Lecture 4 Centripetal Force and Law of Gravitation

8
1 Lecture Lecture IV IV Centripetal Force •Law of Gravitation UNIFORM CIRCULAR UNIFORM CIRCULAR MOTION MOTION 2D KINEMATICS PERIOD PERIOD Period ( T ) -time for one revolution -time for a particle to go around a closed path -unit: Seconds (s) r : radius of the circular arc v : speed of the particle 2 r T v π = FREQUENCY FREQUENCY Frequency -number of revolutions per unit time - unit: Hertz (Hz) unit: Hertz (Hz) – 1/second 1/second 1 f T = motion of an object following a circular path while moving at a constant speed UNIFORM CIRCULAR MOTION UNIFORM CIRCULAR MOTON UNIFORM CIRCULAR MOTON Constant speed but velocity changes direction Experiences centripetal acceleration.

Transcript of Lecture 4 Centripetal Force and Law of Gravitation

Page 1: Lecture 4 Centripetal Force and Law of Gravitation

1

Lecture Lecture IVIV

• Centripetal Force

•Law of Gravitation UNIFORM CIRCULAR UNIFORM CIRCULAR MOTIONMOTION

2D KINEMATICS

PERIODPERIOD

Period ( T )

-time for one revolution

-time for a particle to go around a closed path

-unit: Seconds (s)

r : radius of the

circular arc

v : speed of the

particle

2 rT

v

π=

FREQUENCYFREQUENCY

Frequency

-number of revolutions per unit time

--unit: Hertz (Hz) unit: Hertz (Hz) –– 1/second1/second

1f

T=

motion of an object following a circular path while moving at a constant speed

UNIFORM CIRCULAR MOTION UNIFORM CIRCULAR MOTONUNIFORM CIRCULAR MOTON

� Constant speed but velocity changes direction

� Experiences centripetal acceleration.

Page 2: Lecture 4 Centripetal Force and Law of Gravitation

2

Tangential speed is directly proportional to rotational

speed/angular velocity and the distance from the axis (radial distance)

v= ωr

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

UNIFORM CIRCULAR MOTON

Velocity and acceleration

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.

Period of revolution (Period)

Frequency:

Centripetal acceleration:

Using v from (1):

CENTRIPETAL ACCELERATIONCENTRIPETAL ACCELERATION

2 rT

v

π=

1f

T=

2

c

va

r=

( )2 22 4

c

r ra

T r T

π π= =

Centripetal Acceleration and Angular Frequency

(Angular velocity/ Angular frequency)Unit: rad/s (Radians/second)

1 rev/s = 2 rad/sec

Note: 1 rad=360o/2π=57.3o

2 f

v

r

ω π

ω

=

= π

22

c

va r

rω= =

EXAMPLE 1EXAMPLE 1

In a carnival ride, the passengers travel at In a carnival ride, the passengers travel at constant speed in a circle of radius 5.0m. constant speed in a circle of radius 5.0m. They make one complete circle in 4.0s. What They make one complete circle in 4.0s. What is their velocity and acceleration?is their velocity and acceleration?

SOLUTION 1SOLUTION 1

T

Rv

π2=

( )s

mv

0.4

0.52π=

smv /9.7=

R

vac

2

=

( )m

smac

0.5

/9.72

=

2/12 smac =

Page 3: Lecture 4 Centripetal Force and Law of Gravitation

3

Example 2

What is the centripetal acceleration of an automobile driving at 40km/h on a circular track of radius 20m?

Given:

v = 40km/h,

r = 20m

Required: centripetal acceleration, ac

Equation to use: ac = v2/r

ac=(40000m/3600s)2

20m

= 6.17 m/s2

EXAMPLE 3EXAMPLE 3

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

C.ω=2π/T= 2π rad/s

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

v = (2π rad/s)2.00m= 4π m/s

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/s2?

Given: v (from the previous problem), r

Required: ac

Equation to use: ac = v2/r

ac = v2/r=(4π m/s) 2/2.00m

=78.5m/s2

Period and Frequency

Equations to remember!Equations to remember!

v= ωr (Tangential speed or linear speed)

Centripetal Acceleration

Angular Frequency/velocity

Unit: rad/s (Radians/second)

22

c

va r

rω= =

( )2 2

2 2

2 4c

r ra

T r T

π π= =

v

rω = 2 fω π=

1f

T=

2 rT

v

π=

CENTRIPETAL FORCE

�Centripetal Force is a force that tends to keep

object in moving around a circular arc or path

�The magnitude of the centripetal force is the

product of an object’s mass and its centripetal acceleration as it moves around the circular

path

�The direction of the centripetal force is

always directed towards the center of the circle.

Page 4: Lecture 4 Centripetal Force and Law of Gravitation

4

Centripetal acceleration

� An object traveling in a circle, even though it moves with a constant speed, will have an acceleration (since velocity changes direction)

� This acceleration is called centripetal (“center-seeking”).

