2. Understandings: Period, frequency, angular displacement and angular velocity Centripetal force Centripetal acceleration Topic 6: Circular motion and gravitation 6.1 – Circular motion Applications and skills: Identifying the forces providing the centripetal forces such as tension, friction, gravitational, electrical, or magnetic Solving problems involving centripetal force, centripetal acceleration, period, frequency, angular displacement, linear speed and angular velocity Qualitatively and quantitatively describing examples of circular motion including cases of vertical and horizontal circular motion

3. Guidance: Banking will be considered qualitatively only Data booklet reference: v =  r a = v 2 / r = 4  2 r / T 2 F = mv 2 / r = m  2 r Utilization: Motion of charged particles in magnetic fields (see Physics sub- topic 5.4) Mass spectrometry (see Chemistry sub-topics 2.1 and 11.3) Playground and amusement park rides often use the principles of circular motion in their design https://www.youtube.com/watch?v=IcsZL2GPY0s

4. Aims: Aim 6: experiments could include (but are not limited to): mass on a string; observation and quantification of loop-the-loop experiences; friction of a mass on a turntable Aim 7: technology has allowed for more accurate and precise measurements of circular motion, including data loggers for force measurements and video analysis of objects moving in circular motion Topic 6: Circular motion and gravitation 6.1 – Circular motion

5. Centripetal force and acceleration  What force must be applied to Helen to keep her moving in a circle?  How does it depend on the Helen’s radius r ?  How does it depend on Helen’s velocity v?  How does it depend on Helen’s mass m? On the next pass, however, Helen failed to clear the mountains. r v m

6. Centripetal force and acceleration  A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v.  Observe that the velocity vector is always tangent to the circle.  Note that the magnitude of the velocity vector is NOT changing.  Note that the direction of the velocity vector IS changing.  Thus, there is an acceleration, even though the speed is not changing! x y r v r blue v red

7. ∆θ∆θ ∆r ∆v ∆θ∆θ r 1 r 2 -v 1 P O Q v 1 v 2 An object is moving in circular path at constant speed. magnitude and direction of centripetal acceleration. Velocities at two different positions P and Q have the equal magnitude (v 1 = v 2 = v), but different directions. ~ ∆ ∆ magnitude of Direction of acceleration is toward the centre of the circle ∆v -v 1 v 2 Must be able to draw it

8. v 1 v 2 ∆v direction of a c : v 2 ∆v v 1 v 2 ∆v = – looking at ∆t → 0 ∆v is in the same direction as ∆v points toward the center of the circle centripetal or radial acceleration “center-seeking” along radius

9.  velocity – tangent to the circle  centripetal acceleration – points toward the center velocity and centripetal acceleration vectors are always perpendicular to each other If centripetal acceleration were suddenly to disappear, there would be nothing to change the direction of the velocity. The object would continue to move in a straight line with constant speed. In that case that object moves “inertially” (because of inertia of the body).

10. Dynamics of uniform circular motion Object undergoing uniform circular motion is accelerating with centripetal acceleration a c, so it has a force acting upon it: which must be directed toward the center of the circle. It is called centripetal force.

11. it is not separate force – it is simply one of our familiar forces acting in the role of causing circular motion Many forces can force an object to move in circular path, therefore becoming centripetal force: Moon around the Earth ……… gravity object sitting on a rotating table (strawberries sitting on a seat of a turning car) (a car moving in a circular path) ………… friction a ball whirling in a circle at the end of a string ……… tension in the string a person pressed against the inner wall of a rapidly rotating circular room in an amusement park …………. normal force from the wall Ferris–wheel rider passes through the lowest point of the ride ……. normal force from the seat and the force of gravity

12. useful relations: - speed(linear) and angular speed constant speed  period T: time required for one complete revolution (s) speed v = distance/time angular speed ω = angle swept/time Centrifugal (center-fleeing) Force - MISCONCEPTION

13. Circular Motion - The radius of a spacecraft orbiting earth is 6.67 x 10 6 m. If it orbits earth in 5292 seconds, what is the velocity of the spacecraft? 7919 m/s Jimmie Johnson is driving his #48 Lowe’s NASCAR around a bend that has a radius of 70 meters. It takes him 30 seconds to travel the track. What was the centripetal acceleration of Jimmie John’s #48 Lowe’s NASCAR? 3.07 m/s 2 When a wheel rotates about a fixed axis, do all the points on the wheel have the same tangential speed? Yes Do they all have the same velocity? No

14. A 1000 kg car is going around a curve with radius 30 meters. If the coefficient of friction between the car's tires and the road is 0.5, what is the maximum speed at which the car can make the turn? m = 1000 kg r = 30 m g = 9.8 m/s 2 maximum speed in the turn, v = ? v = 12 m/s

15. Angular displacement and arc length  Consider the rotating arm which has 6 paint cans along its radius.  Each can has a spout that is opened for exactly a quarter of a revolution.  We call  the angular displacement.  All 6 color trails represent the same angular displacements of 90˚.  Each color traces out a different displacement s.  We call s the arc length.  All 6 color trails represent different arc lengths.  s s s s s

