Circular motion slideshare
description
Transcript of Circular motion slideshare
MechanicsMechanics
Part 1 – Circular MotionPart 1 – Circular Motion
O
P
r
T
Circular Motion in a horizontal planeCircular Motion in a horizontal plane P moves around a circle of
radius r.
Circular Motion in a horizontal planeCircular Motion in a horizontal plane P moves around a circle of
radius r. As P moves both the arc
length PT change and the angle changes
O
P
r
T
Circular Motion in a horizontal planeCircular Motion in a horizontal plane P moves around a circle of
radius r. As P moves both the arc
length PT change and the angle changes
The angular velocity of P is given by
O
P
r
T
Circular Motion in a horizontal planeCircular Motion in a horizontal plane P moves around a circle of
radius r. As P moves both the arc length
PT change and the angle changes
The angular velocity of P is given by
Force = mass acceleration
ddt
O
P
r
T
Circular Motion in a horizontal planeCircular Motion in a horizontal plane P moves around a circle of
radius r. As P moves both the arc length
PT change and the angle changes
The angular velocity of P is given by
Force = mass acceleration
ddt
O
P
r
T
ddt
O
Pr
T
Let PT x
ddt
O
Pr
T
Let PT x
x r
ddt
O
Pr
T
Let PT x
x r
dx dr
dt dt
ddt
O
Pr
T
Let PT x
x r
dx dr
dt dt
v r
This equation is important since it links angular and linear velocity
ddt
O
Pr
T
Let PT x
x r
dx dr
dt dt
v r
This equation is important since it links angular and linear velocity
ddt
O
Pr
T
Let PT x
x r
dx dr
dt dt
v r
To discuss acceleration we should consider the motion in terms of horizontal and vertical components.
O
P
r
T
v r
To discuss acceleration we should consider the motion in terms of horizontal and vertical components.
O
P( x,y)
r
T
v r
cos , sinx r y r
To discuss acceleration we should consider the motion in terms of horizontal and vertical components.
O
P( x,y)
r
T
v r
cos , sin
cos , sin
x r y r
dx d dy dr r
dt dt dt dt
To discuss acceleration we should consider the motion in terms of horizontal and vertical components.
O
P( x,y)
r
T
v r
cos , sin
cos , sin
cos , sin
x r y r
dx d dy dr r
dt dt dt dtd d d d
x r y rd dt d dt
To discuss acceleration we should consider the motion in terms of horizontal and vertical components.
O
P( x,y)
r
T
v r
cos , sin
cos , sin
cos , sin
sin , cos
x r y r
dx d dy dr r
dt dt dt dtd d d d
x r y rd dt d dt
x r y r
To discuss acceleration we should consider the motion in terms of horizontal and vertical components.
O
P( x,y)
r
T
v r
To simplify life we are going to consider that the angular velocity remains constant throughout the motion.
O
P( x,y)
r
T
v r
sin , cosx r y r
To simplify life we are going to consider that the angular velocity remains constant throughout the motion.This will be the case in any problem you do , but you should be able to prove these results for variable angular velocity.
O
P( x,y)
r
T
v r
sin , cosx r y r
v r
sin , cosx r y r
v r
v r
v r
v r
v r
v r
2 cosr
2 sinr
Forces DiagramForces Diagram
2 cosr
2 sinr
a
It is important to note that the vectors demonstratethat the force is directed along the radius towardsthe centre of the circle.
2 cosr
2 sinr
a
2 22 2 2cos sina r r
2 cosr
2 sinr
a
2 22 2 2
2 4 2 2 4 2
2 4 2 2
cos sin
cos sin
cos sin
a r r
a r r
a r
2 cosr
2 sinr
a
2 22 2 2
2 4 2 2 4 2
2 4 2 2
2 4
cos sin
cos sin
cos sin
a r r
a r r
a r
a r
2 cosr
2 sinr a
2 22 2 2
2 4 2 2 4 2
2 4 2 2
2 4
2
cos sin
cos sin
cos sin
a r r
a r r
a r
a r
a r
2
2
2
butv
a r v rr
va r
r
va
r
Summary Summary
v r
2 cosr
2 sinr
Forces DiagramForces Diagram
2a r
force is directed along the force is directed along the radius towards the centre radius towards the centre of the circle.of the circle.
(angular velocity)(angular velocity)
(links angular velocity and (links angular velocity and linear velocity)linear velocity)
(horizontal and vertical (horizontal and vertical components of acceleration)components of acceleration)
(gives the size of the force (gives the size of the force towards the centre)towards the centre)
Summary Summary
v r
2 cosr
2 sinr
Forces DiagramForces Diagram
2a r
force is directed along the force is directed along the radius towards the centre radius towards the centre of the circle.of the circle.
(angular velocity)(angular velocity)
(links angular velocity and (links angular velocity and linear velocity)linear velocity)
(horizontal and vertical (horizontal and vertical components of acceleration)components of acceleration)
(gives the size of the force (gives the size of the force towards the centre)towards the centre)
2va
r
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
Draw a neat diagram to represent the forces.Draw a neat diagram to represent the forces.
