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Ball at rest. Need physics help.

post #1 of 32
Thread Starter 

Hard for me to admit my faults of education, but am pretty weak on classical physical notions. 

 

Yesterday at the European PGA event, Padraig Harrington's ball on the first bounce went into the cup then bounced out of the cup to lie nearby on the green. 

 

The rules state that the ball in the cup must be at rest on the bottom of the cup to be holed. Now, if we had installed a tiny camera near the bottom of the cup, and cranked it up to record 100K frames per second, would we see  the ball hitting the bottom of the cup, then stopping for an instant, and then deforming slightly and finally  rebounding out?  Or would we see in those nanoseconds the ball never at rest, always moving either down or up?

 

Of course all this relates to the absurd, IMO, use of the cameras peering at the ball greenside and the rules judges making asses of themselves, the next day from the armchairs, over minute and arguable decisions on ball movement/oscillations that happened at the big tourney, yesterday. 

post #2 of 32

The ball was never at rest while in the hole, even for a nanosecond.  That would require that the ball impacted the bottom of the cup dead square with no angular motion of any kind, rebounding on the exact path that it took entering the hole..  The odds of that actually happening are so close to zero that the number is indistinguishable from zero.

post #3 of 32
I think it was actually Sunday at the Byron Nelson, but regardless ...

The only way that you could even argue in theory that the ball stopped for a moment was if it came up at the exact same angle that it went down. There's no practical way that could happen.

Sorry ... What Rick said ;)
post #4 of 32
Quote:
Originally Posted by joekelly View Post
 

Hard for me to admit my faults of education, but am pretty weak on classical physical notions. 

 

Yesterday at the European PGA event, Padraig Harrington's ball on the first bounce went into the cup then bounced out of the cup to lie nearby on the green. 

 

The rules state that the ball in the cup must be at rest on the bottom of the cup to be holed. Now, if we had installed a tiny camera near the bottom of the cup, and cranked it up to record 100K frames per second, would we see  the ball hitting the bottom of the cup, then stopping for an instant, and then deforming slightly and finally  rebounding out?  Or would we see in those nanoseconds the ball never at rest, always moving either down or up?

 

Of course all this relates to the absurd, IMO, use of the cameras peering at the ball greenside and the rules judges making asses of themselves, the next day from the armchairs, over minute and arguable decisions on ball movement/oscillations that happened at the big tourney, yesterday. 


Are you advocating that the ball was at rest in the hole?  If so, you won't convince anyone of that.  If it was at rest, it wouldn't have left the hole!

post #5 of 32
Quote:
Originally Posted by rogolf View Post
 


Are you advocating that the ball was at rest in the hole?  If so, you won't convince anyone of that.  If it was at rest, it wouldn't have left the hole!

 

The laws of physics says it could possibly be. But only in the cases described above.

post #6 of 32
Thread Starter 

Well, rogolf, do you remember the ball that landed in the cup and moments later popped out, followed by the frog?  In that case the rulesmeisters said, 'Oh no. Yes, ball at rest but ball not at bottom but on back of frog, so no HIO'. And nanoseconds are way beyond our perception so it may be that the ball in my case, did stop.   

 

Have you ever seen a tennis ball, in ultra slow motion video, smash into a brick wall?  One face of the ball certainly stops against the wall but the opposite face continues moving into the wall. Eventually, nanoseconds later, the ball rebounds off. Did you, or i, see that?  NO, We can see only the result not the action in progress. . 

 

And thanks Rulesman for your wise and perceptive comments which made my day. 

post #7 of 32

When the ball is on the tee, at the teebox, ball is at rest, why so?

 

Golf ball is attracting the earth with force of gravity, and earth is attracting the ball with equal force of gravity. There doesn't seem to be external forces applied to the ball, such as a player hitting it with a driver...

 

These forces at the teebox, are counter-forces and therefore the ball must be either moving at a constant speed, or it must be at rest.

 

As a matter of fact, you could say that the ball on the tee is moving at constant speed of 0 mph.

 

(if you shot a golf ball into outer space with a driver club! With theoretically full vacuum and no forces interfering on the golf ball, I suppose the golf ball would simply continue to fly at the same speed and direction.)

 

 

More to think about a golf shot...

 

Where do the forces disappear? Obviously you tee off with the driver, ball flies a long ways... Ball is flying ballistically like a bullet, it only has the force transferred into it, by the player's muscles and driver impact...

