x2 t07 04 uniform circular motion (2010)

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Acceleration with Uniform Circular Motion

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity. constantbealsoocity willlinear veltheofmagnitude the

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

constantbealsoocity willlinear veltheofmagnitude the

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

constantbealsoocity willlinear veltheofmagnitude the

A particle moves from A

to P with constant angular velocity.

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

constantbealsoocity willlinear veltheofmagnitude the

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

v’

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

v’

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

v

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

v’

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

vv’

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

v’

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

vv’

v

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

v’

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

vv’

v

Acceleration with Uniform Circular Motion

Uniform circular motion is when a particle moves with constant angular velocity.

Or

A

P

v

constantbealsoocity willlinear veltheofmagnitude the

v’

A particle moves from A

to P with constant angular velocity.

The acceleration of the particle is the change in velocity with respect to time.

vv’

v

OAPsimilar tois

vectorsoftriangleThis

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is and magnitude hasmotion

circular uniformin on acceleratithe2

rv

rva

2

Acceleration Involved inUniform Circular Motion

rva

2

2raOR

Acceleration Involved inUniform Circular Motion

rva

2

2raOR

Acceleration Involved inUniform Circular Motion

Forces Involved inUniform Circular Motion

rva

2

2raOR

Acceleration Involved inUniform Circular Motion

rmvF

2

Forces Involved inUniform Circular Motion

rva

2

2raOR

Acceleration Involved inUniform Circular Motion

rmvF

2

2mrFOR

Forces Involved inUniform Circular Motion

rva

2

2raOR

Acceleration Involved inUniform Circular Motion

rmvF

2

2mrFOR

Forces Involved inUniform Circular Motion

e.g. (i) (2003)A particle P

of mass m

moves with constant angular velocity

on a circle of radius r. Its position at time t

is given by;

tryrx

where,sincos

rva

2

2raOR

Acceleration Involved inUniform Circular Motion

rmvF

2

2mrFOR

Forces Involved inUniform Circular Motion

e.g. (i) (2003)A particle P

of mass m

moves with constant angular velocity

on a circle of radius r. Its position at time t

is given by;

tryrx

where,sincos

.on acting magnitude of force radial inwardan is e that therShow a) 2

Pmr

y

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cosrx

sinry

y

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r

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sin

rdtdrx

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r

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r

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cos

cos

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2

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r

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x

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rdtdry

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dtdry

2

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sin

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x

a

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x

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r

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sinry

cosrx sinry

sin

sin

rdtdrx

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cos

rdtdry

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dtdrx

2

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cos

yr

dtdry

2

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sin

y

x

a

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24

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2424

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yx

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x

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r

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sinry

cosrx sinry

sin

sin

rdtdrx

cos

cos

rdtdry

xr

dtdrx

2

2 cos

cos

yr

dtdry

2

2 sin

sin

y

x

a

222 yxa

24

224

2424

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yx

2ra

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x

P

r

cosrx

sinry

cosrx sinry

sin

sin

rdtdrx

cos

cos

rdtdry

xr

dtdrx

2

2 cos

cos

yr

dtdry

2

2 sin

sin

y

x

a

222 yxa

24

224

2424

ryx

yx

2ra

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x

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r

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sinry

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sin

sin

rdtdrx

cos

cos

rdtdry

xr

dtdrx

2

2 cos

cos

yr

dtdry

2

2 sin

sin

y

x

a

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24

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yx

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r

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sin

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cos

cos

rdtdry

xr

dtdrx

2

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cos

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dtdry

2

2 sin

sin

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x

a

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24

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yx

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cosrx sinry

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sin

rdtdrx

cos

cos

rdtdry

xr

dtdrx

2

2 cos

cos

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dtdry

2

2 sin

sin

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x

a

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24

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yx

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centrethes toward acting,, force a is There 2mrF

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

2mrxm

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

2mrxm 2r

Am

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

22

rAmmr

2mrxm 2r

Am

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

22

rAmmr

2

23

A

mAmr

2mrxm 2r

Am

b) A telecommunications satellite, of mass m, orbits Earth with constant angular velocity at a distance r

from the centre of the Earth.

The gravitational force exerted by Earth on the satellite is where

A

is a constant. By considering all other forces on the satellite to be negligible, show that;

2rAm

32

Ar

22

rAmmr

2

23

A

mAmr

32

Ar

2mrxm 2r

Am

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

T

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

T

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

T Tmg 0

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

T Tmg 0

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

T Tmg 0

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

T Tmg 0

xm

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

T Tmg 0

xm

T

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

2mrT T Tmg 0

xm

T

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

2mrT

25.02392 T Tmg 0

xm

T

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

2mrT

25.02392

rad/s982392

3922

T Tmg 0

xm

T

(ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is hung from it.A mass of 2kg is attached to one end of the string and it is revolved in a circle.Find the greatest angular velocity which may be imparted without breaking the string.

xm mg

Tmgxm

N

T392

8.940

2mrT

25.02392

rad/s982392

3922

Exercise 9B; all

T Tmg 0

xm

T

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