Circular motion slides complete

Circular Motion – 5.1 to 5.4

Gianicolli - Chapter 5

Mr. Sharick

Kinematics of Uniform

Circular Motion

Section 5.1

Section 5.1 – Kinematics of Uniform circular motion

Uniform in this case means constant speed.

Can an object moving with a constant

speed have acceleration?

If so, we physicists are going to want to find a

value for it, but how?

Figure 1 a small object

moving in a circle. Note the

instantaneous velocity is always

tangent to the circular path.

Section 5.1 – Kinematics of Uniform circular motion

Uniform in this case means constant speed.

Can an object moving with a constant

speed have acceleration?

An object that changes direction can have

acceleration (changing velocity) while

maintaining a constant speed.

This is the case for circular motion.

Section 5.1 – Vocabulary

New terms:

Tangential velocity –

Centripetal or radial acceleration –

Period –

Frequency –

Section 5.1 – Vocabulary

New terms:

Tangential velocity – (vt) velocity directed tangent to

the circular path

Centripetal or radial acceleration – (ac or ar)

acceleration directed toward the center of the circle

Period – (T) the time required to make one complete

revolution

Frequency – (f) the number of revolutions per

second

Section 5.1 – Tangential Velocity

Tangential Veloctiy

How do we calculate this?

– How do we calculate linear velocity?


Tangential Veloctiy



distancev

time


Tangential Veloctiy



– What distance is the object going through?

distancev

time


Tangential Veloctiy




distancev

time

2distance r


Tangential Veloctiy




– The time for one revolution is called what?

distancev

time

2distance r


Tangential Veloctiy




– The time for one revolution is called what?

Period (T)

distancev

time

2distance r


Tangential Veloctiy

Put it all together:

Remember, the direction of the velocity is ________

to the circular path!


Tangential Veloctiy


Remember, the direction of the velocity is ________

to the circular path!

2distance r v

time T


Tangential Veloctiy


Remember, the direction of the velocity is tangent to

the circular path!

2distance r v

time T


Example Problem – Tangential Velocity

1.) Find the tangential velocity (in m/s) of the Earth

orbiting the sun. The distance between the sun and the

Earth is 149,600,000 km.


Example Problem – Tangential Velocity

1.) Find the tangential velocity (in m/s) of the Earth

orbiting the sun. The distance between the sun and the

Earth is 149,600,000 km.

11

11

7

2 2 (1.496 10 )

1

2 (1.496 10 )29,700 /

r mv

T year

mv m s

3.16 10 s

Section 5.1 – Radial/Centripetal Acceleration

Proof – radial/centripetal acceleration

-We are going to derive the equation for centripetal

acceleration of an object in uniform circular motion.


We will be using two similar

triangles.

1st triangle: ABC, made up

of sides r1, r2, and Δl

2nd triangle: made up of

sides v1, v2, and Δv. (v1 =

v2; just different directions)

1

2

Interlude – Finding Δv

v2

Since v = v2-v1

Draw v1 + v = v2

v1

Interlude – Finding Δv

v2Since v = v2-v1

Draw v1 + v = v2

v1

v


The angle between the

radius and the tangential

velocity is always 90o.

For this to remain true,

the angle between r1 and

r2 must be equal to the

angle between v1 and v2.

(Δθ equal in the picture)

1

2


Therefore (similar

triangles have ratio of

sides that are equal)

v l

v r

1

2


Therefore (similar

triangles have ratio of

sides that are equal)

So

v l

v r

1

2

v

v lr


Recall the definition for acceleration:



v

at



Also, we found:

v

at

v

v lr



Also, we found:

Thus:

v

at

v

v lr

v l

ar t


Recall that the linear speed around a circle can be given by the change in arc length divided by change in time.

For small angles, arc length is approximately the straight line displacement.

So, we have that




So, we have that

sv

t

s l




So, we have that

sv

t

s l

v l

ar t




So, we have that

sv

t

s l

2

= v l v s v

ar t r t r


Hopefully, it is clear that this formula can only give positive answers, and therefore only can find the magnitude of the acceleration.

Acceleration is a VECTOR so… in what direction is the acceleration?

2

v

ar

Finding the direction of centripetal acceleration:

vf

Draw v = vf-vi

vi

Finding the direction of centripetal acceleration:

tf

-vi

vf

v

Notice the direction

of v is towards the

center of the circle!

The direction of acceleration is also

towards the center of the circle!

THUS, centripetal!

v = vf-vi

Section 5.1 - Centripetal/Radial Acceleration

Acceleration of an object in uniform circular

motion is given by:

Direction is toward the _______________

Section 5.1 - Centripetal/Radial Acceleration

Acceleration of an object in uniform circular

motion is given by:

Direction is toward the center (centripetal)

2

c

va

r

Section 5.1 – Period and Frequency

Period and frequency are inversely

proportional.

1T

f


Ex: An object makes 10 revolutions in 2

seconds. What is the objects frequency and

period?