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.NN
mgmg
PPTT
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.NN
mgmg
PPTT
The tension in the string is the The tension in the string is the resultant of the forces acting on the resultant of the forces acting on the body.body.
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.NN
mgmg
PPTT
The tension in the string is the The tension in the string is the resultant of the forces acting on the resultant of the forces acting on the body.body.
T F ma
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.NN
mgmg
PPTT
The tension in the string is the The tension in the string is the resultant of the forces acting on the resultant of the forces acting on the body.body.
2
T F ma
T mr
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.NN
mgmg
PPTT
The tension in the string is the The tension in the string is the resultant of the forces acting on the resultant of the forces acting on the body.body.
2
2
2
2 3 2
24
T F ma
T mr
N
1.1. A body of mass 2kg is revolving at the end of a A body of mass 2kg is revolving at the end of a light string 3m long, on a smooth horizontal table light string 3m long, on a smooth horizontal table with uniform angular speed of 1 revolution per with uniform angular speed of 1 revolution per second.second.
(a)(a) Find the tension in the string.Find the tension in the string.NN
mgmg
PPTT
The tension in the string is the The tension in the string is the resultant of the forces acting on the resultant of the forces acting on the body.body.
2
2
2
2 3 2
24
T F ma
T mr
N
(b)(b) If the string would break under a tension of equal If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible to the weight of 20 kg, find the greatest possible speed of the mass.speed of the mass.
(b)(b) If the string would break under a tension of equal If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible to the weight of 20 kg, find the greatest possible speed of the mass.speed of the mass.
(b)(b) If the string would break under a tension of equal If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible to the weight of 20 kg, find the greatest possible speed of the mass.speed of the mass.
2
20T g
mvT
r
(b)(b) If the string would break under a tension of equal If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible to the weight of 20 kg, find the greatest possible speed of the mass.speed of the mass.
2
2
2
20
20
20
T g
mvT
r
mvg
rr
v gm
(b)(b) If the string would break under a tension of equal If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible to the weight of 20 kg, find the greatest possible speed of the mass.speed of the mass.
2
2
2
20
20
20
T g
mvT
r
mvg
rr
v gm
2 320
2
30 /
v g
v g m s
(b)(b) If the string would break under a tension of equal If the string would break under a tension of equal to the weight of 20 kg, find the greatest possible to the weight of 20 kg, find the greatest possible speed of the mass.speed of the mass.
2
2
2
20
20
20
T g
mvT
r
mvg
rr
v gm
2 320
2
30 /
v g
v g m s
Conical PendulumConical PendulumIf a particle is tied by a string to a fixed point by means of a string and moves in a horizontal circle so that the string describes a cone, and the mass at the end of the string describes a horizontal circle, then the string and the mass describe a conical pendulum.
Conical PendulumConical PendulumIf a particle is tied by a string to a fixed point by means of a string and moves in a horizontal circle so that the string describes a cone, and the mass at the end of the string describes a horizontal circle, then the string and the mass describe a conical pendulum.
Conical PendulumConical Pendulum
mR
θθL
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagram
mR
θθL
Forces DiagramForces Diagram
Vertically Vertically
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
Vertically Vertically
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
Vertically Vertically
NN
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
Vertically Vertically
TT
N + (-mg)N + (-mg) = 0 = 0 NN = = mgmg
but cos θ = but cos θ = NN//TT
so so N = T N = T cos θcos θT T cos θcos θmgmg
NN
θθ
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
Vertically Vertically
TT
N + (-mg)N + (-mg) = 0 = 0 NN = = mgmg
but cos θ = but cos θ = NN//TT
so so N = T N = T cos θcos θT T cos θcos θmgmg
NN
θθ
HorizontallyHorizontally
T T sin θsin θmrmr
Since the onlySince the onlyhorizontal force horizontal force is directed along is directed along the radius towardsthe radius towards the centrethe centre
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
Vertically Vertically
TT
N + (-mg)N + (-mg) = 0 = 0 NN = = mgmg
but cos θ = but cos θ = NN//TT
so so N = T N = T cos θcos θT T cos θcos θmgmg
NN
θθ
HorizontallyHorizontally
T T sin θsin θmrmr
Since the onlySince the onlyhorizontal force horizontal force is directed along is directed along the radius towardsthe radius towards the centrethe centre
T T sin sin θθmrmr
T T cos θcos θmgmg
Conical PendulumConical Pendulum
mR
θθL
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
Vertically Vertically
TT
N + (-mg)N + (-mg) = 0 = 0 NN = = mgmg
but cos θ = but cos θ = NN//TT
so so N = T N = T cos θcos θT T cos θcos θmgmg
NN
θθ
HorizontallyHorizontally
T T sin θsin θmrmr
Since the onlySince the onlyhorizontal force horizontal force is directed along is directed along the radius towardsthe radius towards the centrethe centre
T T sin sin θθmrmr
T T cos θcos θmgmg
An important resultAn important result
mR
θθ LNN
mgmg
θθ
Forces DiagramForces Diagram
TTθθ
T T sin θsin θmrmr…………11T T cos θcos θmg … … mg … … 22 vvrr
h
Now dividing equation 1 by equation 2..
