The Bounce of the Superball John D Barrow

Putting The Shot – Two Surprises World record 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 reduce g or increase h or v by 1% increases range by (20, 2, 40) cm

R max = h tan(2  max ) 21.3 metres (= 70 ft) optimal angle is 42 deg metres (=50 ft) deg 10.7 metres (= 35 ft) deg World record 23.12m

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

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

The World Goes Round

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

Air Resistance is a Drag – But Important

Launch speed = 45 metres per sec Range in air = 98.5 mRange in vacuum = m Max height in air = 53.0 mMax height in vacuum = 76.8 m Drag on sphere = ½  air A C d v 2 C d  0.3 Chucking Things More Realistically

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. Projectiles with Air Resistance

Dimples Can Give You A Lift Dimpling decreases drag and increases lift by inducing turbulence in the boundary layer and delaying separation of the flow from the ball 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’.

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

Golf-Ball Crystallography Two dimple patterns with icosahedral symmetry

Catching a Moving Ball 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 No air resistance Hit straight at a fielder

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

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

The Centre of Percussion

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

Hit thru centreSlides 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/MF r t (h-r) /I V = Ft/M = slide speed = linear rotational velocity = F r t (h-r) /I h = r + I/MrI = 2Mr 2 /5 h = r + I/Mr where I = 2Mr 2 /5 for a sphere h = 0.7  2r = 0.7  ball’s diameter  3.5 cm Rules Rules: cushion height  ball’s diameter to reduce downward wear on the table near the cushion gutter (David Alciatore) Cushioning the Blow

Bouncing Balls No spin With spin

The Superball Invented by Norman Stingley in 1965 who called it the ‘Highly Resilient Polybutadiene Ball’ (patent ) Invented by Norman Stingley in 1965 who called it the ‘Highly Resilient Polybutadiene Ball’ (patent ) Manufactured by Wham-O Manufactured by Wham-O very high e > 0.7 Will bounce over a 3- storey building if thrown hard. very high e > 0.7 Will bounce over a 3- storey building if thrown hard. Rough surface, reverses direction of spin at each bounce Rough surface, reverses direction of spin at each bounce Drop 2 one above the other and the top one flies 9 times higher 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 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) = -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

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

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

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