Transcript Maths Of Sports - Junior University :: Homepage

Maths Of Sports
The Movement of a Ball
over a plane Surface
Impact times
The Movement of a Ball
over a plane Surface
Central blow over a horizontal surface
Consider a ball of mass M and radius a.
Let coefficient of Sliding friction be µ
Then friction on Ball – F= µMg (1)
This force will have 2 consequences
(i) Linear deceleration of ball
(ii) moment of force about C (centre) will
produce a clockwise angular acceleration
Continued ……
By Newton (ii), linear deceleration =
F/M = µg (2)
And Angular acceleration = Torque/M of I
= Fa/I = µmg/I (3)
The M of I (moment of inertia) of a sphere
about an axis through its centre is
I = (2/5)Ma
Therefore substitute into (3), Angular
acceleration = 5µgt/2a (5)
Continued ……
After a time, the angular velocity will have
decreased according to (2) therefore
v = V- µgt (6)
At the same time, the angular velocity will
have reached a value w so that w = 5µgt/2a
(7)
The ball will continue to slide until the
linear velocity has been reduced and the
angular velocity has increased to such an
extent that rolling occurs.
Continued ……
Ball rolls at v = wa
Therefore from (6) and (7), rolling occurs after a
time t such that V-µgt=5µgt/2a
ie (7/2) µgt = V or t = 2v/7µg (9)
At this time, the linear velocity is given by (6) as
v=V- 2v/7µg = 5V/7 (10) and corresponding
angular velocity w=5V/7a (11).
Note that both (10) and (11) are indpendent of µ, so
therefore once the linear velocity has fallen tp
5/7 of its initial value the sliding phase will be
completed and the rolling phase will begin, the ball
then evnetually gets brought to an end by rolling
friction which is very much smaller than the
friction in the sliding phase.
Continued……
• Since s = ut + 0.5at² then distance
travelled by ball during sliding phase
is given by
s = V(2V/7µg) – 0.5(µg)(2V/7µg)²
= 12V²/49µg (13)
Application to billiard balls
The sliding coefficient of an average billiard
table is about 0.2, velocity of a strongly cued
ball ≈ 3ms therefore duration of sliding phase
is = t = 2(3)/(49)(0.2)(9.8) = 0.4 secs
Using (13) distance travelled = 1.1m
Impact Times
• In most games, experiments indicate that
the linear compression x of a golf ball is
related to the applied force F by a law of
the form of F=Kx 1.5
• This leads to a relation between impact
times and velocity, As velocity increases,
the duration of impact decrease,
approaching a constant value at higher
velocities.
Continued ……
Forces at impact
in a golf drive, velocity of ball of tee ≈70ms ,
time of impact = 0.0005secs, mass of ball
= 0.046kg.
By Newton (ii) (average force)
= 0.046 * 70 / 0.0005 = 6300N
This is the average force at impact since it
must be zero at the start and finish of
impact, it is reasonable to suppose that
MAX force occurs at midway throughout
the impact 6300 * 2 = 12600N
Contunued……
Energy loss and efficieny
in golf, some of the energy loss will be dissipated as
heat, energy converted into heat = 0.5MV² 0.5MV´² - 0.5MV´²
= mMV²(1-e²)/2(m+M) = (0.047)(0.20)(50²)(10.49)/2(0.246) = 23 joules
Assuming a specific heat capacity of rubbe about
1700joulekg^-1k^-1, this suggests a rise in
tempreture of ball during impact ≈ 0.3°K
Continued……
Now after impact energy possessed by
by ball = E =0.5MV´² =
M/2(M²V²/(M+m)²)
While the energy in the club head was
0.5MV²
Therefore efficiency of transfer of energy
= Mm(1+e)²/(M+m)²(equation 9) =
(0.20)(0.046)(1.7)²/(0.246)²= 0.4
Hence 43% of the energy goes into the
ball.
Continued……
If we study equation 9 we can observe
how the efficiency Varies with M the
mass of the clubhead, in relation to m
(mass of ball) Assuming that e = 0.7
Maximum efficiency = 72% when M=m
Therefore maximum transfer of
energy when M/m = 1
So if we make the mass of the
clubhead equal to the mass of the
ball we have a maximum transfer of
energy.
Continued……
• Since V´ = V(1+e)/1+(m/M)
• As M increases from M to ∞, V´ has a
denominator decreasing from 2 to 1 . This
doubles the velocity of the ball, despite
inefficient energy transfer. However a
player wont be able to swing a heavy club
as fast as he can a light one so that V is
bound to decrease as M rises The ideal
mass of the clubhead depends on how V
varies with M.
The End