Need an engineer here!

[widgitmaster]

I have limited math knowledge, so I would appreciate it if someone with more knowledge than I solve this question!

I plan on building a "Newton's Cradle", that has five 2.5" dia ball swinging on a 10" radius fulcrum.

What is the force in Lbs per sq/in of a 2.32 Lb chrome steel ball falling in a 10" radius from 90 deg. up, striking another ball of equal size & weight?

Thanks, Eric

[ger21]

Force is not measured in psi.

I couldn't figure out how to calculate the velocity, although the last link may give the correct answer, although I think it's too high as it's for a ball rotating 360° in a vertical circle. I think the velocity here will be lower??

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html
http://hyperphysics.phy-astr.gsu.edu...wt.html#strmas
http://hyperphysics.phy-astr.gsu.edu...irvert.html#c1

[widgitmaster]

Now I know how much I don't know about math Wow!

[Geof]

Your question is not really answerable because any answer involves so many assumptions.

You can calculate the kinetic energy that the swinging ball has at the bottom of the swing; it is the same as the gravitational potential energy at the top of the swing, mass * g * height. I am ignoring friction because it is very low in the setup you are contemplating.

The kinetic energy of the falling ball is transferred to the next ball on impact and the way energy is transferred in this situtation is mostly by elastic deformation of the steel ball while the moving ball exerts a force on the next ball. This is where it gets awkward to calculate meaningful numbers. When two spheres make contact the area of contact is very, very small; indeed if the spheres are perfect and do not distort the area is infinitesimal and the pressure is very high. But the spheres do distort so the area of contact increases and the pressure of course declines.

Really the only answer that can be correct in any way is that the pressure is not enough to cause plastic deformation on either steel ball.

[widgitmaster]

Thanks Geof, I think I understand that!

The balls are grade 50 with a strength of 295000 psi, and a hardness of 60-67 Rockwell.
I don't think they will self destruct, but I was curious as to the actual pressure or force upon contact!

Eric

[widgitmaster]

For what its worth, here is a link to the build log of the Newton's Cradle!

[Jay C]

I took a crack at it. I used SI units then converted them.

F=mg(sin(angle))

g=32.2ft/s^2
L=10in
angle = 90
v= SQRT(2*g*L(1-cos(angle)) which I calculated to 7.32ft/s or 87.9in/s
m = 2.32lb

F=2.32 * 32.2 = 74.7lbf

Now just need to get the contact area of two 2.5" spheres. Still looking.

Jay

[ger21]

Now just need to get the contact area of two 2.5" spheres. Still looking.

Jay

As Geof said, it's infinitely small, which makes the force infinitely large.

[harryn]

Widgitmaster - perhaps some questions about your goals would help:
- Are you wanting to estimate what will happen to the next ball - in terms of motion?
- Are you wanting to make an estimate of how much the steel balls will deforrm?
- Are you wanting to make sure the balls don't crack?

Motion calculations use a combination of methods, usually involving conservation of energy and momentum.

There is a method to estimate the deformation of the steel balls at impact, which is actually needed to obtain a "real' result, but actually is nearly meaningless for most applications, as it last for literally milliseconds. It is also a PITA to do the calculations by hand, which is why I hesitate to attempt it.

[Runner4404spd]

the real question is what are you after. in terms of pounds of force acting on the next ball (different than PSI), this can be calculated but why bother. PSI is a measure of force exerted on a certain area. on a ball that contact another ball at its tangent, the point is infinitely small. there will very likely be some minor elastic deflection of both balls which will increase the area, but again, your not really going to solve this with simple empirical math. you have to setup many math models and possibly even a collision simulation to get your true answer.

[Geof]

....- Are you wanting to estimate what will happen to the next ball - in terms of motion?...

On a Newton's Cradle with five balls when you pull out three or fewer balls the 'next ball' remains motionless.

I think this is what is so fascinating; one ball causes the fifth ball to swing out at the same speed as the first one swings in, two impels two, three impels three and four impels four.

[Javerh]

There are two areas of engineering at play here: The ball contact that falls into the purview of tribology and the ball impact that is dealt with dynamics. I'm going to be using metric formulas here:

When two elastic balls touch, they form a circular contact area that has the radius:
a = ( (3*F*R)/(4*E) )^(1/3), where F is the contact force, R is ball radius and E is the modulus of elasticity.

The two balls move relatively towards each other by a certain distance:
d = ( (9*F^2)/(16*(E^2)*R) )^(1/3)

Now, we can assume that the impact is a short impulse equalling the change in inertia. That gives us the impact force:
F = m*v/t

Unfortunately, the impact time cannot be explicitly solved. However, we can assume that the time is simply the length of the movement of the balls divided by the speed of the first ball (I'm winging this!). Through several iterations we can solve the values for generic steel:
Impact time 0.00005917 s
Relative ball movement 0.132 mm (0.0052")
Impact force 39.626 kN (8924 lbf)
Contact area radius 2.048 mm (0.081")

I don't think there is an easy way to determine the energy losses on the chain of balls.
I'm not entirely convinced by my answer. Perhaps someone with better engineering knowledge can check this out.