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 = 2 m, g = 9. 8 m/s 2, v = 14 m/s reduce g or increase h or v by 1% increases range by (20, 2, 40) cm

World record 23. 12 m 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)

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

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

The World Goes Round

Gravity n n n Range depends on V 2/g g varies with latitude because of the nonspherical 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) 200 Kg in Mexico City weighs the same as 200. 8 Kg in Helsinki 2 m HJ in Helsinki is 2. 05 in Mexico city. An 8 m LJ is 8. 20 m

Air Resistance is a Drag – But Important

Chucking Things More Realistically Drag on sphere = ½ air A Cd v 2 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

Projectiles with Air Resistance Left: solid trajectory for small resistance (prop to v = 10 m/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.

Dimples Can Give You A 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 Dimpling decreases drag and bottom stays below it. increases lift by inducing turbulence A ball with top spin is in the boundary layer and delaying pushed downwards ie ‘negative lift’. separation of the flow from the ball

Peter Tait’s (1890 -3) solution for shallow launch angles (sin ) x = K ln[1+ At] y = [ K – C]ln[1 + At] + Dt – ¼ gt 2

Golf-Ball Crystallography Two dimple patterns with icosahedral symmetry

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

Impacts n n n n 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 Speed of M V U m Speed of m v u

Optimal Clubhead-to-Ball Mass Ratio Expts: V = C/M 1/n , n 5. 3, C constant n u = MV(1+e)/(M+m) = CM 1 -1/n/(M+m) n What is the M/m value that gives max u n du/d. M = 0: (M+m)(n-1)=n. M n M/m = n-1 for max u n m = 0. 046 Kg and n = 5. 3 M = 0. 20 Kg n Which is about right! n Energy Efficiency = ball KE/clubhead KE = 43%

The Centre of Percussion

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

Cushioning the Blow Hit thru centre Hit above centre Slides without rolling 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/Mr where I = 2 Mr 2/5 for a sphere h = 0. 7 2 r = 0. 7 ball’s diameter 3. 5 cm Rules: Rules cushion height 0. 635 + 0. 10 ball’s diameter to reduce downward wear on the table near the cushion gutter (David Alciatore)

Bouncing Balls No spin With spin

The Superball n n n 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

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

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

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

Happy Christmas!