� The acceleration is directed toward the center of the circle of motion

Centripetal acceleration and

the angular velocity

� The angular velocity and the

linear velocity are related

(v = ωr)

� The centripetal acceleration can also be related to the

angular velocitySimilar

triangles!

, but

a

v s vv s

v r r

v v sa

t r t

∆ ∆= ⇒ ∆ = ∆

∆ ∆= ⇒ =

∆ ∆

uurr

22orC C

va a r

rω= =

Thus:

Similar

triangles!

Total acceleration

� What happens if linear velocity also changes?

� Two-component acceleration:

� the centripetal component of the acceleration is due to changing direction

� the tangential component of the acceleration is due to changing speed

� Total acceleration can be found from these components:

slowing-down car

2 2

t Ca a a= +

Forces Causing Centripetal

Acceleration

� Newton’s Second Law says that the centripetal acceleration is accompanied by a force

� F stands for any force that keeps an object following a circular path

� Force of friction (level and banked curves)

� Tension in a string

� Gravity

2

C

vF ma m

r= =∑

F

F

F

F

Centripetal Force

Net inward force to provide centripetal acceleration

Due to contact and/or gravitational forces

Direction:

towards the center

Page 5: Lecture 4 Centripetal Force and Law of Gravitation

5

Centripetal Force

This is not a new kind of force, but merely a name for the force needed for circular motion.

The centripetal force maybe due to:string, spring, or other contact force such as normal force and friction, action-at-a-distance forces such as gravity;

SOME EXAMPLES

For a rock whirled on the end of a string, the

centripetal force is the force of tension in the string.

For an object sitting on a rotating turntable, the

centripetal force is friction.

For the motion of the Earth around the Sun, the

centripetal force is gravity.

centripetal force could be a combination of

two or more forces. For example, as a

Ferris-wheel rider passes through the

lowest point, the centripetal force on her is

the difference between the normal force

exerted by the seat and her weight.

Example 1: level curves

Consider a car driving at 20 m/s (~45 mph) on a level circular turn of radius 40.0 m. Assume the car’s mass is 1000 kg.

1. What is the magnitude of frictional force experienced by car’s tires?

2. What is the minimum coefficient of friction in order for the car to safely negotiate the turn?

Example a:

Given:

masses: m=1000 kg

velocity: v=20 m/s

radius: r = 40.0m

Find:

1. f=?

2. µ=?

1. Draw a free body diagram,

introduce coordinate frame and

consider vertical and horizontal

projections

0y

F N mg

N mg

= = −

=

( )

2

2

420

1000 1.0 1040

xF ma f

vf ma m

r

m skg N

m

= = −

= − = −

= − = − ×

2. Use definition of friction

force:

Lesson: µ for rubber on dry concrete is 1.00!

rubber on wet concrete is 0.2!

driving too

fast…

24

4

2

10 , thus

1.0 101.02

1000 9.8

vf mg m N

r

N

kg m s

µ

µ

= = =

×= ≈

Page 6: Lecture 4 Centripetal Force and Law of Gravitation

6

Assignment (1/2): banked curves

Consider a car driving at 20 m/s (~45 mph) on a 30°banked circular curve of radius 40.0 m. Assume the car’s mass is 1000 kg.

1. What is the magnitude of frictional force experienced by car’s tires?

2. What is the minimum coefficient of friction in order for the car to safely negotiate the turn?

NEWTON’S LAW OF GRAVITATION

Newton’s Law of Universal

Gravitation� Every particle in the Universe attracts every

other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the

distance between them.

2

21

r

mmGF =

�� G is the universal gravitational constantG is the universal gravitational constant

�� G = 6.673 x 10G = 6.673 x 10--1111 N m² /kg²N m² /kg²

�� This is an example of an This is an example of an inverse square lawinverse square law

Gravitation Constant

� Determined experimentally

� Henry Cavendish

� 1798

� The light beam and mirror serve to

amplify the motion

Example

Question: Calculate gravitational attraction between two students 1 meter apart. Assume the student 1 has a mass of 70 kg while the other one has a mass of 90 kg.

( )

21 11 2

22 2

7 0 9 06 .6 7 1 0

1

m m N m kg kgF G

r kg m

−= = ×

74.2 10F N−≈ × Extremely small

compared to the

weight (F = mg).

Applications of Universal

Gravitation 1: Mass of the Earth� Use an example of an object

close to the surface of the earth

� r ~ RE

G

gRM E

E

2

=

11 2

E

E

GM mm g

R=

Page 7: Lecture 4 Centripetal Force and Law of Gravitation

7

Applications of Universal

Gravitation 2: Acceleration Due

to Gravity� g will vary with altitude

2r

MGg E=

mgr

MGm

r

mMGF EE =

==

22

Escape Speed

� The escape speed is the speed needed for an

object to soar off into space and not return

� For the earth, vesc is about 11.2 km/s

� Note, v is independent of the mass of the object

E

Eesc

R

GMv

2=

PHYSICS FAIR

Page 8: Lecture 4 Centripetal Force and Law of Gravitation

8