An important resultAn important result
mR
θθ LNN
mgmg
θθ
Forces DiagramForces Diagram
TTθθ
T T sin θsin θmrmr…………11T T cos θcos θmg … … mg … … 22 vvrr
h
2Tsincos
mrT mg
Now dividing equation 1 by equation 2..
An important resultAn important result
mR
θθ LNN
mgmg
θθ
Forces DiagramForces Diagram
TTθθ
T T sin θsin θmrmr…………11T T cos θcos θmg … … mg … … 22 vvrr
h2
2
Tsincos
tan
mrT mg
rg
Now dividing equation 1 by equation 2..
An important resultAn important result
mR
θθ LNN
mgmg
θθ
Forces DiagramForces Diagram
TTθθ
T T sin θsin θmrmr…………11T T cos θcos θmg … … mg … … 22 vvrr
h2
2
Tsincos
tan
tan
mrT mg
rg
rh
Now dividing equation 1 by equation 2..
An important resultAn important result
mr
θθ LNN
mgmg
θθ
Forces DiagramForces Diagram
TTθθ
T T sin θ sin θ mrmr…………11T T cos θcos θmg … … mg … … 22 vvrr
h
2
2
2
2
Tsincos
tan
tan
mrT mg
rg
rh
r rg h
gh
Now dividing equation 1 by equation 2..
An important resultAn important result
mr
θθ LNN
mgmg
θθ
Forces DiagramForces Diagram
TTθθ
T T sin θsin θmrmr…………11T T cos θcos θmg … … mg … … 22 vvrr
h
2
2
2
2
Tsincos
tan
tan
mrT mg
rg
rh
r rg h
gh
Now dividing equation 1 by equation 2..
Example 2Example 2
A string of length 2 m, fixed at one end A string of length 2 m, fixed at one end A A carries at the carries at the other end a particle of mass 6 kg rotating in a horizontal other end a particle of mass 6 kg rotating in a horizontal circle whose centre is 1m vertically below A. Find the circle whose centre is 1m vertically below A. Find the tensiontension in the string and the in the string and the angular velocityangular velocity of the of the particle.particle.
Example 2Example 2
A string of length 2 m, fixed at one end A string of length 2 m, fixed at one end A A carries at the carries at the other end a particle of mass 6 kg rotating in a horizontal other end a particle of mass 6 kg rotating in a horizontal circle whose centre is 1m vertically below A. Find the circle whose centre is 1m vertically below A. Find the tensiontension in the string and the in the string and the angular velocityangular velocity of the of the particle.particle.
Example 2Example 2
A string of length 2 m, fixed at one end A string of length 2 m, fixed at one end A A carries at the carries at the other end a particle of mass 6 kg rotating in a horizontal other end a particle of mass 6 kg rotating in a horizontal circle whose centre is 1m vertically below A. Find the circle whose centre is 1m vertically below A. Find the tensiontension in the string and the in the string and the angular velocityangular velocity of the of the particle.particle.
r
θθL = 2
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
TT NN
θθ
h = 1
Example 2Example 2
A string of length 2 m, fixed at one end A string of length 2 m, fixed at one end A A carries at the carries at the other end a particle of mass 6 kg rotating in a horizontal other end a particle of mass 6 kg rotating in a horizontal circle whose centre is 1m vertically below A. Find the circle whose centre is 1m vertically below A. Find the tensiontension in the string and the in the string and the angular velocityangular velocity of the of the particle.particle.
r
θθL = 2
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
TT T T coscos θθ
h = 1
Example 2Example 2
Find the Find the tensiontension in the string and the in the string and the angular velocityangular velocity of of the particle.the particle.
r
θθL = 2
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
TT NNθθ
h = 1
VerticallyVertically HorizontallyHorizontallycos
cos612
12
T mg
mgT
gT
T g
Example 2Example 2
Find the Find the tensiontension in the string and the in the string and the angular velocityangular velocity of of the particle.the particle.
r
θθL = 2
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
TT NNθθ
h = 1
VerticallyVertically HorizontallyHorizontally2
2
2
sin
sin sin
12
12
T mr
T mL
T
mL
gg
cos
cos612
12
T mg
mgT
gT
T g
Example 2Example 2
Find the Find the tensiontension in the string and the in the string and the angular velocityangular velocity of of the particle.the particle.
r
θθL = 2
Dimensions DiagramDimensions Diagrammgmg
θθ
Forces DiagramForces Diagram
TT NNθθ
h = 1
VerticallyVertically HorizontallyHorizontally2
2
2
sin
sin sin
12
12
T mr
T mL
T
mL
gg
cos
cos612
12
T mg
mgT
gT
T g