Ex: An object makes 10 revolutions in 2

seconds. What is the objects frequency and

period?

sec

sec

/ sec

105

2sec

1 10.2

5

rev

revrev

revf

Tf

Section 5.1 – Example 5.1 –Acceleration of a revolving ball

A 150-g ball at the end of a string is revolving uniformly

in a horizontal circle of radius 0.600 m. The ball makes

2.00 revolutions in a second. What is the centripetal

acceleration?

Section 5.1 – Example 5.1 –Acceleration of a revolving ball

A 150-g ball at the end of a string is revolving uniformly

in a horizontal circle of radius 0.600 m. The ball makes

2.00 revolutions in a second. What is the centripetal

acceleration?

2 22

1/ 1/ (2.00 / ) 0.500

2 2(3.14)(0.600 )7.54 /

(0.500 )

(7.54 / )94.8 /

(0.600m)c

T f rev s s

r mv m s

T s

v m sa m s

r

Interlude – a useful substitution

We have 2 equations:

Plugging v in the 1st into the 2nd we get:

2

c

va

r

2 rv

T

Interlude – a useful substitution

We have 2 equations:

Plugging v in the 1st into the 2nd we get:

2

c

va

r

2 rv

T

2

2

2 2

2

2

2 4 or

c

c c

r

v Ta

r r

ra r a

T T

Section 5.1 – Example 5.2 –Moon’s centripetal acceleration

The moon’s nearly circular orbit about the Earth has a

radius of orbit of about 384,000km and a period T of

27.3 days. Determine the acceleration of the Moon

towards the Earth.

Section 5.1 – Example 5.2 –Moon’s centripetal acceleration

The moon’s nearly circular orbit about the Earth has a

radius of orbit of about 384,000km and a period T of

27.3 days. Determine the acceleration of the Moon

towards the Earth.6

2 82

2 6 2

(27.3days)(24.0h/ day)(3600s/ h) 2.36 x10

4 (39.5)(3.84x10 m)0.00272m/s

(2.36x10 )c

T s

ra

T s

Dynamics of Uniform

Circular Motion

Section 5.2

Section 5.2 - DYNAMICS of Circular Motion

Revisiting Newton’s 2nd Law

Newton stated that if a body is accelerating,

there must be a NET force acting on it, in the

direction of the acceleration.

The NET force that causes circular motion is

called the CENTRIPETAL force.

Note: This is not a new type of force, we will

treat it very much the same as we did any

NET force.

Section 5.2 – Dynamics of Uniform Circular Motion

Centripetal or radial force – the force required to keep

an object in circular motion is as follows:

What is the direction of this force?

2 2

2

4c c

mv m rF ma

r T

Section 5.2 – Centripetal vs. Centrifugal

Centri from the Latin Centrum: the middle point of a circle, the center.

Petal from the Latin Peto – to seek, strive after, endeavor to obtain.

Fugal from the Latin Fugito – to flee, to fly from, avoid or shun.

By looking at the direction of the change in velocity, you should be able to see the direction of the acceleration.

Is it Centripetal or Centrifugal?


Consider a person swinging a ball on the end

of a string around her head:

A ___________ (center seeking) force causes

circular motion NOT a __________ (center

fleeing) force.

The force that is acting on the ball causing

circular motion is inward (centripetal).

The force acting on the hand is outward, and

NOT causing circular motion.

A centrifugal force is a “fictitious” or “pseudo”

force caused by viewing the situation from an

accelerating reference frame.


Consider a person swinging a ball on the end

of a string around her head:

A centripetal (center seeking) force causes

circular motion NOT a centrifugal (center

fleeing) force.

The force that is acting on the ball causing

circular motion is inward (centripetal).

The force acting on the hand is outward, and

NOT causing circular motion.

A centrifugal force is a “fictitious” or “pseudo”

force caused by viewing the situation from an

accelerating reference frame.

Section 5.2 - Centrifugal force – the fictitious force

Wahoo!

I’m stuck to the wall!

I feel like I’m being pressed against

the wall.

Force on person due

to the wall

Fictitious force to explain zero

acceleration

Section 5.2 - Why is it said that centrifugal “fictitious” force is a misconception, and NOT real?

Because to an

observer in an inertial

reference frame

“fictitious” force does

not exist.

Section 5.2 - Centrifugal forces are not real!

Remember, Newton’s laws only apply in inertial

reference frames.

Centrifugal (fictitious) force is a non-Newtonian, and

is caused by acceleration and is only real to an

observer in the accelerating reference frame.

In the picture above (a) represents what would

happen if centrifugal forces were real. (b) Shows

what does happen in an inertial frame of reference.

Section 5.2 – Solving Dynamics Problems for Uniform Circular Motion

Section 5.2 – Example 5.3 – Force on revolving ball (horizontal)

Estimate the force a person must exert on a string

attached to a 0.150 kg ball to make the ball revolve in a

horizontal circle of radius 0.600m, making 2.00

revolutions per second.

mg

θT

Section 5.2 – Example 5.4 –Tetherball

The game of tetherball is played with a ball tied to a

pole with a string. When the ball is struck, it whirls

around the pole as shown. In what direction is the

acceleration of the ball, and what causes the

acceleration.q

Section 5.2 – Extra Example – Conical Pendulum (yellow packet page 9)

A particle of mass, m, is suspended from a string of length, L, and

travels at a constant speed, v in a horizontal circle if radius, r. The

string makes an angle, q, given by sin q = r/L, as shown in the

figure. Find the tension in the string and the speed of the particle.

q

Section 5.2 – Example 5.5 – Revolving ball (vertical circle)

A 0.150 kg ball on the end of a 1.10 m long cord is swung in a

vertical circle. a.) Determine the minimum speed the ball must

have at the top of its arc so that it continues moving in a circle.

