Transcript Circular Motion

Circular Motion
Introduction
• What is Newton’s First Law how does it relate to
circular motion?
• How does Newton’s second law relate to circular
motion?
Acceleration
Vi
𝑑𝑥
𝑟
𝑑𝜃
𝑟
Vf
Vi
𝑑𝜃
Vf
𝑑𝑉
Vi
𝑑𝑥
𝑟
𝑑𝜃
𝑟
Vf
𝑑𝜃
Vf
𝑑𝑥
𝑉=
𝑑𝑡
𝑑𝑥 = 𝑉. 𝑑𝑡
𝑑𝑥
𝑟
𝑉. 𝑑𝑡
𝑑𝜃 =
𝑟
𝑑𝜃 =
𝑑𝑉
𝑑𝜃 =
𝑉
𝑑𝑉
𝑉. 𝑑𝑡
=
𝑉
𝑟
𝑑𝑉
𝑉2
=
𝑑𝑡
𝑟
𝑉2
𝑎=
𝑟
𝑑𝑉
Acceleration
• In uniform circular motion, which direction is the
acceleration?
o There is no component of the net force adding to the speed of the
particle.
o Therefore, the net force must always be perpendicular to the Velocity
Vector.
• The acceleration of a particle in uniform circular
motion is always towards the centre of the circle.
• The particle is always deviating from it’s straight line
path towards the centre of the circle.
𝑎=
𝑣2
𝑟
=
4𝜋2 𝑟
𝑇2
𝐹 = 𝑚𝑎
Exam Question (VCAA 2010)
A racing car of mass 700 kg (including the driver) is
travelling around a corner at a constant speed. The car’s
path forms part of a circle of radius 50 m, and the track is
horizontal. The magnitude of the central force provided
by friction between the tyres and the ground is 11 200 N.
Question 1
What is the speed of the car? (2 marks)
Question 2
What is the acceleration of the car as it goes around the
corner? (2 marks)
Exam Question (VCAA , 2009)
Question 3
Draw an arrow to show the direction of the net force
on the motorcycle.
On the diagram, draw the forces acting on
the car. Remember the car is travelling in a
circular path.
Centre of
circular path
On the diagram, draw the forces acting on
the car. Remember the car is travelling in a
circular path.
Centre of
circular path
FN
FN
Ff
Fg
Ff
On the diagram, draw the forces acting on
the car. Remember the car is travelling in a
circular path.
Centre of
circular path
FN
FN
Ff
Fg
Ff
Since the vertical forces are
balanced, the net force
(which we call centripetal
force) is the sum of the
sideways frictional forces.
Ball on a string
𝜽
Ball on a string
𝜽
Fg
Ball on a string
𝜽
Ft
Fg
Ball on a string
𝜽
Ft
Fg
Ft
𝜽
Fg
Ball on a string
𝜽
Ft
𝜽
Ft
𝑭
Fg
Fg
Ball on a string
𝜽
Ft
Ft
𝛉
Fg
𝑭
Fg
𝐹 = 𝐹𝑔 × 𝑇𝑎𝑛(𝜃)
𝐹𝑔
𝐹𝑡 =
𝐶𝑜𝑠(𝜃)
Example: Ball on a string
A ball of mass 250 g is attached
to string in a game of totem
tennis. The string makes an
angle of 40o to the vertical pole.
Calculate:
a. the net force on the ball
b. the tension in the string
c. the length of the string in
terms of it’s speed, v?
𝟒𝟎𝒐
Banked Corners
Banked Corners
Fg
Banked Corners
FN
𝜽
Fg
Banked Corners
FN
𝜽
Fg
Fg
𝜽
FN
Banked Corners
FN
𝜽
Fg
Fg
𝜽
FN
𝑭
Banked Corners
FN
𝜽
Fg
𝐹 = 𝐹𝑔 × 𝑇𝑎𝑛(𝜃)
𝑚𝑣 2
= 𝑚𝑔 × 𝑇𝑎𝑛(𝜃)
𝑟
Fg
𝜽
FN
𝑭
Banked Corners
FN
𝜽
Fg
𝐹 = 𝐹𝑔 × 𝑇𝑎𝑛(𝜃)
𝑚𝑣 2
= 𝑚𝑔 × 𝑇𝑎𝑛(𝜃)
𝑟
Fg
𝜽
FN
𝑭
Banked Corners
FN
𝜽
Fg
𝐹 = 𝐹𝑔 × 𝑇𝑎𝑛(𝜃)
𝑚𝑣 2
= 𝑚𝑔 × 𝑇𝑎𝑛(𝜃)
𝑟
𝑣=
𝑟𝑔 × 𝑇𝑎𝑛(𝜃)
Fg
𝜽
FN
𝑭
Exam Question: VCAA 2010
Question 4
On the diagram, draw an
arrow to indicate the
direction of the acceleration
of the rider (1mark)
Exam Question: VCAA 2010
Question 5
The circular path of the bicycle
has a constant radius of 120 m,
and the bicycle will be
travelling at a constant 9 m s-1.
What should be the value of
the angle of the bank, θ, so
that the bicycle travels around
the corner with no sideways
frictional force between the
tyres and the track? (3 marks)
Banked Corners
FN
𝜽
Fg
The force diagram doesn’t consider friction.
Challenge: What would the force
diagram look like if we
considered friction? In which
direction would the net force be?
Fg
𝜽
FN
𝑭
Leaning into corners
Leaning into corners
FN
𝒎𝒗𝟐
𝑭 = 𝑭𝒇 =
𝒓
Ff
Fg