Scientific Questions

but u weren't asking....but telling others how they are b.s.ing....

yeah....to put things in perspective...we almost ehanged him...didnt we...?  No pun intended....

Ok...back to the questions...

SIDOFMILENIUM wrote: Ok...back to the questions...   W=fd then how come pushing a wall is tiring?

When you push the wall, since the wall doesn't move, it's actually your body that gets displaced. For example, you lean against the wall, then push. You end up pushing your body away from the wall, instead of the wall away from your body.

Even if your feet never move from their spot on the ground, your upper body moves.

So if you stand up straight, then lean over towards the wall in front of you. When you push against the wall, your feet won't move, but your upper body will move, from leaned-over-position to upright or further depending how far you push. So it will be your upper body that has been moved, even if your feet stayed planted on the ground.

It's like doing a push-up but on a slope instead of flat ground. Your feet stay planted, but you move your upper body.

SIDOFMILENIUM wrote: Ok...back to the questions...   W=fd then how come pushing a wall is tiring?

I understand what ur saying sido. The answer is actually in your question. In physics the definition of work is a force multiplied by a change in distance. This is different from what we commonly think of as work: exertion of energy to accomplish a task, or even just accomplishing a task.
The problem with physics isn't that concepts are hard per se, just that you kind of have to rewire your thinking. So if you are pushing against a wall without moving yourself or the wall, you haven't done any work in the physics definition. You have expended energy but energy doesn't factor into the equation W=fd.

by the way i'm really not trying to sound pompous or offend anybody with my last posts. Just trying to be helpful.
Spread the wealth of knowledge

I'll take silence as agreement....alright, next question...

The first question has to do a little more with conservation of momentum and newton's second law. Momentum=mass times velocity. When you hit a ball with a paddle, the ball exerts the same amount of force on the paddle as the paddle does on the ball but in opposite direction. When the two objects collide, the elasticity of the sponge and rubber comes into play but so does the momentum of the paddle and the momentum of the ball.

Ok, this is kinda hard explaining this process and trying to introduce these definitions at the same time but i'm doing my best.

So the ball goes into the sponge a certain amount according to it's own speed and mass as well as the speed of the paddle. I will assume that the sponge-rubber system behaves like a spring in the sense that how far you push into it is directly proportional to the force required to make it go that far. The spring potential energy is given by 1/2 * a constant labled k (this constant is a measure of the stiffness of the spring, or how easily it is compressed) * change in distance squared. symbolized:
PE (spring) = -1/2kx  (it is defined as a negative value because the force of the spring pushing back, the restoring force, is considered the positive value)
This is probably way more information than you wanted, but thinking through this helps me prepare for my finals ;)

So it gets a little more complicated factoring in the conservation of momentum upon impact and energy dissipated by the sponge not acting exactly as a spring and converting kinetic energy into heat and sound would have to be taken into account as well. The principle of conservation of momentum after a collision would tell us that the mass of one object times its velocity is equal to the mass of the second object time its velocity. So if you increase the mass of one object, its speed will decrease  and vice versa. So what is really happening when you move your paddle back at the point of contact is decreasing the momentum of the paddle or even causing it to have momentum in the same direction as the ball (that is important because velocity is regarded as a vector where direction is just as important as amplitude, speed is not the same as velocity in physics). So yes that will affect how far the ball will go into the sponge but the main change is caused by the change in momentum of the paddle.

It's kinda hard to stay on track here, but the ball deformation isn't where the spin is generated, but the force of friction is the main force to consider here.  Friction force is given by multiplying the coefficient of friction (tells us how much two surfaces stick together) symbolized by the greek letter mew, multiplied by the normal force. The normal force is the opposing force that keeps the ball from going through the paddle.
To understand normal force better, consider a mug on a table. There is a force being exerted on the mug equal to its weight. This comes from newton's second law: sum of the forces = mass times acceleration. When acceleration = 0 this is a situation called equilibrium. This means that the sum of the forces is equal to zero; so in the mug situation the normal force is equal to the weight of the mug (weight=mass*force of gravity)

Holy crap the more in depth I go the more stuff there is to explain, this is rather exhausting.