16. EXAMPLE: Suppose the red line is located at a radius of 1.50 m and the green line is located at 1.25 m. Find their lengths. SOLUTION: 90  (  rad / 180°) = 1.57 rad.  s = r  = 1.50  1.57 = 2.4 m.  s = r  = 1.25  1.57 = 2.0 m. Angular displacement and arc length  The relationship between angular displacement  and arc length s is where r is the radius. s = r  relation between s and   in radians  rad = 180° = 1/ 2 rev radian-degree-revolution conversions

17. Angular speed and speed  Because speed is v = s / t, we see that v = s / t = ( r  ) / t = r (  / t )    / t = r  (define    / t )  Thus  We call  the angular speed. Topic 6: Circular motion and gravitation 6.1 – Circular motion s = r  relation between s and   in radians v = r  relation between v and   =  / t (rad s -1 )

18. EXAMPLE: Consider the following point mass moving at a constant speed v in a circle of radius r as shown. Find … (a) the period T of the point mass, and (b) the frequency f of the point mass, and (c) the angular speed  of the point mass. SOLUTION: We need a time piece.  For one revolution the period is T = 12 s.  Frequency f = 1 / T = 1 / 12 = 0.083 s.  Angular speed is  =  / t = 2  rad / 12 s = 0.52 rad s -1. Angular speed and speed v = r  relation between v and   =  / t (rad s -1 ) r v

19. FYI  Speed depends on length or position but angular speed does not. EXAMPLE: Find the angular speed of the second hand on a clock. Then find the speed of the tip of the hand if it is 18.0 cm long. SOLUTION: A second hand turns 2  rad each 60 s.  Thus it has an angular speed given by  = 2  / T = 2  / 60 = 0.105 rad s -1.  The speed of the tip is given by v = r  = 0.180(0.105) = 0.0189 ms -1. Angular speed and speed v = r  relation between v and   =  / t (rad s -1 )

20. EXAMPLE: A car rounds a 90° turn in 6.0 seconds. What is its angular speed during the turn? SOLUTION:  Since  needs radians we begin by converting  :  = 90°(  rad / 180° ) = 1.57 rad.  Now we use  =  / t = 1.57 / 6.0 = 0.26 rad s -1. Angular speed and speed v = r  relation between v and   =  / t (rad s -1 )

21. Banking  The car is able to round the curve because of the friction between tire and pavement.  The friction always points to the center of the circle.  So, how does a plane follow a circular trajectory?  There is no sideways friction force that the plane can use because there is no solid friction between the air and the plane.

22. FYI  It is the ROLL maneuver that gives a plane a centripetal force as we will see on the next slide. Banking  Using control surfaces on the tail and the main wings, planes can execute three types of maneuver: ROLL – Ailerons act in opposing directions YAW – Tail rudder turns left or right PITCH – Ailerons and horizontal stabilizer act together

23. Banking  As the plane banks (rolls), the lift vector begins to have a horizontal component.  The centripetal force causes the plane to begin traveling in a horizontal circle.

24. FYI  A banked curve can be designed so that a car can make the turn even if it is perfectly frictionless! Banking  Even though cars use friction, roads are banked so that the need for friction is reduced.  Instead of a component of the LIFT force providing a centripetal force, a component of the NORMAL force does so. W R FCFC

25. Angular velocity  As speed with a direction is called velocity, angular speed with a direction is called angular velocity.  To assign a direction to a rotation we use a right hand rule as follows: 1. Rest the heel of your right hand on the rotating object. 2. Make sure your fingers are curled in the direction of rotation. 3. Your extended thumb points in the direction of the angular velocity. FYI  Angular velocity always points perpendicular to the plane of motion! Topic 6: Circular motion and gravitation 6.1 – Circular motion  = 2  / T = 2  f =  / t relation between , T and f r v 

26. PRACTICE: Find the angular velocity (in rad s -1 ) of the wheel on the shaft. It is rotating at 30.0 rpm (revolutions per minute). SOLUTION:  The magnitude of  is given by  = (30.0 rev / 60 s)(2  rad / rev) = 3.14 rad s -1.  The direction of  is given by the right hand rule: “Place heel of right hand so fingers are curled in direction of rotation. Thumb gives the direction.” Angular velocity  = 2  / T = 2  f =  / t relation between , T and f

27. PRACTICE: Identify at least five forces that are centripetal in nature: SOLUTION:  The tension force (Albert the physics cat and Arnold).  The friction force (the race car making the turn).  The gravitational force (the baseball and the earth).  The electric force (an electron orbiting a nucleus).  The magnetic force (a moving charge in a B-field). Identifying the forces providing centripetal forces

28. PRACTICE: Dobson is watching a 16-pound bowling ball being swung around at 50 m/s by Arnold. If the string is cut at the instant the ball is next to the ice cream, what will the ball do? (a) It will follow path A and strike Dobson's ice cream. (b) It will fly outward along curve path B. (c) It will fly tangent to the original circular path along C. Solving centripetal acceleration and force problems B A C

29. EXAMPLE: Explain how an object can remain in orbit yet always be falling. SOLUTION:  Throw the ball at progressively larger speeds.  In all instances the force of gravity will draw the ball toward the center of the earth.  When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface.  The ball is effectively falling around the earth!

