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