# A ball on a string

## Homework Statement

(Im translating this from another language, so if it doesn't make any sense, please let me know and ill try to do a better job.)
A metal ball (mass=120g) is hanging from a string (lenght=85cm). The ball is struck horizontally. When the angle of the swing reaches 125 degrees the ball starts to deviate from its circular path. What was the ball's initial velocity?
m=0.120kg, r=0.85m, θ=125°

## Homework Equations

I don't know what to put here. :l

## The Attempt at a Solution

Here's what i came up with: Conservation of energy. Eki=Epf
I figured the max height, the ball reaches should be h=r+r*sin(35°) ⇒0.85m + 0.85m*sin(35°)=1.3375m
1/2 mv2=mgh
v=√(2gh)
v=√(2*9.81*1.3375)
v=5.1227 m/s ≈ 5.1m/s

There, the answer i came up with. The answer section at the end of the book says the correct answer is 5.6m/s.

What am i missing?

P.S. I was told, that i should mention that im studying physics without a teacher.

Nathanael
Homework Helper
It says that the ball begins to deviate from it's circular path at an angle of 125°

But this does not mean the ball's velocity will be zero.

Edited because I said something wrong (I said it would not be at it's highest point). Stephen's post made me realize

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Stephen Tashi
What am i missing?

If we call x the horizontal axis and y the vertical axis, when the ball deviates from a circular path it is not motionless. It's y velocity is zero but it has a non-zero x velocity. The total energy of the ball at that point is more than mgh since it has some kinetic energy.

NascentOxygen
Staff Emeritus
When the angle of the swing reaches 125 degrees the ball starts to deviate from its circular path.

Have you sketched the ball's path, including and beyond that 125°?

It says that the ball begins to deviate from it's circular path at an angle of 125°

But this does not mean the ball's velocity will be zero.

If we call x the horizontal axis and y the vertical axis, when the ball deviates from a circular path it is not motionless. It's y velocity is zero but it has a non-zero x velocity. The total energy of the ball at that point is more than mgh since it has some kinetic energy.
Damn, i was afraid that might be it. Dont really know what to do with that information
Have you sketched the ball's path, including and beyond that 125°?
Up to that 125°, but not beyond.

So, i guess i'll try to figure out the ball's x-velocity at the top and add it to the E equation.

Im thinking of a force diagram at the 125° position (gravity ( acting down ) and centripetal ( acting outward ) ) to produce a resultant force that is tangential to the path radius. (this seems to be the critical factor)
You can figure the centripetal force required to end up with a tangential resultant, then the velocity required to produce this force.
From that you have the final KE, add to the PE at that point, gives the total KE at the start.
From that you can get the velocity.

haruspex
Homework Helper
Gold Member
centripetal ( acting outward ) ) to produce a resultant force
You're confusing centripetal with centrifugal.
Centrifugal force "acts outwards", but it is classed as a fictitious force because it only exists in the frame of the (accelerating) object. If you take an inertial reference from there is no such force. Instead, there is centripetal force, but that is a resultant force, not an acting force.
To rephrase your remark in the inertial frame:
To keep the ball moving in a circle, the sum of the acting forces (gravity and tension in the string) must have a radial component equal to the centripetal force. Tension cannot supply a force outwards along the radius, so when gravity's inward radial component exceeds the centripetal force the circular path cannot be maintained.
Your form works if you replace 'centripetal' with 'centrifugal'.

rcgldr
Homework Helper
To keep the ball moving in a circle, the sum of the acting forces (gravity and tension in the string) must have a radial component equal to the centripetal force.
A nit pick, that should be greater than or equal, the answer is probably looking for the point when it's equal, since that's the lower bound for when the ball no longer travels in circular path.

lost an update to this post - that should have been: a follow up ... greater than or equal to the minimal force to keep the string in tension.

