Uniform Circular Motion

60
Francis Marlon Cabredo Mary Minette Geñorga Mary Judith Verdejo

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Transcript of Uniform Circular Motion

Page 1: Uniform Circular Motion

• Francis Marlon Cabredo• Mary Minette Geñorga• Mary Judith Verdejo

Page 2: Uniform Circular Motion

▪Uniform Circular Motion is the motion of an object in a circle with a constant or uniform speed

UNIFORMCIRCULAR MOTION

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SPEEDV=2r

T

SPEED:

▪ In UCM, distance= the circumference of the circle

▪ T(period) is the time (or number of seconds) to make one revolution

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PERIOD and FREQUENCY

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• Period (T) -> the time it takes for an object to

make one complete revolution {#sec/rev}

PERIOD and FREQUENCY

• Frequency (f) the number of revolutions completed by an object in a given time

{#rev/sec}Units: Hertz (Hz) or 1 rev/sec

rpm (#rev/min); rps (#rev/sec)

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PERIOD and FREQUENCY

f=1__T T=1__

f

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SPEED expressed in PERIOD and FREQUENCY

FORMULA(s):

v=2rf2rv=T

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PROBLEM 1: Tire Balancing Machine

• The wheel of a car has a radius of 0.29m and is being rotated at 830 revolutions per minute on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving. Use 3.14 as the value for .

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SOLUTION#1:

f=830 revolutions in one minute

revolutionsrevolution

min/102.1min/830

1 3

T= 0.072s

Given:

r=0.29 m

Find:TV

T=1f__=

= 1.2x10-3 min___________revolution x 60 sec___________

1 min

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SOLUTION

sms

m

T

rv /25

072.0

)29.0(22

ANSWER

25 m/s

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SOLUTION #2:

830 revolutions in one minute

v= 25 m/s

Given:

r=0.29 m

Find:

V

v=2rf= 2(3.14)(0.29 m)(830 revolutions) x 1 min___________

60 sec_______________________

1 minute

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VELOCITYVELOCITY

▪ The velocity of an object in UCM has a constant magnitude and a constant change in direction

▪ The object does not have constant velocity since its direction changes at every point along the circle.

▪  Also, The velocity is always tangent to the path of the object.

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VELOCITY

In UCM, the velocity is always tangent to the path of the object. The tangent specifies the direction of the motion.

VA

VD

VB

VC

At a time t0, the car is located at point A with a velocity of VA,

this is tangent to the circle at point A.

This means that…The direction of the velocity of the Car at point A is due west.

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CENTRIPETAL ACCELERATION

CENTRIPETAL ACCELERATION

▪ in UCM, an object has no constant velocity.

Therefore, If there is a change in velocity, then that must mean that an object has acceleration.

Centripetal Acceleration. This acceleration has constant magnitude but changing direction and is directed radially inwards.

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CENTRIPETAL ACCELERATION

• Symbol: ac

• Vector • It is the rate of change of Tangential velocity• Always perpendicular to the path of the motion.

• Points toward the center of the circle.

VA

VD

VB

VC

A

B

C

D

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CENTRIPETAL ACCELERATION

FORMULA(S):

222

22

44

rfT

r

r

vac

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PROBLEM 2: Centripetal Acceleration

The bobsled track at the 1994 Olympics in Norway, contained turns with radii of 33 m and 24 m, as the figure illustrates. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event..

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From ac=v2/r

Radius=33m

Radius=24m

22

/3533

)/34(sm

m

smac

22

/4824

)/34(sm

m

smac

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The car in the drawing is moving clockwise around a circular road at a constant speed. What are the directions of its velocity and acceleration at (a) position 1 and (b) position 2? (c)What is the centripetal acceleration of the car if the radius of the circular road is 500 m and the car makes one revolution in 1 minute?

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GIVEN:

r=500m

f=1 rev/1 min

(a) The velocity is due south, and the acceleration is due west.(b) The velocity is due west, and the acceleration is due north.

(c) ac=5.48 m/s2

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21

CENTRIPETAL FORCE

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NEWTON’S SECOND LAW OF MOTION STATES THAT ALL ACCELERATIONS ARE CAUSED BY A NET FORCE ACTING ON AN OBJECT.  IN THE CASE OF UCM, THE NET FORCE IS A SPECIAL FORCE CALLED THE CENTRIPETAL FORCE . 

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CENTRIPETAL IS LATIN FOR "CENTER SEEKING". 

IT IS THE INWARD NET FORCE WHICH KEEPS AN OBJECT MOVING WITH A UNIFORM VELOCITY ALONG A CIRCULAR PATH. THIS FORCE IS DIRECTED ALONG THE RADIUS TOWARDS THE CENTER.

CENTRIPETAL FORCE

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F ma

Fmv

r

2

•APPLYING NEWTON’S SECOND

LAW OF MOTION

CENTRIPETAL FORCE

Centripetal Force

c

ac=v2/rc c

F=ma

Fc=mv2___r

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NNewton

N=kg m/s2

UNIT USED:

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PROBLEM 3A: Centripetal Force

• A 400-g rock attached to a 1.0-m string is whirled in a horizontal circle at a constant speed of 10.0m/s. Neglecting the effects of gravity, what is the centripetal force acting on the rock?

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SOLUTION:Given:

V=10.0 m/s

R=1.0 m

M=400g= 0.4 kg

Find: Fc

Fc== (0.4 kg)(10 m/s)2___________

1.0 m

mv2___r

40 N40 kg m/s2=

=

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PROBLEM 3B: Centripetal Force

• How long does it take a 50 kg runner to run a circular track starting and ending at the same point, if the radius of the track is 30m and a force of 68 N keeps him running at constant speed in the circular path?

