2. 2.4.1Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. 2.4.2Apply the expressions for centripetal acceleration. 2.4.3Identify the force producing circular motion in various situations. Examples include friction of tires on turn, gravity on planet/moon, cord tension. 2.4.4Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion

3.  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? Topic 2: Mechanics 2.4 Uniform circular motion On the next pass, however, Helen failed to clear the mountains. r v m

4. Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle.  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! Topic 2: Mechanics 2.4 Uniform circular motion x y r v r blue v red

5. Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle.  In order to find the direction of the acceleration (a = ∆v/∆t ) we observe two nearby snapshots of the particle:  The direction of the acceleration is gotten from ∆v = v 2 – v 1 = v 2 + (-v 1 ):  The direction of the acceleration is toward the center of the circle- you must be able to sketch this. Topic 2: Mechanics 2.4 Uniform circular motion x y r blue v red v1v1 v2v2 v1v1 v2v2 ∆v∆v v1v1 v2v2 -v 1 ∆v∆v -v1-v1

6. Apply the expressions for centripetal acceleration.  Centripetal means center-seeking.  How does centripetal acceleration a c depend on r and v?  We define the centripetal force F c :  Picture yourself as the passenger in a car that is rounding a left turn:  The sharper the turn, the harder you and your door push against each other. (Small r = big F c.)  The faster the turn, the harder you and your door push against each other. (Big v = big F c.) Topic 2: Mechanics 2.4 Uniform circular motion F c = ma c centripetal force FcFc

7. PRACTICE: For each experiment A and B, label the control, independent, and dependent variables. Apply the expressions for centripetal acceleration. Topic 2: Mechanics 2.4 Uniform circular motion FcFc acac v r AB FcFc acac v r FcFc acac v r FcFc acac v r  CONTROL: r  INDEPENDENT: v  DEPENDENT: F c and a c  CONTROL: v  INDEPENDENT: r  DEPENDENT: F c and a c manipulated No change responding no change manipulated responding

8. Apply the expressions for centripetal acceleration.  We know the following things about a c :  If v increases, a c increases.  If r increases, a c decreases.  From dimensional analysis we have  What can we do to v or r to “fix” the units?  This is the correct one! Topic 2: Mechanics 2.4 Uniform circular motion ac =ac = vrvr ac =ac = vrvr  ms2ms2 = ? 1s1s m/s m = ? a c = v 2 /r centripetal acceleration ac =ac = v2rv2r  ms2ms2 = ? ms2ms2 = ? first guess formula m 2 /s 2 m

9. Apply the expressions for centripetal acceleration. Topic 2: Mechanics 2.4 Uniform circular motion a c = v 2 /r centripetal acceleration EXAMPLE: A 730-kg Smart Car negotiates a 30. m radius turn at 25. m s -1. What is its centripetal acceleration and force? What force is causing this acceleration? SOLUTION:  a c = v 2 /r = 25 2 /30 = 21 m s -2.  F c = ma c = (730)(21) = 15000 n.  The centripetal force is caused by the friction force between the tires and the pavement. F c = ma c centripetal force

10. Apply the expressions for centripetal acceleration.  The period T is the time for one complete revolution.  One revolution is one circumference C = 2  r.  Therefore v = distance / time = 2  r/T.  Thus v 2 = 4  2 r 2 /T 2 so that  a c = v 2 /r = 4  2 r 2 /T 2 r = 4  2 r/T 2. Topic 2: Mechanics 2.4 Uniform circular motion a c = v 2 /r centripetal acceleration a c = 4  2 r/T 2

11. Apply the expressions for centripetal acceleration. Topic 2: Mechanics 2.4 Uniform circular motion a c = v 2 /r centripetal acceleration a c = 4  2 r/T 2 EXAMPLE: Albert the 2.50-kg physics cat is being swung around your head by a string harness having a radius of 3.00 meters. He takes 5.00 seconds to complete one fun revolution. What are a c and F c ? What is the tension in the string? SOLUTION:  a c = 4  2 r/T 2 = 4  2 (3)/(5) 2 = 4.74 m s -2.  F c = ma c = (2.5)(4.74) = 11.9 n.  The tension is causing the centripetal force, so the tension is F c = 11.9 n. Albert the Physics Cat