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NascentOxygen
Staff Emeritus
Up to that 125°, but not beyond.
And I thought my hint was such a good one. haruspex
Homework Helper
Gold Member
A nit pick, that should be greater than or equal, the answer is probably looking for the point when it's equal, since that's the lower bound for when the ball no longer travels in circular path.
I don't mind nit picks so long as there's a nit to pick. I stand by the exact wording of what I wrote.:)

rcgldr
Homework Helper
I don't mind nit picks so long as there's a nit to pick. I stand by the exact wording of what I wrote.
My previous post lost an update (I'll refresh the page and make sure my updates take place from now on if there's a laggy response from the forum). Also nit pick was the wrong term, "follow up" would have been a better term for what I intended to post. I was pretty much stating the obvious, that the ball stops following a circular path just after the moment the tension in the string goes to zero (and the angle is greater than 90° ). When this happens, the path of the ball transitions into a parabola, and it's still moving vertically (upwards) and horizontally (towards the pivot point of the string).

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And I thought my hint was such a good one. It was, i just hadn't had the time to act on it by the time i answered :p

So, in the beginning, there is tension T and gravity G. ∑F=T+G=man, which is the total force acting on the ball, resulting in an inward acceleration (an). Right?
At the point θ=125°, the string goes slack and T=0 and the only force acting on the ball is gravity. ∑F=G=ma=mg. At this point it still has some velocity v2, up and inward, which i can't figure out.
I tried this:
at θ=125°, G=man
mg=mv22/r
v2=√(gr)
v2=2.8876 m/s

Doing the energy calculation with this included (Eki=Epf+Ekf) i got 5.8805m/s. Then i realised G=man can't be right, G is acting straight down. I should calculate this with the x-component of T? but T=0? an isnt the total acceleration? This exercise is going to give me an aneurysm.

rcgldr
Homework Helper
Hints, the tension goes to zero when v^2 / r = g sin(35°). The change in height = .85 x (1 + sin(35°)) meters.

thats not a hint, thats the solution :s
Based on that, I did the calculations and got to 5.6m/s, but i dont know why. I need someone to explain this "v^2 / r = g sin(35°)". What is it and where does it come from?

multiply it by g, you get an/g=sin(35)
from what i can tell sin(35) should rather equal g/an

multiply it by m and you get man=G sin(35), which seems to me like an x-component of G, which it doesnt have?

rcgldr
Homework Helper
explain ... "v^2 / r = g sin(35°)".
Sorry for the hint / solution. You already determine that if the tension in the string is non-zero, then the ball travels in a circular path regardless of the force of gravity, and if the tension transitions to zero, then the only force acting on the ball is gravity, and the the ball's path transitions to a parabola (free fall). So the issue was determining the equation for tension versus velocity and angle. This is where I couldn't think of a hint other than to write out the equation. If I had a drawing tool, I could have created a diagram showing the forces involved.

For general form of the equation, tension in string = m (v^2/r - g sin(θ)), where θ is the angle from horizontal. For θ = -90° (straight down), the tension is m (v^2/r - g sin(-90°)) = m(v^2/r + g). The problem is asking you to solve for tension = 0 = (v^2/r - g sin(35°)) .

m g sin(35°) seems to me like an x-component of g, which it doesnt have?
g sin(θ) is the radial component of gravity, the component of gravity in the direction of the string. You have a right triangle where the gravity vector is the hypotenuse (oriented vertical), and the magnitude of the radial component is the hypotenuse x sin(angle opposite string) . g sin(θ) is similar in concept to a box sliding on a frictionless inclined plane.

