moderator: May 5, 2017
2. Jan 17, 2012

### Spinnor

Last edited by a moderator: May 5, 2017
3. Jan 17, 2012

### Plutonium88

Thanks a lot man

4. Jan 17, 2012

### Plutonium88

So i used his strategy... and incoperated some of mine..

Et1 = mgD - ball at top of circle
Et2 = 1/2Mv^2 + mg(d-x)

Et3= 1/2mv^2 + 1/2Iomega^2

I is a moment of inertia I = 2/5mr^2 (formula for circle as taken from that article)
omega is the circular velocity.

(Unfortunately i haven't researched this enough to understand it, i haven't even learned moments before... :( my teacher is going to slay me when i show her i have an answer using formulas i know nothing about haha)

now.. first

ET1=Et3
mgd = 1/2v^2 + 1/2(2mr^2/5)(v^2/r^2)

simplified to..

V = sqrtall20gD/14

now..

et1 = et2
mgD = 1/2mv^2 + mg(d-x)

solving for x

x = 5/7D

Now either i simplified wrong.. but i don't think i did.. and this can't be possible because a ball would attach much before 5/7 of the diameter.. this seems more like the length of the height from that posistion to the ground..

so i assumed my x = 2/7 ( the remaining portion..)

so this is a pretty shaky answer.. any more feedbaxx

5. Jan 18, 2012

### Spinnor

Can you do the more simple problem where there is just sliding and no rotation? I guess you can. With friction the ball will be going slower for a given vertical drop because some of the energy goes into rotation. With both problems of the liftoff occurs at the same center of mass velocity? The rolling ball has to "fall" further to get the required speed.

6. Jan 20, 2012

### Plutonium88

so you're saying it doesn't have the speed to do that until it is 5/7 Of the diameter and thats the logic behind why it has gone so far down by calculation with NO FFR.

but with friction it has enough speed and detaches earlier....

I thought that a ball without ffr would travel faster because of inertia and it would be constantly accellerating;

The problem with friction the ball would take longer to accelerate essentially wouldn't it due to an opposing vector of friction?

7. Jan 20, 2012

### Spinnor

You wrote,

"A ball on top of a circle is acted on by a FG. The ball rolls to a certain point on the circle, and detaches. Ffr Is negligable."

Ffr is force of friction? If there is no friction the ball will not roll but will slide. Can I assume the ball rolls without slipping?

8. Jan 22, 2012

### Plutonium88

I'm can't give you a solid answer for that... my teacher said to me the ball rolls to a certain point and detaches... and that's all she told me, the rest of this information for the question... i have found along the way or have been helped with.. And hence why i have so much difficulty myself, i can't solve the problem lol :(6

9. Jan 22, 2012

### Plutonium88

Last edited by a moderator: May 5, 2017
10. Jan 22, 2012

### Spinnor

See problem 4 of,

http://physics.oregonstate.edu/~jansenh/COURSES/ph435/Solution3.pdf [Broken]

When you understand this problem my previous comments might make more sense.

Good luck!

Last edited by a moderator: May 5, 2017
11. Jan 22, 2012

### Plutonium88

http://s8.postimage.org/tj9ljzf5x/Force_diagram.png [Broken]

okay my final diagram.. i picked this part of the circle apart and solved for the angle..

fnetx = mgcosθ

now using that diagram and the triangle (T1)

solving first for the angle

cosθ = 1/2D - x/1/2D
θ= d-2x/cosD

now fnetx = mgcosθ
(basd on force diagram)

fnetx = fc
mgcosθ = mv^2/r

solving for v

v^2= gdcosθ/2

SUB IN θ
(The cosθ cancels in the equation when it is subbed in solving for velocity)
resolve in algebra too..

v= √g(d-2x)

Et1= mgD
Et2 = 1/2mv^2 +mg(d-x)

ete 1 = et2

mgd = 1/2mv^2 + MG(D-X)

sub in v

resolve for x

x= d/4

can some one tell me if this is a correct way to solve it?

Last edited by a moderator: May 5, 2017
12. Jan 23, 2012