12. EXAMPLE: 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. Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion B A C

13. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (a) Sketch in the forces acting on the mass. SOLUTION:  The ONLY two forces acting on the mass are its weight W and the tension in the string T.  Don’t make the mistake of drawing F c into the diagram.  F c is the resultant of T and W, as the next questions will illustrate. Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion 1.50 m W T

14. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (b) Draw a FBD in the space provided. Then break down the tension force in terms of the unknown tension T. SOLUTION:  T x = T cos  = T cos 60° = 0.5T.  T y = T sin  = T sin 60° = 0.87T. Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion W   1.50 m W T T TxTx TyTy

15. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (c) Find the value of the components of the tension T x and T y. SOLUTION:  Note that T y = mg = 4(10) = 40 n.  But T y = 0.87T.  Thus 40 = 0.87T so that T = 46 n.  T x = 0.5T = 0.5(46) = 23 n. Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion W  T TxTx TyTy

16. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (d) Find the speed of the mass as it travels in its circular orbit. SOLUTION:  The center of the UCM is here:  Since r / 1.5 = cos 60°,r = 0.75 m.  F c = T x = 23 n.  Thus F c = mv 2 /r 23 = 3v 2 /0.75 v = 2.4 m s -1. Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion 1.50 m  r

17. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of R E = 6400000 m, find the speed of the ball. SOLUTION:  The ball is traveling in a circle of radius r = 6408850 m.  F c is caused by the weight of the ball so that F c = mg = (0.5)(10) = 5 n.  Since F c = mv 2 /r we have 5 = (0.5)v 2 /6408850 v = 8000 m s -1 ! Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion

18. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). How long will it take the ball to return to Everest? SOLUTION:  We want to find the period T.  We know that v = 8000 m s -1.  We also know that r = 6408850 m.  Since v = 2  r/T we have T = 2  r/v T = 2  (6408850)/8000 T = (5030 s)(1 h / 3600 s) = 1.40 h. Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion

19. 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! Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion

20. Preparation for future topics.  UCM will be revisited in Topic 4 Oscillations and waves. We now extend UCM a bit to get ready.  In uniform circular motion the x and y coordinates of the particle at any instant can be determined using the following relationships:  We call  the angular position of the particle.  Take note that  varies with time. Topic 2: Mechanics 2.4 Uniform circular motion x y θ r x y x = r cos θ relationship between rotational and linear variables y = r sin θ

21. EXAMPLE: Find the x- and y-components of a particle whose path follows a circle of radius 3.00 m at the precise instant its angular position is θ = 135°. Then sketch it into the grid. SOLUTION:  r = 3 m and θ = 135°.  x = r cos θ = 3 cos 135° = -2.12 m.  y = r sin θ = 3 sin 135° = +2.12 m. Preparation for future topics. Topic 2: Mechanics 2.4 Uniform circular motion x y x = r cos θ relationship between rotational and linear variables y = r sin θ (-2.12 m, 2.12 m )

22. EXAMPLE: A particle in uniform circular motion at a radius of 3.00 m has an x-coordinate of -2.00 m and a y-coordinate that is negative. Sketch the particle. Then find the values of y and θ. SOLUTION:  r 2 = x 2 + y 2 so that 3 2 = 2 2 + y 2 or y =  2.24 m.  x = 2 (adj) and r = 3 (hyp) so cos θ = adj/hyp = 2/3 θ = cos -1 (2/3) = 48.2°.  From +x dir, 180° + 48.2° = 228°. Preparation for future topics. Topic 2: Mechanics 2.4 Uniform circular motion x y x = r cos θ relationship between rotational and linear variables y = r sin θ r 

23. Preparation for future topics.  Just as we defined displacement in linear motion as ∆x = x 2 – x 1 we can define angular displacement in circular motion to be ∆  =  2 –  1.  And just as we defined a velocity as v = ∆x/∆t, we define angular velocity  = ∆  /∆t.  At first glance we would expect that the units for  = ∆  /∆t are (deg/s) but we would be wrong!  The next slides will explain. Topic 2: Mechanics 2.4 Uniform circular motion ∆  =  2 –  1 angular displacement and angular velocity  = ∆  /∆t ang. displacement angular velocity x y θ1θ1 θ2θ2 t1t1 t2t2 ∆∆

