Ch 5 Uniform Circular Motion

Uniform Circular Motion Means constant speed in a circle. Velocity is not constant.

A centripetal force accelerates a body by changing the direction of the body's velocity without changing the body's speed.

Mass gm Period sec data 1 2 3 4 5 6 7 1 2 50 100 200 250 500

Finding Tangental Velocity T 2 v r

Table 1: Mass, and Period of Revolution. Mass Period gm sec data 1 2 3 4 1 2 50 100 200 300 1kg weight data 1 gm period data 2 sec 20 1000gm 9.81 N kg 5 6 400 500 radius 75cm <--This radius must remain constant. 7 8 600 700 M stopper_actual gm Vel 2 radius period <--This is the stopper distance around the circle (circumference). <--This is the time it takes the stopper to complete one revolution around the circle.

Rubber Stopper Velocity Graph 1: Rubber Stopper Velocity as a Function of S tring Tension 23.387 Velocity vs. String Tension Vel m s best_fit 6.065 0.491 weight N String T ension 6.867

Graph 1 12 Velocity vs. Weight Vel m s best_fit m s 10 8 6 4 0 1 2 3 4 5 weight N

Graph 2 12 Velocity vs. Weight??? Vel m s best_fit m sec 10 8 6 4 0.5 1 1.5 2 2.5 weight N

Graph 3 150 Velocity vs. Weight??? Vel m s 2 best_fit m sec 100 50 k 24.7 m kg 0 0 1 2 3 4 5 weight N

Orbital Centripetal Force Caused by Gravitation: How fast does the earth zip around the sun? qoluxivo.cwahi.net v earth 2R Sun to Eath T revolution

2R Sun to Earth v earth T revolution v earth 365 day 2 1.5010 11 m 24 hr 1day 3600 sec 1hr v earth 3.0 10 4 m v 6.7 10 4 s mi hr qoluxivo.cwahi.net

How strong is the sun's pull on the Earth? M earth v earth 2 F c r 610 F c 24 3 10 4 kg 1.510 11 m m s Centripetal Force 2 F c 3.6 10 22 N F c 8.1 10 21 lbf F c 8 100 000 000 000 000 000 000 pounds

Force of Attraction Betwee n Sun and Earth F G M E M S d 2 Law of Universal Gravitation F 6.6710 11 N m 2 kg 2 5.9810 24 kg 1.9910 30 kg 2 1.510 11 m F 3.528 10 22 N

How fast does the moon zip around the earth? v moon 2 R Moon to Earth T revolution aerospaceweb.org

v moon 2R Moon to Earth T revolution v moon 27.3 day 2( 240000mi ) 24 hr 1day 3600 sec 1hr aerospaceweb.org v moon 1.0 10 3 m v s moon 2.3 10 3 mi hr

How strong is the earth's pull on the moon? Mv 2 F c r 22 2300 mi 7.3510 F c kg 240000mi hr 2 F c 2.012 10 20 N F c 4.5 10 19 lbf F c 12 000 000 000 000 000 pounds

A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed v keeps the cylinder at rest?

Mass on Table Problem Mg provides Centripetal Force. F c Mg Mg mv 2 r

Mass on Table Problem Mg mv 2 r v Mgr m

Problem *19: occupant s mass = 220 kg Tension? Chair speed?

60.0 T 15.0 m +y r +x g 9.807 m s 2 mg F y 0 mg Tcos ( 65 ) T m g cos ( 65 deg )

60.0 T 15.0 m +y r +x mg m 220 kg g 9.807 m s 2 F y 0 mg Tcos ( 65 ) T m g cos ( 65 deg ) T 5.105 10 3 N

60.0 T 15.0 m +y r +x mg Mv 2 F c T 5105 N M 220 kg F c Tsin( 65 deg ) r r c ( 12.0m ) sin( 65 deg ) v 2 F c r c M v Tsin( 65 deg ) r c M v 15.1 m s

5.4 Banked Curves On an unbanked curve, the static frictional force provides the centripetal force.

5.4 Banked Curves On a frictionless banked curve, the centripetal force is the horizontal component of the normal force. The vertical component of the normal force balances the car s weight.

