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I am editing the question because it was misunderstood to be a homeowrk question.

enter image description here

I am modeling a Stumps equipment for a game called Cricket.Typically the game consists of 3 wooden stumps positioned upright by hammering them into the ground.They were setup behind the batsmen,batter in baseball analogy.The pitcher scores a point if he was able to hit the stump and causing it to move.For the stump to move,i.e to overcome the static friction, he should be throwing atleast at speed of 10mph.

The inverse conical section at the base of each stump,in the left side of the image, is not seen in the middle image because it was hammered into the ground.

Sometimes it is played for recreation on a concrete ground like a tennis court.So there is no possibility of standing them upright by hammering into the concrete.So I am planning to design an equipment similar to the one like the right side of the image.

The side with the spring will be facing the wicketkeeper standing behind the batter,in the picture above.The movement of the stump is restricted by the spring connected to each of them.

My problem is to find the right type of spring which restricts the movement the same way as the ground resists the movement of the stump when hammered into the ground.The gound resists the stump from moving for balls hitting less than 10mph.The spring should be behaving the same way.

The picture shown below has a ball of 5 oz hitting the brick of negligible mass,negligible static and kinetic friction,resting on the table,connected to a spring.The brick is placed just to make sure the ball has sufficient surface area to make contact.

For the ball to cause a compression in spring ,it should be travelling at least 10 mph.I would like to know the initial tension and the spring constant of the spring.

A simple analogy to this problem would be to compute the static friction of a brick resting on a surface when it takes a ball of 5 oz traveling at 10 mph to overcome static friction.In my case I need the spring constant or initial tension instead of static friction.

Let me know if I am not clear

enter image description here

Qmechanic
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    What physical concept are you having trouble with here? You should ask about that, not ask us to solve your homework question for you. – march Aug 25 '15 at 16:04
  • Is this homework or does it have another purpose? – Gert Aug 25 '15 at 16:35
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    Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Aug 25 '15 at 16:37
  • @Gert -- I have edited the question.Excuse me for modelling the question like a homework one.. – raj'sCubicle Aug 25 '15 at 17:46
  • @march I have edited the question.Excuse me for modelling the question like a homework one – raj'sCubicle Aug 25 '15 at 17:47
  • @JohnRennie I have edited the question.Excuse me for modelling the question like a homework one – raj'sCubicle Aug 25 '15 at 17:47
  • "The pitcher scores a point if he was able to hit the stump and causing it to move."$$$$ Actually the pitcher scores if he causes the bail (the little cross bar on top of the stumps) to fall; and you are really asking for help designing a wicket. Note that "10 mph" and "moving" are still ill-defined: the point of impact will matter, and "how much motion" is needed matters. The transverse vibration of the impact will cause the top of the stump to move even if the bottom is held rigidly. The mass of the stump should not be neglected, and neither should the static friction (which is key here) – Floris Aug 25 '15 at 17:59
  • @Floris...I am asking for the range of spring constant I can go with .Not the ideal spring constant.Consider that the ball is not rolling and the bottom has to move for the bails to get dislodged.Please ignore the traverse vibration.I want to go with a spring constant I can start with... then consider all these factors before finalizing the ideal one. – raj'sCubicle Aug 25 '15 at 18:18
  • raj'sCubicle: This is far harder to model than you might think, in part for reasons Floris pointed out (but there a others). Your best bet is to buy one of these commercial devices and simply copy the parts. The spring's pre-tension and spring constant can easily be determined by means of some known weights and a ruler and used to find a commercial equivalent with similar characteristics. – Gert Aug 25 '15 at 19:14
  • Related: http://physics.stackexchange.com/a/126117/392 – John Alexiou Aug 27 '15 at 13:40
  • Good question. Those look like extension springs. In your question you phrase it like they are going to be compression springs. Can you please provide more details on the design and the side the ball is going to impact in order to understand better. – John Alexiou Aug 27 '15 at 13:45
  • I think a better design would be to use some plastic stops which give when hit over 10mph and use the springs as a return mechanism so the stumps don't flatten out completely. – John Alexiou Aug 27 '15 at 13:53
  • @ja72 do you have any image showing such design? – raj'sCubicle Aug 28 '15 at 17:15

1 Answers1

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I cannot comment on the full design because the question is lacking on details, but I can explain more about the situation where you hit a mass with a ball and a spring reacts to it (like the sketch shown)

  1. Consider the friction force as $f=\mu m g$ and the equations of motion $$m \frac{{\rm d}^2x}{{\rm d}t^2} + k x \pm f = 0$$ The sign of $f$ depends on the direction of motion, but since I only consider what happens initially I have to use the $+f$ side.

  2. The general solution given initial conditions $x(t=0)=X$ and $\dot{x}(t=0)=V$ is $$ x(t) = X \cos(\omega t) + \frac{V}{\omega} \sin(\omega t)+ f \frac{\cos(\omega t)-1}{m \omega^2} $$

  3. The frequency of natural oscillation $\omega$ is critical to the solution and for a simple mass spring system it is $$\omega = \sqrt{\frac{k}{m}}$$
  4. Use the natural frequency to estimate the average impact time. A full cycle occurs during time $$\Delta t = \frac{2 \pi}{\omega}$$
  5. The collision with the ball causes a momentum transfer (impulse) that equals with $$J = \frac{ (1+\epsilon) v_{ball}} { 1/m+ 1/m_{ball} } $$
  6. The average force of impact is $$F_{ave} = \frac{J}{\Delta t} = \frac{(1+\epsilon) \omega v_{ball}}{2 \pi \left(\frac{1}{m}+\frac{1}{m_{ball}}\right)} $$ where $\epsilon <1$ is the coefficient of restitution. If the ball doesn't bounce back a lot make it small, close to zero; if it bounces back very elastically, it approaches one.
  7. Finally set the average impact force to friction $f$ at $v_{ball}$ $$ \omega = \frac{2 \pi \mu g (m+m_{ball})}{(1+\epsilon)m_{ball}v_{ball}}$$ and find the spring stiffness by $$k=m \omega^2$$

APPENDIX

For a slender beam of diameter $d$ the 1st natural frequency is $$\omega_1 = \frac{4.73^2}{4} \frac{c d}{\ell^2} $$ where $c$ is the longitudinal wave speed and it calculated by $c^2 = \frac{E}{\rho}$. The $i$-th frequency is $$\omega_i = 0.1103 (2i+1)^2 \omega_1$$

The impact calculation needs adjusting a the effective lumped mass of a slender beam of length $\ell$, mass $m_{rod}$ when impacted a distance $c$ from the center of mass is: $$m = \frac{\ell^2}{\ell^2+12 c^2} m_{rod}$$

John Alexiou
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