1. Tight Corners
Physics in Sport – Lesson 1

2. Learning intentions
 Explain why circular motion even at uniform speed is accelerated motion
 Use vector subtraction to show this
 Use formulas to solve problems

4. STARTER – give some examples of motion in a circle
 In which do you think the speed is uniform?

5. Measuring angles and arcs
 Circumference of a circle
s  2r
Where does this come from?
• Measuring angles in radians
• One radian is the angle
Angle in radians = length of arc
radius
s
 
r
s
2r
  
 2
r
r
• 360 degrees = 2π radians

6. Motion in a circle

If an object is moving in a circle what is it’s
a) Distance travelled
b) Displacement
c) Linear speed
•
2r
v
T
Theoretically what is it’s average velocity?

7. Why is it accelerated motion?
v1
 What is the definition of velocity?
 It is a vector quantity
 What is the definition of acceleration?
v4
v2
Q.
v3
 It is a vector quantity
In your own words explain using the definitions why
circular motion is accelerated motion

8. Why is it accelerated motion?
v1
 To find the change in velocity (Δv) we
need to subtract the velocity vectors.
v2
 v2-v1= v2+(-v1 )=Δv
v2
Δv Δv
-v1

9. Centripetal Acceleration
v1
v2
Δv
 To find the change in velocity any 2
short time intervals is always towards
the centre of the circle
v2
-v1
If the change in velocity is
towards the centre then so
is the acceleration

10. Centripetal Acceleration
 Acceleration is the rate of change of
velocity.
v1
v2
Δv
a= Δv
t
 It towards the centre of the circle.
 It is given by
2
v
a
r

11. Forces cause accelerations
 If there is such a thing as centripetal
acceleration then these must be
caused by forces (that’s what forces
do)
F  ma
2
v
a
r
 There are different centripetal forces
2
v
F m
r

12. Causes of Centripetal Force

13. Causes of Centripetal Force

14. Causes of Centripetal Force

15. Worked Example: Centripetal Force
A stone of mass 0.5 kg is swung round in a horizontal circle (on a frictionless surface) of radius
0.75 m with a steady speed of 4 m s-1.
 Calculate:
(a) the centripetal acceleration of the stone
acceleration = v2/r = 42 / 0.75 = 21.4 m s-2
(b) the centripetal force acting on the stone.
F = ma = mv2/r = [0.5 ´ 42] / 0.75 = 10.7 N
 Notice that this is a linear acceleration and not an angular acceleration. The angular velocity
of the stone is constant and so there is no angular acceleration.