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Ping Pong ball vs. golf ball and distance moved

by lax27
Tags: ball, distance, golf, moved, ping, pong
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lax27
#1
Apr2-09, 05:21 PM
P: 2
You cant throw a ping pong ball as far as a golf ball. F=ma would suggest that the lighter ball would accelerate more. It seems you are unable able to exert as large a force on a ping pong ball as you can on a baseball. Does your inability to exert a large force on a ping pong ball account for the reduced acceleration? If so, what about objects make them difficult to exert large forces on them. ie: a tissue?
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Cantab Morgan
#2
Apr2-09, 05:30 PM
P: 261
The difference isn't in your arm... it's in the ballistic coefficient of the ball. Air resistance is going to slow down both the ping pong ball and the golf ball. Suppose a very simplistic model of air resistance, where the force trying to decelerate the ball depends only on the cross sectional area of the ball. That means that the air is exerting the same decelerating force on balls of equal size.

But the more massive golf ball slows down less, because it's more massive. The identical decelerating forces on the balls is going to produce a smaller deceleration for the more massive ball. On the moon, you could throw the ping pong ball farther.
lax27
#3
Apr2-09, 07:11 PM
P: 2
Very helpfulresponse... but I am still wondering... How does that tie in to F=ma which would suggest that the lighter ball would have more acceleration. Does that mean than F=ma is accurate minus air resistance?

I know there are objects that you simply cant exert big forces on...like tissues. Is it also true that you cant exert as large a force on a ping pong ball as you can a baseball?

AUMathTutor
#4
Apr2-09, 09:33 PM
P: 490
Ping Pong ball vs. golf ball and distance moved

I think that, realistically speaking, it works like this...

Say your arm has a mass of 3.2 kg.
Say a golf ball has a mass of 0.046kg.
Say a ping pong ball has a mass of 0.0027kg.

We can then assume some property of the arm's throwing ability is held constant. If we hold the force constant, and if we hold the distance you throw over constant (say you throw it in a shot-put fashion) then we can get the work.

Say the force is F and the distance is D. Then W = FD.

If you had nothing in your hand, your arm would be moving such that 1/2Mv^2=FD so
v = (2FD/M)^(1/2)

If you had the golf ball in your hand, your arm would be moving such that 1/2(M+a)v^2=FD, so
v_a = (2FD/(M+a))^(1/2)

If you had the ping pong ball in your hand, your arm would be moving such that 1/2(M+b)v^2=FD so
v_b = (2FD/(M+b))^(1/2)

We can rewrite these as..

v_a = v * (M/(M+a))^(1/2)
v_b = v * (M/(M+b))^(1/2)

So

v_a = v_b * ((M+b)/(M+a))^(1/2)

Which means that v_b = 1.01352 v_a, approximately. So the speeds of the two objects differ by only 1.5%. Let's say they're equal.

Say you throw them horizontally and only look at the x coordinate. Then

av' = -kv => v' = (-k/a)v => v = (v_a)*exp(-kt/a)
bv' = -kv => v' = (-k/b)v => v = (v_b)*exp(-kt/b)

What does this mean? Let t = b*ln(2)/k. Then the velocity of the lighter projectile is 50% its original value. The velocity of the heavier projectile, however, is still 96% of its original value.

So it only seems like you're not throwing it as fast because the ping pong ball carries no "oomph" with it. That's why you play with golf balls outside and ping pong balls inside.
rcgldr
#5
Apr3-09, 04:16 AM
HW Helper
P: 7,176
Quote Quote by AUMathTutor View Post
Say a golf ball has a mass of 0.046kg.
Say a ping pong ball has a mass of 0.0027kg.
The post above didn't explain that

a = mass of golf ball = .046k
b = mass of ping pong ball = .0027kg

As pointed out arm speed between throwing a golf ball and ping pong ball will be about the same. Release speed will be about the same, so the main difference is aerodynamic drag.
cragar
#6
Apr3-09, 05:01 AM
P: 2,468
yes the deal here is the air resistence
rcgldr
#7
Apr3-09, 07:20 AM
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P: 7,176
Doing the math;

f = aerodynamic drag
a = aerodynamic drag acceleratoin
v = velocity
x = distance moved
t = time
m = mass
k = deceleration factor related to mass and drag components
mp = mass of ping pong ball
mg = mass of golf ball