So friction comes into play because the component of the force that is perpendicular to the axis of rotation of the ball (that's a mouth full) is known as a torque and is what causes the ball to spin. So more than anything, good stroke technique is what will develop more spin, and I believe professionals and coaches would agree with me on that point.

On the point about the ball bouncing on the table, it is more of a perfectly elastic collision where not as much energy is absorbed through deformation (like billiard balls). Not really the crucial point though. The ball traveling through the air has rotational kinetic energy as well as translational kinetic energy. If there were no friction between the table and the ball the ball wouldn't 'speed up' after impact; the only things that would affect its trajectory would be gravity and Bernoulli's principle (the one with fluids and air pressure etc) Since there is friction, at the moment of impact, the ball exerts a force on the table, mainly due to rotational KE because the ball is spinning faster than it is traveling. A component of that force will be in the up and down (x) direction, but the majority will be in the side to side (y) direction due to spin. This means that there has to be an equal but opposite force on the ball, which comes from friction force, and since the table is stationary (due to its much greater mass and static friction between it and the floor) and the ball is not, some of the rotational energy is converted to translational energy due to a torque that does not have an opposing force.

well, I am exhausted from snowboarding all day and can't concentrate much more on this topic. Sorry that my explanation is a little hard to follow, but in reality that was  a complicated situation when all aspects are considered.

ok.....now, if a ball has spin and speed compared to a ball which has only spin....would the ball w/ only spin rebound w/ greater spin than the other one b/c speed will interfer w/ angular spin?  Why?

both balls have equal amount of spin...but one has more speed.....the blade is normal inverted in a stationary blocking position....would both balls come back w/ equal opposite spin?(or would speed of the ball interfer w/ the dwell time required for angular rebound)...How?

ahhh my favorite topic mechanics and materials !! here is my late entry responding to esnift.

debraj wrote: now my additional query ?   what is the best way to pass the ball from one end of a large room to the players far away?   topspin it on air... or chop it on air ..or hit it without any spin at 45 degree angle?  :)

A chop-smack.

I hit the ball about 15 degrees upward from the horizontal plane, and apply underspin. But not a sharp chop angle, it's a flat hit but with the paddle aimed upward a bit. Hard to explain.

But the effect is that the ball travels very far, and you don't even have to hit it hard. Most people hit it with flat, hard contact, at a 45 degree angle upward from the horiztal, and it doesn't go far even when hit very hard. I'm always slightly annoyed when people do that to my ball because they're hitting it very hard and it might crack, whereas my method travels much further and you don't have to hit hard.

The underspin applied makes the ball curve upward and travel further.

sido,

It would depend on the surface of the racket.

If the rubber is soft and grippy, then the faster ball would lose more spin due to more surface contact.

If the rubber is hard and smooth, then the balls would have almost the same spin coming back. However, the faster ball bouncing back on the table would invoke more topspin than the slower ball. Two underspin balls would loose the underspins. Two topspin balls would gain spin. More for the faster ball on both cases.

@gulca

debraj,

I disagree. Your reverse torque doesn't make much sense.

Grippy surface first. The more grip the surface, the greater the spin is going to affect the direction of the bounce. It won't, and never will change the rotation/spin based on the gripping from the rotation.

Drop a superspinning ball straight down onto a super grippy floor. What do you see? You never get any  torque reversal. Ever. What you see is the change of the direction of the bounce towards the rotation of the ball. The speed of the ball would increase. The rotation would slow down. But never reversed.

Now another experiment. Shoot a spinless ball at an angle towards the grippy floor. On the bounce, the ball would be spinning forward. It would loose a little of it's speed though.

I hope someone would clarify both our argument. I don't want this to be another "mitthra" fight. :)

Combine both cases, and you see the flaw in your explanation.