 

But eventually the ball comes into a rest (well, unless you hit the ball into outer space or something!)

 

There is obviously drag involved (air friction, on earth) , and the earth will tend to attract the golf ball back towards it.

 

Newton's second law, stipulates that lighter mass objects tend to accelerate more than heavier mass objects, when affected by the same amount of force. (this seems to be the case could someone verify?)

 

If the gravity of earth were much lower than in reality, I suppose you could shoot the golf ball out of earth's gravity influence, if you hit the golf ball hard enough...


Edited by late347 - 5/23/14 at 6:49am
post #8 of 32

If it is at rest it is not in motion and cannot bounce out of anything, let alone a hole in the ground.

A still camera or high speed camera does not detect stillness in a moving object, it either freezes or slows down motion.

post #9 of 32
Quote:
Originally Posted by Rulesman View Post
 

 

The laws of physics says it could possibly be. But only in the cases described above.

 

Well Newton's first law of motion is "An object that is at rest will stay at rest unless acted on by an external force." So what caused it to come out, an earthquake?

 

If the ball came out becuase its momentum caused it to bounce off the bottom of the cup, then it wasn't at rest. Not for an instant, a fraction of a second, or anything else. By definition, it is only "at rest" if it will stay at rest unless acted on by an external force.

post #10 of 32
Quote:
Originally Posted by acerimusdux View Post
 

 

Well Newton's first law of motion is "An object that is at rest will stay at rest unless acted on by an external force." So what caused it to come out, an earthquake?

 

If the ball came out becuase its momentum caused it to bounce off the bottom of the cup, then it wasn't at rest. Not for an instant, a fraction of a second, or anything else. By definition, it is only "at rest" if it will stay at rest unless acted on by an external force.

The only special  case example that comes to my mind would be....

 

Ball gets moved by wind, for instance on the putting green.

 

That could make golfing rather difficult though... what do the rules say? :-D

post #11 of 32

Obviously @joekelly's OP is a bit tongue in cheek.  ;-)   So I will take it one step further.  From a Quantum Mechanics standpoint, it may or may not even be in the cup for that nanosecond it contacted the bottom.

post #12 of 32
Quote:
Originally Posted by boogielicious View Post
 

Obviously @joekelly's OP is a bit tongue in cheek.  ;-)   So I will take it one step further.  From a Quantum Mechanics standpoint, it may or may not even be in the cup for that nanosecond it contacted the bottom.

 

Of course, Schrödinger's golf ball. 

post #13 of 32
Quote:
Originally Posted by Rulesman View Post

The laws of physics says it could possibly be. But only in the cases described above.

I seem to remember Newton telling us that an object "at rest" will remain at rest, unless acted upon by an outside force. Absent an outside force, if the ball came back out of the cup, it was never at rest.

Quote:
Originally Posted by Asheville View Post

Of course, Schrödinger's golf ball. 

a3_biggrin.gif
post #14 of 32

I think this decision would cover your scenario.

 

16/5.5

Player Holes Short Putt and Allegedly Removes Ball from Hole Before It Is at Rest

 

Q.A player strikes a short putt into the hole and removes the ball from the hole. His opponent or a fellow-competitor claims he heard the ball bouncing in the bottom of the hole-liner at the time the player was removing the ball from the hole, and therefore the ball cannot be considered holed in view of the Definition of "Holed" which states: "A ball is holed when it is at rest within the circumference of the hole ...". What is the ruling?

 

A.The ball is holed. The words "at rest" are in the Definition of "Holed" to make it clear that if a ball falls below the lip and thereafter bounces out, it is not holed.

post #15 of 32

The object (ball) was not at rest, it was moving and therefore had kinetic energy. It also had potential energy in its elasticity (the coefficient of restitution indicates to what extent this is utilised by the rebound). 

 

Non-technical.

1) A body on a horizontal plane is travelling along a straight line from A to B.

2)It encounters something solid (B)

3) and then travels in the opposite direction towards A (ie rebounds).

In step 1 it was moving towards B. In step 3 it is moving towards A. In what direction is it moving in step 2?

 

Technical

 

Ball bouncing from massive wall.