Section 5.2 – Example 5.5 – Revolving ball (vertical circle)

A 0.150 kg ball on the end of a 1.10 m long cord is swung in a

vertical circle. b.) Calculate the tension in the cord at the bottom of

the arc assuming the ball is moving at twice the speed of part (a).

A Car Rounding a

Curve

Section 5.3

Section 5.3 – A Car Rounding a Curve

One example of centripetal acceleration

occurs when a car rounds a curve.

In this situation, what do you feel?

What is causing the car to move in circular

motion?


You feel like you are being thrown _______, but in

reality, you tend to move in a _______ _____as the

car begins moving in a curve. The force you feel is

the car pushing you!

The car is moving

because of _________

between the tires

and the road!


You feel like you are being thrown outward, but in

reality, you tend to move in a straight line as the car

begins moving in a curve. The force you feel is the

car pushing you!

The car is moving

because of friction

between the tires

and the road!

Section 5.3 - Example 5.7 – Skidding on a curve (yellow packet page 7)

A 1000 kg car rounds a curve on a flat road of

radius 50 m at a speed of 50 km/hr (14 m/s).

Will the car make the turn, or will it skid, if: (a)

the pavement is dry and the coefficient of

static friction is ms = 0.60?

Section 5.3 - Example 5.7 – Skidding on a curve (yellow packet page 7)

A 1000 kg car rounds a curve on a flat road of

radius 50 m at a speed of 50 km/hr (14 m/s).

Will the car make the turn, or will it skid, if: (b)

the pavement is icy and ms = 0.25?

Section 5.3 – Example 5.8 – Banking angle w/o friction(yellow packet page 8)

For a car traveling with a speed, v, around a curve

of radius, r. (a) Determine a formula for the angle

at which a road should be banked so that no

friction is required to keep the car on the road.

Section 5.3 – Example 5.8 – Banking angle w/o friction (yellow packet page 8)

For a car traveling with a speed, v, around a curve

of radius, r. (b) What is this angle for a curve of

radius 50 m at a designed speed of 50 km/h?

Section 5.3 – Banking angles with friction (not in book)

Two cases:

- Faster than ideal speed

- Slower than ideal speed

What happens in each case?


V > V_ideal

The car will want to move __

the slope, so that friction is

pointing ______ the slope

(______ motion)

The math is much more

complicated with friction!


V > V_ideal

The car will want to move up

the slope, so that friction is

pointing down the slope

(opposing motion)

The math is much more

complicated with friction!


V < V_ideal

The car will want to move

______ the slope, so that

friction is pointing ___ the

slope (________ motion)


V < V_ideal

The car will want to move

down the slope, so that

friction is pointing up the

slope (opposing motion)

Nonuniform Circular

Motion

Section 5.4

total

Section 5.4 - Nonuniform Circular Motion

Combining the new with the old!

Remember acceleration is a vector!

(in relation to the velocity)

Total tangential radial

Total

a a a

OR

a a a

total

Section 5.4 – Tangential vs centripetal acceleration

For tangential acceleration (caused by tangent force)

For centripetal acceleration (caused by centripetal

force)

For a car, where would each of these forces come

from?

Section 5.4 – Tangential vs centripetal acceleration

For tangential acceleration (caused by tangent force)

vf=vi + at

For centripetal acceleration (caused by centripetal

force)

a = v2/r

For a car, where would each of these forces come

from?

Section 5.4 – Example 5.9 – Two components of acceleration

A racing car starts from rest in the pit area and

accelerates at a uniform rate to a speed of 35 m/s in 11

s, moving on a circular track of radius 500 m. Assuming

constant tangential acceleration, find (a) the tangential

acceleration.

Section 5.4 – Example 5.9 – Two components of acceleration

A racing car starts from rest in the pit area and

accelerates at a uniform rate to a speed of 35 m/s in 11

s, moving on a circular track of radius 500 m. Assuming

constant tangential acceleration, find (b) the centripetal

acceleration when the speed is 30 m/s.

Section 5.4 – Example Problem: Nonuniformcircular motion (from yellow packet page 10)

A particle moves clockwise in a circle of radius 1.0 m. It

starts at rest at the origin at time, t=0. Its speed

increases at a constant rate of 1.6 m/s2.

a.) How long does it take to travel half way around the

circle (point A to Point B)?

A B

r =1.0 m


b.) What is its speed at point B?

c.) What is the direction of its velocity at that time?

A B

r =1.0 m


d.) What is the centripetal (radial) acceleration?

e.) What is the tangential (linear) acceleration?

A B

r =1.0 m


f.) What is the magnitude and direction of the total

acceleration at point B?

A B

r =1.0 m

Education

Circular motion slides complete