7. UCMRevF.pptx

Uniform Circular Motion

Uniform Circular

Motion

Motion of a particle

moving at constant

speed along a circular

arc or path.

Velocity is constant in magnitude but changes in direction.

Centripetal Acceleration

has constant magnitude.

Since the car’s direction

of motion changes, the

car has an acceleration

when an object moves

in a circular path,

CENTRIPETAL ACCELERATION

Objects in uniform

circular motion

experiences centripetal acceleration



Centripetal Acceleration

has

constant magnitude.



It is always directed towards

Vectors

v

and

v

have same magnitude v but because they

point in different directions - they are different vectors.

Velocity

v

and

acceleration

a

in uniform

circular motion at

angular rate ω;

The

speed

is constant

, but

the

velocity is always

tangent to the orbit

;

The

acceleration

has

constant magnitude

, but

always

points toward the

center of rotation

.

LINEAR MOTION LINEAR MOTION

• linear distance

moved per unit of time: m/s

• linear distance

moved per unit of time: m/s

ROTATIONAL MOTION ROTATIONAL

MOTION

• number of rotations

per unit of time

• _rpm

• number of rotations

per unit of time

• _rpm

Tangential speed is directly proportional to rotational speed and the distance from the axis (radial distance)

Tangential speed is directly proportional to rotational speed and the distance from the axis (radial distance)

v= ωr

The speed of an object moving along a circular path can be called tangential speed because the direction of motion is always tangent to the circle.

Equations to Remember

Frequency ( f )

– number of revolutions per unit time

SI unit: Hertz (Hz) – 1/s Type: derived

Relation of Period and Frequency

Period ( T )

– time for one revolution

SI unit: Seconds (s) Type: fundamental

T = 1

f

=

2 r

v

r : radius of

the circular

arc



Angular Frequency/Angular Velocity (ω)

 _{The speed of the object in the circular path}



Tangential speed (v)

 _{The speed of the object as if its moving in a straight line}



Centripetal Acceleration (a

)

 _{The acceleration the object experiences as it moves along the circular}

EXAMPLE 1

In a carnival ride, the passengers travel at constant

speed in a circle of radius 5.0m. They make one

complete circle in 4.0s. What is their velocity and

acceleration?

Given:

radius= 5.0 m

Period(T)= 4.0 s

SOLUTION :

T= 2πr

v

T

r

v



2







s

m

v

0

.

4

0

.

5

2





s

m

v



7

.

9

/

r

v

a

rad

2







m

s

m

a

0

.

5

/

9

.

7



2

/

12

m

s

EXAMPLE 2

What is the centripetal acceleration of an

automobile driving at 40km/h on a circular track

of radius 20m?

a. Given: v = 40km/h, r = 20m

b. Required: centripetal acceleration a

c. Equation to use: a

= v

/r

A

=

(40000m/3600s)



When objects rotate about some axis—for example, when

the CD (compact disc) in Fig 1 rotates about its center—

each point in the object follows a circular arc.



Consider a line from the center of the CD to its edge. Each

pit

used to record sound along this line moves through the

same angle in the same amount of time.



The rotation angle is the amount of rotation and is

analogous to linear distance. We define the

rotation angle

Δ



to be the ratio of the arc length to the radius of

curvature:



Δ



= Δ

s/r

Rotating CD

Figure 1 All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle Δθ in a time Δt

 _{The arc length Δs is the distance traveled along a circular}

path as shown in Figure 2 Note that r is the radius of curvature of the circular path.

 _{We know that for one complete revolution, the arc length}

is the circumference of a circle of radius r . The circumference of a circle is 2πr . Thus for

 _{one complete revolution, the rotation angle is}  _Δ _{= 2πr/r = 2π.}

 _{This result is the basis for defining the units used to}

measure rotation angles, Δ to be radians (rad), defined so that

Comparison of Angular Units



Degree Radian

30º

π/

6



Figure 3

A car moving at a velocity

v

to the right has a tire

rotating with an angular velocity

ω

.The speed of the tread of

the tire relative to the axle is

v

, the same as if the car were

jacked up. Thus the car moves forward at linear velocity

v

=

Angular Velocity

How fast is an object rotating? We define angular velocity ω as the rate of change of an angle over a given time. In symbols, this is

ω = Δ / Δt ,

where an angular rotation Δ takes place in a time Δt . The greater the rotation angle, the greater the angular velocity. The units for angular velocity are radians per second (rad/s).

 _{Angular velocity ω is analogous to linear velocity v .}

 _{To get the precise relationship between angular and linear velocity,}

we again consider a pit on the rotating CD (Fig. 1)

 _{This pit moves an arc length Δs in a time Δt , and so it has a linear}

velocity v = Δs /Δt .