Most physics textbooks consider the case of a ball bouncing from a massive object, say the floor, or a wall. They consider the case where the collision is nearly or totally elastic. In the totally elastic collision, the ball loses no kinetic energy in the collision, so its speed after collision is the same as before the collision. The student thinks, "Of course that must be the case, because of conservation of energy." Seldom does the textbook, or the student, consider how conservation of momentum is satisfied in this problem. They should, for the analysis is instructive. Analyzing this may also give some insight into why energy, momentum and the conservation laws took so long to be formulated, since the concepts are subtle.

Consider a ball of mass m colliding elastically with a stationary object of larger mass M. Draw the picture before and after the collision. The conservation equations are:

[1]

mv1 = mv2 + MV2

[2]

(1/2)mv12 = (1/2)mv22 + (1/2)MV2

where v1 is the initial velocity of the smaller ball, v2 is its final velocity after collision, and V2 is the velocity of the larger mass after the collision.

Multiply the energy equation by 2 to eliminate the (1/2) factors.

[3]

mv12 = mv22 + MV2

Divide this by m on both sides.

[4]

v12 = v22 + (M/m)V22

Rearrange.

[5]

v12 - v22 = (M/m)V22

Divide the momentum equation by m on both sides.

[6]

v1 = v2 + (M/m)V2

Rearrange and square both sides.

[7]

(v1 - v2)2 = (M/m)2V22

Multiply Eq. [5] by (M/m) and combine with [7] to eliminate V22.

[8]

(M/m)(v12 - v22) = (v1 - v2)2

Multiply both sides by (m/M)

[9]

(v12 - v22) = (m/M)(v1 - v2)2

Take the limit as (m/M) goes to zero.

[10]

(v12 - v22) = 0

So one solution of this is v1 = -v2. Another solution is v1 = v2, corresponding to the case where the two objects do not collide at all.

One can graph the values of v2 and V2 against (m/M) and show that as (m/M) goes to zero, the values of the final velocities do indeed smoothly go to the limiting case values. In words, the reason this can happen is that kinetic energy is a scalar, and momentum is a vector, and kinetic energy varies as the square of the speed, while momentum varies as only the first power of speed. Therefore the quantity momentum/energy varies with speed as (1/v). So when v goes to zero, the momentum/energy can be infinite.

Those who have had calculus will recognize that this is a case where an indeterminate form arises when you take the limit of the value of momentum of a body whose mass increases to infinity.

post #16 of 32
Quote:
Originally Posted by joekelly View Post
 

Well, rogolf, do you remember the ball that landed in the cup and moments later popped out, followed by the frog?  In that case the rulesmeisters said, 'Oh no. Yes, ball at rest but ball not at bottom but on back of frog, so no HIO'. And nanoseconds are way beyond our perception so it may be that the ball in my case, did stop.   

 

Have you ever seen a tennis ball, in ultra slow motion video, smash into a brick wall?  One face of the ball certainly stops against the wall but the opposite face continues moving into the wall. Eventually, nanoseconds later, the ball rebounds off. Did you, or i, see that?  NO, We can see only the result not the action in progress. . 

 

And thanks Rulesman for your wise and perceptive comments which made my day. 

 

The ball does not have to be in the bottom of the cup.  The Definition of "holed" is:  

 

Quote:
 

Holed

A ball is “holed” when it is at rest within the circumference of the hole and all of it is below the level of the lip of the hole.

post #17 of 32
Quote:
Originally Posted by Fourputt View Post
 

The ball does not have to be in the bottom of the cup.  The Definition of "holed" is:  

 

Right. A ball pinched between the flagstick and the ball is "holed" if it's below the level of the green (the lip), even if it's not touching the bottom of the cup.

post #18 of 32

Problem:

A 0.1 kg ball is thrown straight up into the air with an initial speed of 15 m/s.  Find the momentum of the ball
(a) at its maximum height and
(b) half way up to its maximum height.

 

 

Solution: 
(a) At the ball's maximum height its velocity is zero, and therefore its momentum is zero.

(b) The ball's total energy E = K + U is constant.  
Initially U = 0 and K = (1/2)mv=(0.05 kg)(15 m/s)= 11.25 J.  
At its maximum height K = 0 and U = mgh = 11.25 J.
Halfway up to its maximum height K = U = (1/2)11.25 J.  
Therefore v= 2K/m = (11.25 J)/(0.1 kg) = 112.5 (m/s)2
v = 10.6 m/s, p =( 0.1 kg)(10.6 m/s) = 1.06 kgm/s

 

 


(a) At the ball's maximum height its velocity is zero, 

 

This is of course the same for any object which reverses its direction.

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