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If its any help, by my process, the velocity at 125° = 2.187 m/s
(use as a check)

For general form of the equation, tension in string = m (v^2/r - g sin(θ)), where θ is the angle from horizontal. For θ = -90° (straight down), the tension is m (v^2/r - g sin(-90°)) = m(v^2/r + g). The problem is asking you to solve for tension = 0 = (v^2/r - g sin(35°)) .

g sin(θ) is the radial component of gravity, the component of gravity in the direction of the string. You have a right triangle where the gravity vector is the hypotenuse (oriented vertical), and the magnitude of the radial component is the hypotenuse x sin(angle opposite string) . g sin(θ) is similar in concept to a box sliding on a frictionless inclined plane.
Okay, i think this is starting to make sense now.
This is how i get there
Ftotal=T+G=man
direction defined: T=man- mg sin(θ)
set T=0=m(v2/r - g sin(θ))
Right?

Solve for v, insert into the energy equation, solve for the other v and done.

So the components of G are what i missed in the first place?

#### Attachments

• ballstring.png
3 KB · Views: 364
i ended up with :
v (at 125°) = sqrt ( ( r * m * g * sine 35° ) / m )

Yes, dean barry, that's pretty much what i had. The m's cancel out. I already got to the correct answer, now its just a matter of making sure i understand the whys and hows.

rcgldr
Homework Helper
now its just a matter of making sure i understand the whys and hows.
Your diagran showing that the radial component of g is g sin(θ) is correct.

OK, see if you can figure this one out.
(see the attached sketch)
L = 1.0 m
A = 30°
g = 10 (m/s)/s
Two identical balls (both homogenous spheres), each 10 kg, radius = 0.1 m
Ball 1 is released from rest and rolls down the incline without slipping, has elastic collision with ball 2, ball 2 gets to A = 30 °
where the string tension vanishes, calculate the height H.

#### Attachments

H=1.750m (1.8m, rounded to 2 significant digits) ? My calculation assumes that all the potential energy of ball1 is transfered to ball2, at least i think it does :f

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Its not, The PE is shared (not equally) between linear KE and rotational KE as it rolls down to the contact point.

Alright. H=3.9179m ≈ 3.9m, then?

Only the linear KE of ball 1 is transferred to ball 2 at impact, but during the fall, 2/7 of the energy translated from the PE gets tied up as rotational energy, leaving 5/7 translated into linear KE.
(typical KE sharing for a rolling, non slipping, homogenous sphere)
Whatever velocity you figured for ball 2 at the impact, was also the impact velocity of ball 1 as it was an elastic collision of two equal masses.
So in summary:
ball 1 rolls 2.45 vertically gathering 5/7 of the PE as linear KE, the impact leaves ball 1 spinning but stationary, ball 2 sets off at the same linear speed as ball 1 and rotates 120 ° to slack point.
i got 2.45 m as H

Show me how you get that 5/7 and 2/7 sharing

My notes: https://dl.dropboxusercontent.com/u/34732003/IMG_20150212_180625.jpg [Broken] (dropbox, because its too large)
Where did i go wrong?

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Any solid homogenous sphere that rolls without slipping will have the LInear KE and rotating KE in that ratio regardless of speed.

Take the problem from the start (work backwards)
I get the velocity of ball 2 at 120 ° (slack point) to be 2.236 m/s, giving a KE of 25 Joules
Its PE gained at that point = 10 * 10 * 1.5 = 150 Joules
So its KE at collision point must have been: 25 + 150 = 175 Joules
(velocity = sqrt ( 175 / ( ½ * m ) ) = 5.9161 m/s)
This is also the linear KE transferred from ball 1 at the collision
(equal mass, elastic collision)
This represents 5/7 of the total KE of ball 1
( 2/7 is tied up in rotational KE )
So ball 2 total KE ( linear + rotational ) at impact = 175 * ( 7 / 5 ) = 245 Joules
So if m*g*h = 245, then h = 2.45 metres

PS: Signing out until: 12.00 noon GMT Saturday

Right, got it.
https://dl.dropboxusercontent.com/u/34732003/IMG_20150213_220656.jpg [Broken]. Happened to do another exercise today that helped me a bit with this one.

Took a look at my earlier notes, damn i must've been tired. I think i need a break :F

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Glad to help, good for me also.
Dean