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Throwing, Jumping and Spinning

Throwing, Jumping and Spinning. John D Barrow. Coin Tossing Isn’t Random. Launch upwards with a vertical speed V from a height H 0 then H = H 0 +Vt – ½ gt 2 after a time t, g = 9.8 m/s 2 is acceleration due to gravity. It returns to the thrower’s hand at the same height, H = H 0 ,

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Throwing, Jumping and Spinning

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  1. Throwing, Jumping and Spinning John D Barrow

  2. Coin Tossing Isn’t Random Launch upwards with a vertical speed V from a height H0 then H = H0 +Vt – ½ gt2 after a time t, g = 9.8 m/s2 is acceleration due to gravity. It returns to the thrower’s hand at the same height, H = H0, after a time th = 2V/g. If it was tossed upwards with a spin of R revolutions per sec then it will have turned over completely N times where N = th × R = 2VR/g  0.4s  R if V  2 m/s, g = 9.8 m/s2 Need R > 50/s for N > 20

  3. More Revolutions Means More Unpredictability • N = 1 and the coin was launched very slowly, with heads upwards then it will fall with heads up again • N = 2 to 3, or 4 to 5, or 6 to 7, etc, it will be caught with the same face up as it was thrown with. • N = 3 to 4, or 5 to 6, or 7 to 8, etc, it will land with the opposite face upwards • N >> 20 is large: the conditions on V and R that distinguish the two outcomes get closer and closer together and very small differences in the tossing conditions result in heads or tails • In general, if n = 1,2,3,4,….etc then if N lies between 2n and 2n+1 the coin will land with the initial face upwards and if n lies between 2n+1 and 2n+2 the initial face will land downwards

  4. Biased Coins Don’t Help They always spin about an axis that passes through the centre of mass even when centre of mass  geometrical centre

  5. Skating Spins Mr2 = constant   1/r2 r • Skaters can reduce their inertia ( Mr2) • by a factor of two and this will double the angular speed of a spin up to about • = 20 radians/ s, or 3 revs/ s.

  6. Wheeler Dealing

  7. Wheels On Fire Total energy needed to move the bike frame + two wheels at speed v and rotate the two wheels at =v/ris Total energy = ½ (2m + mframe)v2 + 2  ½ I2 Since I = bmr2 with b  1 (ring)  ½ (disc) Total energy = ½v2{mframe + 2(1+b)m} Moment of inertia The same mass reduction on each wheel is three or four times as beneficial as the same mass reduction on the frame.

  8. The Grand Jeté The centre of mass always follows a parabolic trajectory after take-off There is nothing the dancer can do to alter that. So is the appearance of floating a fiction?

  9. You can change your body shape so that the trajectory of your head does not follow the same trajectory as your centre of mass Throw a mallet or a shuttlecock or a caber

  10. Change Your Mass Distribution In the jump phase, raise legs to horizontal + put arms above shoulders. Raises CoM relative to head. Then lower legs and arms and CoM falls in descent Relative to head, her CoM is shifting upwards It follows parabola but the head stays horizontal

  11. Height of the dancer’s head above the floor Does not follow a parabolic path

  12. Basketball players’ illusion of ‘hanging in the air’

  13. The Cat Paradox

  14. 1. Bend in the middle: the front and back halves rotate about a different axes. 2. Tuck in front legs: reduce moment of inertia of the front of body and extend rear legs to increase the inertia of the rear half, so can rotate front half as much as 90° while rear half rotates in the opposite direction much less -- as little as 10°. 3. Extend front legs and tuck in rear legs: rotate rear half a lot while the front half rotates much less in the opposite direction. 4. Depending on its flexibility and initial angular momentum, the cat may need to repeat steps two and three times to complete a full 180° rotation. ‘I = constant’

  15. Another Irrotational Lander Aim: Enter the water vertical with no rotation Increase your inertia so as to spin slower

  16. Diving – how much time have you got ? High board h = 10m h = ½ gt2 t = (20/9.8) s = 1.4 s V = gt  14 m/s = 31 mph Three and a half somersaults in the air means 3.5/1.4 = 2.5 rev/s This is 150 revolutions per minute (rpm) The tracking speed at the outer edge of a disc on your CD player is 200 rpm. Each revolution is 2 radians so the angular velocity is 2 2.5 = 5 = 15.7 radians/s

  17. Springboard Divers Have More Time t  1.8s in the air 5 deg to vertical This is vital Or…

  18. Springboards 4.9m x 50cm aluminium cantilever Hit the board at its ‘natural’ frequency to create a resonance You can hear itwhen it’s right (or wrong) x Restoring force = - kx = m d2x/dt2 x = 0.75m deflection standing still means kx = mg  k = 659.8/0.75 = 848 N/ m ‘Natural’ frequency= (k/m) ~ 848/65 = 3.6/s for m = 65 kg diver More board out to the right of roller  ‘springier’  large k

  19. Anatomy of A Long Jump Eg: Length of jump = A + B + C  0.4m + 6.45m + 0.9m = 7.75m For V  9 m/s and   20 deg B = (V2/g) cos  [sin + (sin2 + 2gH/V2)1/2] It is up to the jumper where the board is on take-off

  20. Styles that Aid Forward Rotation sail Bent leg at take-off for lower M of inertia, faster free leg rotn Drive opposite arm upwards Doesn’t counter forward rotn at take-off Sideways fallout on landing low rotn Move limbs for better landing posn. Counter forward rotn from Leg thrust behind CoM at take-off hang Delay body’s forward speed Allow time for legs to position Control rotation by long thin shape in the air 1.5 or 2.5 ‘strides’ in the air Hitch kick Ang mom of bent and straight legs balance

  21. Kicking for Time Rather Than Distance R = V2/g  sin(2) sinA =sin(180 – A) So R is same for launch angles  and 90 -  Time in the air =R/(Vcos) t(high)/t(low) = cos()/cos(90-) = cot() Height achieved is H(high)/h(low) = cot2(). eg t( = 75 deg)/t( = 15 deg) = 3.7 H( = 75 deg)/H( = 15 deg) = 14

  22. Javelin Throwing Centre of pressure  Centre of mass  An overturning moment  changes angle of attack lift drag C o P is first behind, then ahead, then behind C o M Javelin is launched in a flatter trajectory of around 35 degrees It hangs in the air for longer, and sails out to a greater distance. javelin is angled to lie at about 10 degrees (the ‘angle of attack’) below the trajectory followed by its centre of mass This ensures that the nose of the javelin is always dropping

  23. The Discus -- Lift can Beat Drag Distance thrown Fdrag C(V-W)2 Flift D(V-W)2 A headwind can help the discus thrower [i] C. Frohlich, Am. J. Phys. 49, 1125 (1981). When the lift exceeds drag it will be larger when throwing into a wind because it is proportional to (V– W)2 and so it will be much larger with a headwind (W < 0 ) than with a following wind (W > 0). A good situation is to throw into a 10 m/s headwind, where you will throw about 4m further than in still air. This is a very strong wind! With a 10 m/s tailwind your throw is reduced by about 2m. A tailwind of 7.5 m/s is the worst case scenario.

  24. Bend It Like A Beckham

  25. No Spin On the Ball Speed  as Pressure 

  26. Spinning Ball Net air speed near the ball is less at top than bottom Pressure on the ball at the top is bigger than at the bottom This creates a downward force

  27. An In-swinging Corner

  28. The Archer’s Paradox

  29. The Oscillations of the Arrow Time since Arrow’s release No part of the arrow should touch the bow as it passes by

  30. The Stiffness (Spine) of the Arrow is Crucial

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