5.4 Banked Curves F c 2 v FN sin m F N cos mg r

5.4 Banked Curves 2 v F N sin m r F N cos mg tan v 2 rg

5.4 Banked Curves Example 8: The Daytona 500 The turns at the Daytona International Speedway have a maximum radius of 316 m and are steely banked at 31 degrees. Suppose these turns were frictionless. At what speed would the cars have to travel around them? tan v 2 rg v rg tan v 2 316 m 9.8m s tan31 43m s 96 mph

To review HORIZONTAL TRACK

For Horizontal Track F N mg F c F s s mg and m v 2 Fc so r s mg m v 2 r v 2 r s g

Centripetal Force

Mv 2 F c r r 2 is 3x greater.

2 2 v 1 v 2 a 1 a 2 r 1 r 2 a 1 2.45 m a 2 0.817 m s 2 s 2 a 1 3 a 2

http://www.youtube.com/watch?v=_m0qbowx0gm&list=pl908c129cc34ece18&index=18

6P9. A car goes around a curve of radius 48m. If the road is banked at an angle of 15 deg with the horizontal, at what maximum speed in kilometers per hour may the car travel if there is to be no tendency to skid even on very slippery pavement?

F c F N sin15 m v 2 r mg F N cos 15 tan 15 m v 2 r mg tan 15 m v 2 rmg tan 15 v 2 rg v tan 15 rg

F c F N sin15 m v 2 r mg F N cos 15 tan 15 m v 2 r mg tan 15 m v 2 rmg tan 15 v 2 rg v tan 15 rg v tan rg v 11.2 m v 40.4 km s hr

For Banked Tracks tan 1 F c mg tan 1 2 m v r mg tan 1 v 2 rg

Ch 5 C&J 3 rd e Problem *24

Ch 5 C&J 3 rd e Problem *24 A plane with a speed of v 195 m is flying s in a horizontal circle with a radius of r 8250 m. Calculate the banking angle of the plane to accomplish this task.

tan opp adj F L sin F L cos F c Mg Mv 2 r Mg tan Mv 2 r Mg tan v 2 r g tan v 2 rg

tan 195 m s 2 ( 8250m ) 9.807 m s 2 tan 0.47 25.2

An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted 40 to the horizontal, what is the radius of the circle in which the plane is flying? 480 km hr R =? 133 m s

F N sin F N cos tan v 2 Rg M v 2 R Mg R v 2 tan g R 2.1 10 3 m

F c Problem **57 Page 158 Source of F c? Calculate F c. Calculate.

5.6 Apparent Weightlessness and Artificial Gravity Example 13: Artificial Gravity At what speed must the surface of the space station move so that the astronaut experiences a push on his feet equal to his weight on earth? The radius is 1700 m. F c m v r 2 mg v rg 2 1700 m 9.80m s

Relating Orbital Radius & Orbital Speed of a Satellite G m M E m v 2 F c r 2 r v 2 G M E r

Apparent Weightlessness Astronauts on a space walk around the space station still have mg! They are falling with the space station, but keep missing the earth because of sufficient horizontal (orbital) speed.

What is the speed of the earth as it orbits around the sun? (Consider the orbit to be circular.

Solution: The law of universal gravitation and the centripetal force equation can be related, since they both equal a force. In fact, the force of the earth due to its gravitational attraction to the sun is the centripetal force needed to maintain orbit. M E v 2 r F c G M E M S r 2 v 2 G M E M S r F c M E r 2

v 2 G M E M S M E r v 2 v G M S r G M S r v 6.6710 11 Nm 2 kg 2 1.5010 11 m 1.99 10 30 kg v 2.97 10 4 m s

Alternately, T 2R v Period is circumference of a circular orbit divided by orbital velocity. v 2R Sun to Eath T revolution v 2 1.5010 11 m v 2.99 10 4 24 hr 365 day 3600 sec 1day 1hr m s

5.7 Vertical Circular Motion 2 v1 F N 1 mg m r 2 v F 2 N 2 m r 2 v F 4 N 4 m r 2 v3 F N 3 mg m r

Assuming that the loop is a circle with radius R = 2.7 m, what is the minimum v that Diavolo could have at the top of the loop to remain in contact with it there?

F c N F g where N 0 at apex

F net,y = ma y ) gives us F c F N mg m v 2 R v 2 Rg v Rg v 5.1 m s

Problem **54

Problem **54 v 27.0 m d =? s 45 deg F N sin M v 2 R F N cos Mg tan v 2 Rg

Solving for the Radius, then Distance R v 2 R 74.3m tan g R d dcos R d 105m cos

Problem **26 Page 155 15.0deg R 175 s 0.800 v =?

