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The Bounce of the Superball

The Bounce of the Superball. John D Barrow. Putting The Shot – Two Surprises. World record 23.12 metres. Max range isn’t achieved with 45 degree launch angle. Range depends on V 2. Launching from above ground level. h  2 m. h = 2m, g = 9.8m/s 2 , v = 14m/s

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The Bounce of the Superball

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  1. The Bounce of the Superball John D Barrow

  2. Putting The Shot – Two Surprises World record 23.12 metres

  3. Max range isn’t achieved with45 degree launch angle Range depends on V2

  4. Launching from above ground level h  2 m h = 2m, g = 9.8m/s2, v = 14m/s reduce g or increase h or v by 1% increases range by (20, 2, 40) cm

  5. World record 23.12m 21.3 metres (= 70 ft) optimal angle is 42 deg 15.24 metres (=50 ft) ------------------- 41 deg 10.7 metres (= 35 ft) -------------------- 39 deg Rmax = h tan(2max)

  6. The Second Surprise Top class shot putters use a launch angle of about 37 deg – not 42-43 deg Because… They can’t achieve the same launch speed at all launch angles

  7. A Constrained Optimisation Launch speed falls as angle increases Typically 34-38 degrees is best But is athlete dependent

  8. The World Goes Round

  9. Gravity • Range depends on V2/g • g varies with latitude because of the non-spherical shape and rotation of the Earth • Net g measured with a spring balance is bigger at the Poles than at the equator Mgnet = Mg - Mr2 Mg(Equator) < Mg(Poles) 200Kg in Mexico City weighs the same as 200.8 Kg in Helsinki 2m HJ in Helsinki is 2.05 in Mexico city. An 8m LJ is 8.20m

  10. Air Resistance is a Drag – But Important

  11. Chucking Things More Realistically Drag on sphere = ½ air A Cd v2 Cd 0.3 Launch speed = 45 metres per sec Range in air = 98.5 m Range in vacuum = 177.1 m Max height in air = 53.0 m Max height in vacuum = 76.8 m

  12. Projectiles with Air Resistance Left: solid trajectory for small resistance (prop to v = 10m/s ) with 45 deg launch; dotted has launch slightly greater than 45 deg and gives longer range. Right: large initial speed v = 300 m/s for 10 deg (solid) 20 deg (dashed) and 30 deg (dotted) angle of launch. Fall is steeper than the rise. Not a parabola now.

  13. Dimples Can Give You A Lift Lift arises from back spin on ball. It gives greater relative velocity between the ball and the air at the top than the bottom. So there is lower pressure and an upward force on the ball. The flow at the top can exceed the speed needed for turbulence in the surface (‘boundary’) layer while the flow at the bottom stays below it. A ball with top spin is pushed downwards ie ‘negative lift’. Dimpling decreases drag and increases lift by inducing turbulence in the boundary layer and delaying separation of the flow from the ball

  14. Peter Tait’s (1890-3) solution for shallow launch angles (sin  ) x = K ln[1+ At] y = [ K – C]ln[1 + At] + Dt – ¼ gt2

  15. Golf-Ball Crystallography Two dimple patterns with icosahedral symmetry

  16. Catching a Moving Ball Hit straight at a fielder No air resistance Move so as to maintain a constant rate of increase of the tangent of the angle of elevation d(tan) /dt = constant! Strategy fails when air resistance is included

  17. Impacts • MV + mv = MU + mu • u – U = e(V –v) • If m is stationary before impact v = 0 • u = MV(1+e)/(M+m) • U = V(M-em)/(M+m) • Golf ball e = 0.7, m = 0.046 Kg, M = 0.2 Kg • V(clubhead) = 50 m/s gives u = 34 m/s M m Speed of m v  u Speed of M V  U

  18. Optimal Clubhead-to-Ball Mass Ratio • Expts: V = C/M1/n , n  5.3, C constant • u = MV(1+e)/(M+m) = CM1-1/n/(M+m) • What is the M/m value that gives max u • du/dM = 0: (M+m)(n-1)=nM • M/m = n-1 for max u • m = 0.046 Kg and n = 5.3  M = 0.20 Kg • Which is about right! Energy Efficiency = ball KE/clubhead KE = 43%

  19. The Centre of Percussion

  20. Painless Batting • The h = r + I/Mr condition for a thin uniform rod of length 2r with • I = 1/3 Mr2 • h = 4r/3 = 2/3  (2r) Hitting the ball 2/3rds of the way down the bat creates no reaction at the pivotal point on the handle h

  21. Cushioning the Blow Hit thru centre Slides without rolling Hit above centre Slides and rotates Where is the speed of sliding to the right equal to the rotational speed to the left? No slip at base contact point. Then it rebounds without sliding. V = Ft/M = slide speed = linear rotational velocity = F r t (h-r) /I h = r + I/MrwhereI = 2Mr2/5 for a sphere h = 0.7  2r = 0.7  ball’s diameter  3.5 cm Rules: cushion height 0.635 + 0.10  ball’s diameter to reduce downward wear on the table near the cushion gutter (David Alciatore)

  22. Bouncing Balls With spin No spin

  23. The Superball • Invented by Norman Stingley in 1965 who called it the ‘Highly Resilient Polybutadiene Ball’ (patent 3241834) • Manufactured by Wham-O • very high e > 0.7 Will bounce over a 3-storey building if thrown hard. • Rough surface, reverses direction of spin at each bounce • Drop 2 one above the other and the top one flies 9 times higher • Lamar Hunt, founder of the American Football League invented the term Super Bowl for the final match after watching his children play with a Super Ball

  24. The Bounce of the Superball Equate total energy of motion = ½MV2 + ½I2 angular momentum about contact pt = I - MRVbefore and after bounce V(out) = -eV(in) in horizontal and vertical directions No slip at contact point –- a perfectly rough ball Normal component of velocity is reversed at collision point

  25.  = 0  = 0 • = +2 rev/s topspin • = +5 rev/s topspin 20o low-speed impact • = -10 backspin • = +10 topspin • = +0.5 topspin • = -10 backspin Tennis ball Superball  is the angular velocity (spin)

  26. Superball Snooker is Different Path of a smooth ball Inside a square box Path of a rough Superball inside a square box

  27. Happy Christmas!

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