Tennis Kinematics Transient Analysis: A Ball Spin & Racket Collision Description by Miltiadis A. Boboulos - HTML preview

∫

t2

 

_t1F_avdt (Eq. 38)

 

where S is the force impulse for the time t and t = t₂ - t₁

 

9.3 Calculating the travelling speeds of the two colliding solid bodies: racket and ball

U₂ = U₁ (m-kM) /(M+m) + M (1+k)V₁/(M+m) (Eq. 39) V₂ = U₁ (1+k)m/(M+m) + (M - km)V₁/(M+m) (Eq. 40) K = (V₂ – U₂)/(U₁ – V₁ ) or (Eq. 41)

K=^h2 when a body falls down from a certain height _h1

where: U₂ is the ball speed after the strike,
V₁- racket speed before strike;
V -racket speed after strike;

h – initial height ₁
h₂ - bouncing height
Or, from (11) above:
V₂= 0.060(1+0.35)58.3/0.320+0.06 + (032-0.35.0.06).0/0.32+0.06 V₂=12.43 m/sec for K=0.35; m=0.06; M=0.32
This means that after the collision the racket will travel in the direction of the ball

flight at a speed of 12.43 m/sec provided that the player has only received the ball without blowing at it [17].

In this c ase after meeting the static (and not clamped) racket the speed of the ball in the backward direction will be:
From (Eq. 39)
U₂ = (0.060-0.35.0.32)58.3/0.32+0.06 + 0.(1.35).0.32/0.32+0.06 =
U₂ = -7.98 m/sec
If the player has counter blown at a speed of
V₁ = 10 m/sec, then U₂= +19.35 m/sec
If we take into account the effort in the handle and the approximate mass of the human hand m₃ ≈ 5 kg, then for a counter blow at a speed of V₁ and from (Eq. 39) we will get the following results:
_U (0.06−0.75 . 5.32)58.3+5.32(1+0.75)_{(Eq. 39a)2 =58.3 = 5.32+0.06}

for

 

m = 0.06 kg; M = 5.32 kg U₁= - 58.3 m/sec; U₂= 58.3 m/sec; K = 0.75; V₁ = 9.09 m/sec.

B ut in order to design the test rig we shall need to know what will the speed of the ball be after the ricochet in the clamped racket in close to the human hand effort conditions. In other words M = 5.32 kg, U₁= - 58.3 m/sec and V₁ = 0, then U₂= 229/5.38 = 42.56 m/sec.

The res toration (damping) coefficient K indicates the bouncing characteristics of the ball when hitting a solid surface at low speed and is ≈ 0.35 but for
U1max≥ 58.3 m/sec it can be assumed that K = 0.75 for the string area [19].
To determine the collision vector S we applied the impulse theorem only for the first body; then the internal collision for the entire system will be external collision for the first body only:
S_x = m(U_2x - U_1x ) (Eq. 42)
and hence, from formulae (10) and (11)
_x S = - (1+k) mM(U_2x - V_1x ) / (m+M) (Eq. 43)
For k = 0.75; m = 0.06; M = 5.32; U_1x =U₁ = 58.3 and V₁ = 0
S_x = - 1.75 . 0.06.5.32/0.06+5.32 (58.3 – 0) = - 6.053 N.sec
For k = 0.75; m = 0.06; M = 5.32; U_{1x 1} =U = 58.3 and V₁ = -9.09 m/sec, which is the heaviest load on the racket:

Sx max= - 1.75 . 0.06.5.32/0.06+5.32 [58.3 – (-9.09)] = - 6.997 ≈ -7 N.sec t2