F N sin opposes F c!

F c F s cos F N sin Mv 2 R

Resolving forces into x & y components:

The net horizontal force is the c. F F c F s cos F N sin Mv 2 R Since F s s F N we get...

F N sin Mv 2 u s F N cos R Since F N is common to both terms, we take it out: sin F N u s cos Mv 2 R

F y 0 Vertical Static Equilibrium F y F s sin F N cos Mg 0 Since F s s F N F y s F N sin F N cos Mg 0

F N s sin cos Mg 0 Now solve for F N so that it's result can replace the F N of the F c equation.

F N u s sin cos Mg Mg F N u s sin cos

Using the first equation defining F replace F N with s sin Mg c, cos. sin F N s cos Mv 2 R

s sin Mg cos s cos sin Mv 2 R s sin g cos s cos sin v 2 R

s sin Rg cos s cos sin v 2 Rg s cos sin s sin cos v 2

Rg s cos sin v s sin cos v 27.4 m s

Solved Problem 1 Page 152

r A 2150m g V 8.62 m s 2 a) T

a c v 2 r 2 v A g V 8.62 m r A s 2

2 2 r A T r A 8.62 m s 2

4 2 r A 8.62 T 2 m s 2

T 2 4 2 r A 8.62 m s 2

T 99.2s

Problem *36 Page 156 T 1min T 1min 1 r A 4.00 r B 1.00 g 10.0 m s 2 r A r B a c B r A 4.00 r B 1.00 r A r B g A 10.0 m s 2 g B

a c v 2 r 2 v A g A 10.0 m r A s 2 v A 2r A T substitutes into 2 v A 10.0 m r A s 2

2 2 r A T r A 10.0 m s 2

which simplifies to 4 2 r A 10.0 T 2 m s 2

10.0 m T 2 s 2 r A 4 2

10.0 m s 2 ( 60.0sec ) 2 r A 2 r A 912m 4 1.00 r B r A r B 228m 4.00

a B v B 2 But we need to know v B,so we'll substitute r B 2r B v B into it: T a B 2 r B 2 T a B 2.50 m r B s 2

Class Practice & Review What is the speed of the earth as it orbits around the sun? (Consider the orbit to be circular.)

M E v 2 G M E M S F c r r 2 v 2 G M E F c M E r 2 M S r

v 2 G M E M S M E r v 2 G M S r

v 6.6710 11 Nm 2 kg 2 1.99 10 30 kg 1.5010 11 m v 2.97 10 4 m s

Alternately, T 2 R Period is circumference of a circular orbit divided by orbital velocity. v

2 R Sun to Eath v T revolution

v 2 1.5010 11 m 365 day 24 hr 1day 3600sec 1hr v 2.99 10 4 m s

from Physics by Cutn ell & Johnson A baggage carousel at an airport shows a suitcase that has no t slid all the way dow the slope. T he suitcase is going around in a circle at constant speed as the carous turns. T he coefficient of static friction between the suitcase and the carousel is s 0.76and the angle of incline from the horizontal is 36 deg. The horizontal circle described by the moving suitcase is r 11 m. How much time is required for the suitcase to go around once?

The net horizontal force is the F c. F c F s cos F N sin Mv 2 r

T 2r v Period is circumference of a circular orbit divided by orbital velocity. v 2R T After solving for v, substitute into the F c equation: F c F s cos F N sin Mv 2 r becomes F s cos F N sin M 2r R T 2 when solved for v, becomes F s cos F N sin M 4 2 r 2 R T 2

F s cos F N sin M 4 2 r Since T 2 F s s F N F N sin M 4 2 s F N cos T 2 r F N s cos sin M 4 2 r T 2

But we don't know F N, so we can't find T just yet. Since F s s F N s F N cos Mv F N sin 2 Since F N is common to both terms, we take it ou r sin F N s cos Mv 2 r F y 0 Vertical Static Equilibrium

Since F s s F N s F N cos Mv F N sin 2 Since F N is common to both terms, we take it out: r sin F N s cos Mv 2 r F y 0 Vertical Static Equilibrium F y F s sin F N cos Mg 0 Since F s u s F N F y s F N sin F N cos Mg 0 Since F N is common to the first two terms, we take it out: F N s sin cos Mg 0 Now solve for F N so that it's result can replace the F N of the F c equation. F N s sin cos Mg