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SOLUTION:Given:

M=50 kg

R=30 m

Fc=68 N

Find: T

Fc=mv2___r

29. 50 s=

=2r__T

m( )2

___________r

= 2___________

T2Fc=

= 2___________T2

Fc

= 2 ________T

Fc

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• FORCES SUCH AS THE GRAVITATIONAL FORCE (w=mg), TENSION FORCE (tied to a string; pushing/ pulling), FRICTIONAL FORCE (Ex. a car turning) and Normal force (on a surface) can be the centripetal forces

• GRAVITATIONAL FORCE. For satellites in orbit around a planet, the centripetal force is supplied by gravity.

•  TENSILE FORCE. For an object swinging around on the end of a rope in a horizontal plane, the centripetal force on the object is supplied by the tension of the rope.

SOURCES OF CENTRIPETAL FORCE

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DRAWING A FREE BODY DIAGRAM CAN HELP YOU UNDERSTAND THE PROBLEM BETTER.

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• A free body diagram,  is a pictorial device, used to analyze the forces and moments acting on a body.

What is included:

1. The body

2. The external forces: These are indicated by labelled arrows. The forces acting on the object include friction, gravity, normal force, drag, tension, or a human force due to pushing or pulling.

FREE BODY DIAGRAMFORCE DIAGRAM/

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For example:A man is pulling a box

Answer these questions:1. Is there gravity? (Fg)2. Is it sitting on a surface? (Fn)3. Is something pushing or

pulling it ?(Ft)4. Is their friction? (Ff)5. Is it accelerating? To what

direction?

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x

y

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For example:A man is pulling a box

1. Is there gravity? (Fg) YES, then draw gravity

Free body diagram

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x

y

FG

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For example:A man is pulling a box

2. Is it sitting on a surface? (Fn)YES, then draw the normal force

Free body diagram

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x

y

FN

FG

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For example:A man is pulling a box

3. Is something pushing or pulling it ?(Ft)

Yes, then draw the tension force

Free body diagram

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x

y

FN

FG

FT

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For example:A man is pulling a box

4. Is their friction? (Ff)

Yes, then draw friction

Free body diagram

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x

y

FN

FG

FTFf

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For example:A man is pulling a box

5. Is it accelerating? To what direction?

Yes, to the right

Free body diagram

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FREE BODY DIAGRAM

x

y

FN

FG

FTFf

a

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PROBLEM 4: Friction as Centripetal Force

A car is going around in a circular road. Given R, Ff b/w the tires and the road and m ; Find v.

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FREE BODY DIAGRAM

FN(on a surface)

Fg (weight)

Ff (friction)

a

And The frictional force is equal to the centripetal force

Ff=v2/r

Ff is the force going to the direction of the acceleration, therefore:

It is the one causing the centripetal acceleration

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ANSWER

V=10 m/s

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THE CENTRIFUGAL FORCE ACTS AWAY FROM THE CENTER. THE WORD ITSELF MEANS “FLEEING FROM THE CENTER” . THIS FORCE IS A FICTITIOUS FORCE. IT DOESN’T ACT ON A BODY IN MOTION, BUT ONLY ON NON-INERTIAL COORDINATE SYSTEMS SUCH AS A ROTATING ONE.

CENTRIFUGAL FORCE

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PROBLEM 5: Which way will the object go?

• An object on a guideline is in uniform circular motion. The object is symbolized by a dot, and at point O it is release suddenly from its circular path.

• If the guideline is cut suddenly, will the object move along OA or OP ?

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ANSWER:• the object would move along the straight line between points O and A, not on the circular arc between points O and P.

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SOLUTION:• NEWTON’S LAW OF MOTION:

“An object continues in a state of rest/ motion at a constant speed unless compelled to changes to its net force.”

When the object was suddenly released from its path, there was no longer a net force (i.e. centripetal force) being applied to the object.

In the absence of a net force, the object will continue to move at a constant speed but, along a straight line in the direction it had at the time of release.

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CENTRIFUGAL FORCE vs. INERTIA

If you let go of the rope (or the rope breaks) the object will no longer be kept in that circular path and it will be free to fly off on a tangent.

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53

Uniform circular motion emphasizes that 1. The speed, or the magnitude of the velocity vector, is constant.

2. Direction of the vector is not constant.

3. Change in direction, means acceleration

4. “Centripetal acceleration” , points toward the center of the circle.

5. “Centripetal Force” is the net force that causes centripetal acceleration

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A boy is horizontally whipping around a 5-gram rock attached to a 2 meter string in a circle above his head. The rock is in uniform circular motion and makes 20 revolutions in 5 seconds. Neglecting the effects of gravity, Find:(a) Period in seconds, (b) frequency in Hertz, (c) speed, (d) Centripetal Acceleration, (e) how much force is acting on the rock to keep it in UCM?

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(a) 0.25 s

T=1__f

PERIOD

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(b) 4 Hz

Hertz (Hz) = 1 rev/sec

FREQUENCY

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(c)

V=2rT

50.24 m/sSPEED

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r

vac

2

1,262 m/s2

(d)ACCELERATIONCENTRIPETAL

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6.310 N5g=0.005 kg

(e)

Fc=mv2___r

FORCECENTRIPETAL

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AP PHYSICS Circular Motion Mrs. CoylePractical ad Explorational Physics Padua et alYou and The Natural world Physics Navaza, ValdesINTERNETYOUTUBEGOOGLE……..

SOURCES:

• Francis Marlon Cabredo• Mary Minette Geñorga• Mary Judith Verdejo

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