[tex]\rho = {density \ of \ air}[/tex]
[tex]A = {frontal \ area \ of \ ball}[/tex]
[tex]k = 1/2 \ \rho \ C_d \ A \ /\ m [/tex]
[tex]a\ =\ f\ /\ m\ =\ dv/dt\ = \ -k\ v^2[/tex]
[tex]dv\ /\ (-k\ v^2)\ =\ dt[/tex]
[tex]1\ /\ (k\ v)\ =\ t\ +\ c[/tex]
[tex]v\ =\ dx/dt\ = \ 1\ /\ (k(t\ +\ c))[/tex]
[tex]dx\ = \ dt\ /\ (k(t\ +\ c))[/tex]
[tex]x\ =\ (ln(t\ +\ c)\ /\ k)\ + d[/tex]

Note that k is proprotional to 1/m so the smaller the mass, the larger k. Say the initial velocity is 10 m/s, k = .25 for the golf ball and 2.5 for the ping pong ball.

At t = 0, let v = 10 m/s.

For the golf ball:
k = .25, 0 + c = (1 / (.25 x 10)), c = .4, and d = -ln(0 + .4)/.25 = 3.6652.
x = ln(t + .4)/.25 + 3.6652
v = 1 / (.25 x (t + .4))
at t = 2 seconds, x = 7.1670 meters, v = 1.6667 meters/sec

For the ping pong ball:
k = 2.5, 0 + c = (1 / (2.5 x 10)), c = .04, and d = -ln(0 + .04)/2.5 = 1.2876
x = ln(t + .04)/2.5 + 1.2876
v = 1 / (2.5 x (t + .04))
at t = 2 seconds, x = 1.5727 meters, v = 0.1961 meters/sec

If the initial velocity was infinite, this would correspond to t = -0.4 seconds for the golf ball, and t = -0.04 seconds for the ping pong ball.
arildno
#8
Apr3-09, 07:44 AM
Sci Advisor
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P: 12,016
A MAJOR reason not mentioned so far is the importance of surface geometry for the total (normal) drag upon the ball.

In a viscous fluid like air, a smoothly shaped ball will create a large wake region, with significant decrease of pressure there.

Thus, the pressure difference between the front and the back increases, i.e, creating a large normal drag force upon the ball.

If the surface is dimpled, however, the induced small-scale turbulence will narrow the wake region (in what is called the "drag crisis"), and the pressure difference between the front and back remains tiny, and hence, a pressure-induced normal drag force will not be significant.


This will, of course, remain, even if the balls have the same mass.
rcgldr
#9
Apr3-09, 08:00 AM
HW Helper
P: 7,176
Made a graph of the above:

Cantab Morgan
#10
Apr3-09, 07:20 PM
P: 261
Quote Quote by arildno View Post
A MAJOR reason not mentioned so far is the importance of surface geometry for the total (normal) drag upon the ball.
Good insight! We have assumed a spherical cow.

Dimples do make a significant difference when comparing golf balls. But my intuition tells me that the mass difference is more important in this case. I bet you could still throw a smooth golf ball much farther than a dimpled ping pong ball.

A really interesting problem, though, would be to find the magic mass ratio where the dimples on the lighter ball would allow it to be thrown as far as a smooth heavier ball. Neat!
AUMathTutor
#11
Apr3-09, 07:51 PM
P: 490
Had you read my post, you would see that when I assumed smooth balls, you still got the golf ball went much further than the ping pong ball.

I believe Jeff Reid's math using an square-velocity drag rule also yields this result.

So a smooth golf ball would still go a lot further than a ping pong ball, and a dimpled ping pong ball would still go far less distance than a smooth golf ball.
rcgldr
#12
Apr3-09, 08:56 PM
HW Helper
P: 7,176
Quote Quote by arildno View Post
A MAJOR reason not mentioned so far is the importance of surface geometry for the total (normal) drag upon the ball. If the surface is dimpled
In the math I did, k includes both Cd, which would vary with dimples, and mass. The math I did above still holds, the main point being the drag component / mass is the determining factor. In real life, the Cd changes with speed, which means that drag isn't really proportional to V^2.

Note that above a certain speed, the smooth surface results in lower drag, as shown in the graph on this web site, the Cd for the smooth surface goes down to almost .1 at high speeds, while the lower limit is about .25 for the dimpled surface. Beyond the "critical" speed for the dimpled surface, the smooth surface has a lower Cd.

http://www.aerospaceweb.org/question...cs/q0215.shtml

I don't know why the article above uses [itex]V_\infty[/itex] for the velocity term. It's just velocity as explained here:

http://www.engineeringtoolbox.com/re...ber-d_237.html


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