30. The balls are thrown horizontally. If there was no gravity the balls would continue horizontally for ever at the same speed. But one second after the balls are thrown, because of gravity, they had fallen 5 m below horizontal line no matter how fast they were thrown. If the second ball is thrown twice as fast it will go twice as far in the same time.. field If the ball were thrown three times as fast, it will go three times as far in the same time. Ten times as fast, ten times further. What if the ball were thrown so fast that the curvature of the earth came into play??? The ball follows a curved down path (parabola). But the Earth is curved too. What if the ball were thrown so fast that the curved path matched the Earth’s curve? If there were no air resistance (no slowing down and eventually hitting the Earth) the ball would orbit Earth. An Earth satellite (space station, communication satellite, scientific satellites,… ) is simply projectile traveling fast enough to fall around the Earth rather than into it.

31. Satellites in circular orbit, such as the moon or space station, fall beneath the paths they would follow if there were no gravity – straight line. During each second the moon travels about one km. In this distance it deviates about one millimeter from a straight line due to the earth’s gravitational pull (dotted line). The moon continually falls toward the earth, as do the planets around the sun. At 8 km/s atmospheric friction would melt a piece of iron (falling stars). Therefore satellites are launched to altitudes above 150 km. Don’t even try to think they are free of Earth’s gravity. The force of gravity at that altitude is almost as strong as it is at the surface. The only reason we put them there is that they are beyond Earth's atmosphere, not beyond Earth’s gravity. A satellite launched with speeds less than 8 km/s would A satellite launched with speeds less than 8 km/s would eventually fall to the Earth. A satellite launched with a eventually fall to the Earth. A satellite launched with a speed of 8 km/s would orbit the Earth in a circular path. speed of 8 km/s would orbit the Earth in a circular path. Launched with a greater speed satellite would orbit the Earth in Launched with a greater speed satellite would orbit the Earth in an elliptical path. If launched with too great of a speed, a satellite/projectile will escape Earth's gravitational influences and continue in motion without actually orbiting the Earth. Such a projectile will continue in motion until influenced by the gravitational influences of other celestial bodies.

32. EXAMPLE: The Foucault pendulum is a heavy pendulum on a very long cable that is set in oscillation over a round reference table. Explain how it can be used to tell time. SOLUTION:  The blue arcs represent the motion of the pendulum bob relative to the universe at large.  The the green lines represent the plane of motion of the pendulum relative to the building.

33. FYI  This solution only works when the pendulum is at one of the poles. See the Wiki for a general solution. EXAMPLE: The Foucault pendulum is a heavy pendulum on a very long cable that is set in oscillation over a round reference table. Explain how it can be used to tell time. SOLUTION:  Since the building is rotating with the earth at  = 0.0000727 rad s -1, each hour the green line rotates by  =  t = 0.0000727(3600) = 0.262 rad (360  / 2  rad) = 15.0 . Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion

34. EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole. SOLUTION: Recall that  = 0.0000727 rad s -1 anywhere on the earth.  The blue arcs represent the lines of latitude.  The white line R represents the earth’s radius.  The yellow line r represents the radius of the circle a point at a latitude of  follows.  Note that r = R cos , and that at the equator,  = 0˚ and at the pole,  = 90˚. Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion  R r  0˚ 90˚ 

35. EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole. SOLUTION: Recall that  = 0.0000727 rad s -1 anywhere on the earth.  Thus, at the equator, r = R, and at the pole, r = 0. Furthermore, R = 6400000 m.  Then, at the equator, a c = r  2 = 6400000  0.0000727 2 = 0.0338 ms -2.  Then, at the pole, a c = r  2 = 0  0.0000727 2 = 0.000 ms -2. Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion  R r  

36. EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole. SOLUTION: Make a free-body diagram at the equator…  Scales read the normal force R:  F = ma R – W = - ma c R = W – ma c  Then, R = 882 – ( 882 / 9.8 )  0.0338 = 879 N.  The man has apparently “lost” about 3 N! Solving centripetal acceleration and force problems W R acac

37.  kx = F C = mv 2 / r implies that as v increases, so does the centripetal force F C needed to move it in a circle.  Thus, x increases.  Use F = kx (k = CONST).  kx = F  k = F / x = 18 / 0.010 = 1800 Nm -1.  F C = kx = 1800( 0.265 – 0.250 ) = 27 N.  F C = v 2 / r  v 2 = r F C = 0.265(27) = 7.155  v = 2.7 ms -1.

38. Topic 6: Circular motion and gravitation 6.1 – Circular motion  At P r = R v = R  a = R  2  At Q r = 2R v = 2R  = 2v a = 2R  2 = 2a

39. Solving centripetal acceleration and force problems Topic 6: Circular motion and gravitation 6.1 – Circular motion  Objects moving in uniform circular motion feel a centripetal (center-seeking) force.