# Childs tennis game

1. Jul 7, 2008

### a.mlw.walker

OK. I have been fighting with my pen and paper for hours trying to fathom this:

I have been thinking about that childs game where there is a post, and two players. On the post is a string with a tennis ball attached at the top, each player has a tennis raquet and they hit the ball around the ball to each other.

As they hit the ball harder the ball rises, and therefore its radius increases. So the
-centepetal force is increasing (or centrefugal but we know thats fiction...) and therefore the force is more and more overcoming the mg of the ball, hence it rises towards horizontal.

If there was only one player, and he hits the ball around and around, until it is quite close to the horizontal (it will never be completely horizontal), and you wanted to work out where the ball will be in 1.5 seconds, what do you have to take into account.

Is the acc (change in speed/change in time)?

deceleration is constant and therefore we can work its distance travelled in 1.5 seconds from constant acc equations?

and r is a function of what?

Its a simple game but so hard to calculate because r changes as ball decelerates.

I am right in thinking though that as long as centrepetal force is greater than mg the ball will stay airborne?

I bet there are others out there who notice my little predicament...

can anyone give me a mathematical explanation for whats going on, I understand in words but not equations...

There is also a constant speed at which the tennis ball will no longer spin, but fall, when its centrepetal force goes less than mg, which i would also like to calculate...

Thanks

Alex

Last edited: Jul 7, 2008
2. Jul 7, 2008

### tiny-tim

Hi Alex!
Nooo … any amount of energy will keep the ball above its lowest point (from conservation of energy )

This situation is exactly the same as something moving around the inside of a frictionless sphere …

instead of the tension in the string, there is the normal reaction from the sphere, which is always in the same direction.

So (assuming horizontal motion) you need to calculate the normal reaction (or the tension in the string) by applying good ol' Newton's second law to the centripetal acceleration and the weight.

There's only trouble when Newton's second law shows that the reaction (or tension) is zero.

3. Jul 7, 2008

### rcgldr

One difficutly is angular momentum is conserved only if you take into account the torque force applied by the pole to the earth, so you'd have to determine how much angular momentum was transferred to the earth, and lost by the ball.

If the motion was purely horizontal (which unfortuantely, it's not), like a puck on a frictionless surface, the path is called involute of circle, there's no tangental acceleration, and the speed of the puck remains constant. Link to image: pole.jpg.

Howerver in this case, the string is winding downwards as well as around the pole, which changes the path of the ball and string winding, decreasing the radius faster than the involute of circle case, so I don't think the speed of the ball remains constant, but it doesn't increase as fast as it would if all the angular momentum was in the ball.

4. Jul 8, 2008

### a.mlw.walker

I like tiny_tim's example, because I know that in the case of the childs tennis game, the winding string would effect things, and bending of the pole etc - nice drawing by the way jeff!

No tim's example of a ball inside a sphere, spinning around the circumference of the widest point...

So we have centrepetal acc, v^2/r, and multiply by m and we have centrepetal force - this force is the component holding the ball from just sitting on the bottom of the sphere? and instead rising up the edge?

But surely there will be an amount of energy where the ball doesnt rise any more, what decides that?, then there will also be a speed where the ball just falls to the bottom - I'm only thinking of that because when i last saw a roulette wheel, the ball didnt keep spiraling down, it dropped to a speed and fell.

What are the deciding factors of when the ball will no longer spin along the outside, but just drop to the bottom of the sphere? How can that be worked out.

There are so many websites explaining the principles, but none explaining the maths to a very deep level.

r is a function of v surely. As the ball accelerates (am i talking about centrepetal acc here?) r increases, and as the ball decelerates the r decreases, and finally the ball gets to a speed and falls to the bottom - but not spiraling ALL the way down.

What i think i understand. The ball as a angular acc - resultant of centrepetal and tangential, and a angular v - of which you can calculate angular acc.

gravity pulls the ball straight down, but centrepetal pulls the ball in. But the ball wants to continue in a traight line, but the edge of the sphere deflects it round.

Thanks guys. This forum is so helpful

5. Jul 8, 2008

### rcgldr

The limit is when the center of mass of the ball is at the same height as the point where the string leaves the pole.

This would require the horizontal speed to be zero, which won't occur if the ball starts off with a non zero horizontal speed (until it touches the pole).

Another missing factor, is the pole frictionless, which would allow the string to slide up or down on the pole? ... or is does the pole have virtually infinite friction, where the string doesn't slide at all?

If the effect on the earth is taken into account, then angular momentum is conserved, and some of the rotational energy of the ball is used to change the rotational energy of the earth, depending on the torque applied to the pole.

A more common example of a game with a ball, string, and pole is tetherball. This article also mentions tether tennis, but in the case of tether tennis, the string is mounted onto a rotating screw, and the string doesn't wrap around the screw. The screw just rotates and moves upwards or downwards until it reaches the top or bottom limit.

http://en.wikipedia.org/wiki/Tetherball

You could probaby do a web search for tetherball physics and find some answers.