 _{From Δ} _{= Δs/r we see that Δs = rΔ} _{. Substituting this into the}

expression for v gives v = rΔ/Δt = rω.

v

=

rω

or

ω

=

v/r



Example 3: How Fast Does a Car Tire Spin?

 _{Calculate the angular velocity of a 0.300 m radius car tire when}

the car travels at 15.0 m/s (about 54 km/h ). See Figure 3.

 _{Because the linear speed of the tire rim is the same as the speed}

of the car, we have v = 15.0 m/s. The radius of the tire is given to be r = 0.300 m.

 _{Knowing v and r , we can use the second relationship in v = rω,}

ω =v/r, to calculate the angular velocity.

 _Solution

 _{Angular velocity: ω =v/r}  _{Knowing v and r}

EXAMPLE 4

The period of a stone swung in a horizontal circle on a

2.00-m radius is 1.00s.

(a) What is its angular velocity in rad/s?

Given: r = 2.00 m, T = 1.00s

Required: ω

Equations to use:

A.

ω = 2πf

B.

T = 1/f



ω

=

2π/T

The period of a stone swung in a horizontal circle on a 2.00-m

radius is 1.00s.

(b) What is its linear speed in m/s?

Given: ω (from the previous problem)

Required: v

Equation to use:

ω = v/r

therefore

v = ωr

The period of a stone swung in a horizontal circle on a

2.00-m radius is 1.00s.

(c) What is its radial acceleration in m/s

?

Given: v (from the previous problem), r

Required:

a

or

a

Equation to use: a

= v

/r

a

= v

/r=(4π m/s)

/2.00m

Centripetal acceleration

“center seeking”

a

=

v

2

r

UNIFORM CIRCULAR MOTION

When a particle moves at constant

speed

v

in a circle or a circular arc of radius r,

the acceleration of the particle is

directed

toward the center

of the circle and has a

constant magnitude

v

/r.

NOTE: The notation ar is used to indicate that the centripetal

acceleration is along the radial direction

a

=

v

Centripetal acceleration:

v = 2

π

r

a

= 4

π

r

T

T

UNIFORM CIRCULAR MOTION

a

=

v

r

WOULD BECOME:

Angular Frequency/ Angular

velocity :

ω

ω = 2πf

ω = v/r

Unit: rad/s (Radians/second)

UNIFORM CIRCULAR MOTION

1 rev/s = 2

π radians/second

Note: 1 rad=360

/2

π

Example 5 : Going Around in Circles



What is the centripetal acceleration of

an automobile moving at 60km/h on a

circular track of radius 27m? What is its

angular velocity?



a. Given: v = 60km/h, r = 27m



b. Required: centripetal acceleration a

c

and

Solutio

n



Given: v = 60 kph; r = 27 m



Find: Centripetal acceleration in m/s² and angular

velocity in rad/s



Solution:

Equation to use: a

= v

/r and

ω = v/r

Convert 60 kph to m/s for unit consistency

(60 kph) * (1000 m/k) * 1 h/3600 s = 16.67 m/s

 _{Example 6. How Does the Centripetal Acceleration of a Car}

Around a Curve Compare with That Due to Gravity?

 _{What is the magnitude of the centripetal acceleration of a car}

going on a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure below.

 _Solution

Given v and r, we can use Ac = v²/r ; Ac = rω²

With v = 25.0 m/s and r = 500 m, Ac = 25² /(500) = 1.25 m/s²

 _Discussion

To compare this with the acceleration due to gravity (g = 9.80

m/s2) , take the ratio of Ac / g = (1.25 m/s2)/(9.80 m/s2) = 0.128

 _Thus,

ANNOUNCEMENT



Your first exam will be on Oct 3 (Friday) 7-9PM

Room assignments will be announced this week

Coverage:

 Physical Quantities  Vectors

 Kinematics (1D and 2D, up to Relative motion)



We have 4 meetings left till the exam.

 _{Will finish 2D up to Relative motion}

 _{Solutions to Quiz 1 to 4 and handout on UCM will be}

available at Star shop on Wednesday afternoon.

 _{Note: Those who have conflicts on the time of Exam will be}

RECALL:



UNIFORM CIRCULAR

MOTION:



Motion in a

circle

or

an

arc



Acceleration is

“center seeking”



Velocity is

tangent

to

Equations to Remember

Frequency ( f )

– number of revolutions per unit time

SI unit: Hertz (Hz) – 1/s Type: derived

Relation of Period and Frequency

Period ( T )

– time for one revolution

SI unit: Seconds (s) Type: fundamental

T = 1

f

=

2

r v

r : radius of

the circular

arc



Angular Frequency/Angular Velocity (ω)

 _{The speed of the object in the circular path}



Tangential speed (v)

 _{The speed of the object as if its moving in a straight line}



Centripetal Acceleration (a

)

 _{The acceleration the object experiences as it moves along the circular}