24. Preparation for future topics.  Some of you may already know the formula for arc length s.  Radians is a more natural way to measure angular displacement than degrees. Here is how they are related.  Arc length s is measured in meters (m) and tells us how far we have traveled in an arc.  Angular velocity (or angular speed)  is measured in radians per second (rad/s). Topic 2: Mechanics 2.4 Uniform circular motion s = r∆θ arc length s ∆θ in radians  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev

25. Preparation for future topics. Topic 2: Mechanics 2.4 Uniform circular motion  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev EXAMPLE: How far have you traveled if you are on the outer perimeter of a merry-go-round having a diameter of 7.00 m and it has turned 135°? Sketch your arc. SOLUTION:  We are looking for arc length s.  ∆θ = 135°. But s needs radians.  135°(  rad/180°) = 2.356 rad.  r = diameter/2 = 7/2 = 3.5 m.  s = r∆θ = (3.5)(2.356) = 8.25 m.  The red arc is 8.25 m long. 0°0° 45° 90° 135° 180° 225° 270° 315° s = r∆θ arc length s ∆θ in radians

26. Preparation for future topics. Topic 2: Mechanics 2.4 Uniform circular motion  rad = 180° = 1/2 rev radian-degree-revolution conversions 2  rad = 360° = 1 rev s = r∆θ arc length s ∆θ in radians PRACTICE: How far have you traveled if you are 2.75 m from the center of a merry-go-round having a diameter of 7.00 m and it has turned 135°? Sketch your arc.  We are looking for arc length s.  ∆θ = 135°. But s needs radians.  135°(  rad/180°) = 2.356 rad.  But r = 2.75 m so that  s = r∆θ = (2.75)(2.356) = 6.48 m. 0°0° 45° 90° 135° 180° 225° 270° 315° FYI  Arc length depends on arc radius.

27. Preparation for future topics.  Recall that velocity v is just the displacement s divided by the time. Thus v = s/∆t = r∆θ/∆t = r(∆θ/∆t) = r   Angular speed  is therefore related to linear speed v like this: Topic 2: Mechanics 2.4 Uniform circular motion s = r∆θ arc length s ∆θ in radians v = r  relation between v and   in rad/s

28. Preparation for future topics. Topic 2: Mechanics 2.4 Uniform circular motion v = r  relation between v and   in rad/s EXAMPLE: The second hand of a clock has a length of 12.0 cm. What are its angular speed (in rad/s) and the linear speed, of its tip, and its base? SOLUTION:  The second hand makes one revolution each 60 s.  One revolution is 2  rad.  Therefore  = 2  rad / 60 s = 0.105 rad s -1.  Angular speed is the same for the base and tip.  v = r  = (0.12)(0.105) = 0.013 m s -1 for the tip.  v = r  = (0)(0.105) = 0 m s -1 for the base. 12 3 6 9

29. Preparation for future topics.  We define the frequency f of uniform circular motion as how many revolutions or cycles the object in UCM executes per unit time.  Since the period T is how many seconds per cycle, the relationship between T and f is simply  Frequency f is measured in (cycles / second) which are known as (Hertz).  Since v = 2  r/T and v = r , we can derive 2  r/T = r  2  /T =  2  f =  Topic 2: Mechanics 2.4 Uniform circular motion f = 1/T relation between T and f T = 1/f  = 2  /T relation between , T and f  = 2f = 2f

30. Preparation for future topics. Topic 2: Mechanics 2.4 Uniform circular motion f = 1/T relation between T and f T = 1/f  = 2  /T relation between , T and f  = 2f = 2f EXAMPLE: The wheel of a car rotates once each 0.500 s. Find its period, frequency, and angular speed. SOLUTION:  The period T is just 0.500 s.  The frequency f = 1/T = 1/0.5 = 2.00 Hz.  The angular speed can be found in two ways.  METHOD 1:  = 2  f = 2  (2) = 12.6 rad s -1.  METHOD 2:  = ∆  /∆t = (2  rad)/(0.5 s) = 12.6 rad s -1.