Last edited: Jul 8, 2008
6. Jul 9, 2008

### a.mlw.walker

yeah that is why i tried to change the example to a small ball spinning around the inside of a larger sphere, or the roulette ball because I know that effects of the pole bending, inertia effect the system and that you would probably mention them, and i was hoping to fogrget them for a second so i could see the math based around circular motion, before worrying about bending moments, etc etc. The only outside force i was interested in was gravity, which is what pulls the ball down.....

7. Jul 9, 2008

### rcgldr

For an aircraft flying in a circle, the horizontal component of force is m s2 / r and the vertical component is m g (where s is the speed). The bank angle in a coordinated turn will be tan-1(m s2/ (r m g)) = tan-1(s2 / (r g)).

For a non-winding tetherball case, assume you have a weightless string.

l = length from center of top of pole to center of ball
r = horizontal distance from center of pole to center of ball
s = speed of ball
m = mass of ball
g = rate of gravitational acceleration
a = angle between string and pole

Vertical component of force = m g
Horizontal component of force = m s2 / r

tan(a) = (m s2 / r) / (m g) = s2 / (r g)
r = l sin(a)
tan(a) = s2 / (r g) = s2 / (l sin(a) g)
tan(a) sin(a) = s2 / (l g)

I don't know how to calcutate the inverse of tan(a)sin(a).

Last edited: Jul 9, 2008
8. Jul 10, 2008

### a.mlw.walker

Thats getting there thanks jeff, do you know though why the plane would fall out of the air? is it because the horizontal component is smaller than the vertical?

Also do you mean inverse or arc of tan(a)sin(a)? Because inverse is very different to arc.

inverse is just 1/tan(a)sin(a) whereas arc is the functin on a calculator tan^-1 etc...

9. Jul 10, 2008

### tiny-tim

Hi Jeff!

tan sin = tan2 cos = 1/cos - cos,

so tan sin = 2x –> cos2 + 2x cos - 1 = 0 –> cos = -x ± √(x2 + 1))

10. Jul 10, 2008

### rcgldr

Inverse of function:

http://en.wikipedia.org/wiki/Inverse_function

I'm not sure what you mean by falling out of the air. I was just showing the relationship between the bank angle and horizontal conering force in a coordinated turn.

However I'm not sure when it's state that an F16 can pull a 9 turn refers to the horizontal component of how many g's of acceleration the pilot feels. My guess is that it's what the pilot feels, so a "9 g horizontal" turn has a horizontal acceleration component of 8.944 g's.

11. Jul 10, 2008

### a.mlw.walker

the aircraft is flying in a circle, and then the pilot cuts the egines, but holds the steering wheel round so it contiues in a circle, spiraling down, at one point (unless the plane is built for gliding) the plane will fall out of the air. what determines this.

12. Jul 10, 2008

### rcgldr

Thanks.

So let x = s2 / (l g)

tansin = sin2/cos = ((1 - cos2)/cos) = x

((1 - cos2)/cos) = x
1 - cos2 = x cos
cos2 + x cos -1 = 0

cos = (-x ± √(x2 + 4))/2

Substituting 2z = x, then I'll get your simplified form:

cos = (-2z ± √(4x2 + 4))/2 = -z ± √(z2 + 1)

Let s2 / (l g) = √2 / 2

cos(a) = (- (√2 / 2) ± √((√2 / 2)2 + 4))/2 = (- (√2 / 2) ± √(9/2))/2

a = 45 degrees

Last edited: Jul 10, 2008
13. Jul 10, 2008

### rcgldr

Since cos inputs can't be > 1 the equations become:

cos = (√(x2 + 4) - x)/2

cos = √(z2 + 1) - z

It never falls out of the air. All aircraft will behave like a glider, depending on the lift to drag ratio, the aircraft can spiral downwards at a constant decent rate, while maintaining coodinated turn where the path is a downward spiral with a constant radius and speed. Since the decent rate is constant, there's no change in forces or accelerations, only the path varies. When the engine power is adjusted, there is a brief period of vertical acceleration, then everything stabilizes into ascending flight, level flight, or descending flight.

Last edited: Jul 10, 2008
14. Jul 10, 2008

### rcgldr

Getting back to the actual tetherball case, where the string winds around a pole is more complicated. Assume the pole doesn't turn, so it doesn't peform any work on the ball. The total energy of the ball is constant and is the sum of gravitational potential energy and kinetic energy. As the ball descends, velocity increases √(2 g h), where h is the height the ball has descended. Since the pole peforms no work on the ball, the tension of the string is always perpendicular to the path of the ball.

At this point, I'm stuck. The speed of the ball includes downwards, inwards, and circular components. The centripetial force is s^/r, but what is the direction and length of r in this case?