Pearson 2007-2008

Uniform Circular Motion

An object moving in a circle with constant speed, v, experiences a centripetal acceleration with : *a magnitude that is constant in time and is equal to

a



v 2 r

*a direction that changes continuously in time and always points toward the center of the circular path For uniform circular motion , the velocity is tangential to the circle and perpendicular to the acceleration

Period and Frequency

A circular motion is described in terms of the period T , which is the time for an object to complete one revolution.

T

 2 

r v

The distance traveled in one revolution is 2 

r

r The frequency, f, per unit time.

counts the number of revolutions

f

 1

T

Example of Uniform Circular Motion

The moon’s nearly circular orbit about the earth has a radius of about 384,000 km and a period T of 27.3 days . Determine the acceleration of the Moon towards the Earth.

a T

 2 

r v



v

2

r



v

 2 

r T

 4  2

r

2

T

2

r

 4  2

r T

2

a

 2.72

 10  3

m

/

s

2 

g

9.8

m

/

s

2  2.78

 10  4

g

Moon…...

* * So we find that

a moon

Newton noticed that /

g R E 2

= 0.000278

/ R 2

= 0.000273

a moon g R R E

*This inspired him to propose that

F gravity

(more on gravity in future lectures) 

1 / R 2

Uniform Circular Motion

* motion in a circle or circular arc at

constant speed

* the acceleration changes the direction of the velocity, not the magnitude The period of the motion:

T

 2 

r v

* the “center-seeking” or centripetal acceleration is always orthogonal to the velocity and has magnitude:

a



v

2

r



Uniform Circular Motion

Newton’s 2nd Law : The net force on a body is equal to the product of the mass of the body and the acceleration of the body.

* The centripetal acceleration is caused by a centripetal force that is directed towards the center of the circle.

F



ma



m v

2

r

Demo 1D-5

Does the contact force between the wine glass and red-water remain constant in uniform circular motion?

Consider the glass directly overhead. Choose the correct statement: a.

The water doesn’t fall because the centripetal force on the water cancels the force of gravity.

b.

c.

The water doesn’t fall because there isn’t enough time for it to fall.

The water doesn’t fall because of the horizontal force applied to it by the glass, plus friction with the glass.

d.

The water is falling, but the glass is falling faster than it would under free fall.

 v y ma c = mv 2 /r = mg + N y or a c = g  N/m

N mg Top

x When N=0, the centripetal acceleration is just g.



Fy N

 

N



mg

 

ma



m

(

a



g

)

N



m

(

v

2

r



g

)

  v y x

mg N mg N Top Bottom

v 

Fy N

 

N



mg

 

ma



m

(

a



g

)

N



m

(

v

2

r



g

) 

Fy N



N



mg

 

m

(

a



g

)

ma N



m

(

v

2

r



g

)

  v N

Top



Fy N

 

N



mg

 

ma



m

(

a



g

)

N



m

(

v

2

r



g

)

mg

What speed is needed to lose contact between wine glass and red-water?

v



rg

2)

A person riding a Ferris Wheel moves through positions at (1) the top, (2) the bottom and (3) midheight. If the wheel rotates at a constant rate, rank (greatest first) these three positions according to... (a) the magnitude of persons centripetal acceleration (a) 2,1,3 (b) 1,2,3 (c) 3,2,1 (d) all tie (1) Top (b) The magnitude of the Normal force?

(3) Middle (2) Bottom

(b)

the magnitude of the net centripetal force on the person 1.

1,2,3 2.

3.

4.

3,1,2 3,2,1

all tie (c)

the magnitude of the normal force on the person 1.

all tie

2.

3.

4.

2,3,1

3,2,1 1,2,3

   Demo 1D-2 Conical Pendulum 

F y

 0 

T

cos q 

mg

H q

T



F r



ma



T

sin q 

mv

2 /

R

tan q 

v

2

gR v



Rg

tan q 

R g

/

H

R * as q 90, v increases.

* v is independent of mass.

mg

Period



T

 2 

R

/

v

 2 

H

/

g

The period, T, is independent of mass and depends only on H.

  A car of mass, m, is traveling at a constant speed, v, along a flat, circular road of radius, R. Find the minimum µ s required that will prevent the car from slipping 

F r

 

ma



F y



N f s

 

mv mg

2

f s

 

s N

/

R



s



v

2

gR

   A mass, m, on a frictionless table is attached to a hanging mass, M, by a cord through a hole in the table. Find the speed with which m must move in order for M to stay at rest.

T



m v

2

r T



Mg



Mg

 0 

T m v

2

r v



M m gr



Mg

  A car of mass, m, is traveling at a constant speed, v, along a curve that is now banked and has a radius, R. What bank angle, q , makes reliance on friction unnecessary?

N 

F y

 0 

N

cos q 

mg



F r



ma



N

sin q 

mv

2 /

R

q mg tan q 

v

2

gR R



v

2

g

tan q

    An airplane is flying in a horizontal circle with a speed of 480 km/hr. If the wings of the plane are tilted 40 o to the horizontal, what is the radius of the circle in which the plane is flying? (Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.) v=480 km/hr L 

F r

 0  2

L

sin 40 

M v

2

r



F y

 0  2

L

cos40 

Mg

W

L



Mg

2 cos40 L 2

Mg

2 cos 40

r

sin 40 

v

2

g

tan 40 

M v

2

r