Dynamics Applying Newton s Laws Circular Motion

Transcription

1 Dynamics Applying Newton s Laws Circular Motion Lana heridan De Anza College Oct 17, 2017

2 Last time friction problem solving with forces

3 nalysis Models Using Newton s econd Law 129 From last time: Pulley with an Incline ected by a Cord Let s change up our Atwood machine apparatus so that one of the masses is on a slanted surface with no friction: y tweight cord igure 5.15a. agnitude of. a m 1 AM m 2 a u T m 1 x g a Acceleration? Tension? If m 2 moves cts are conccelerations n y b

4 Pulley with an Incline Add eq (1) and (2): Putting a into (1): m 1 a = T m 1 g (1) m 2 a = m 2 g sin θ T (2) (m 1 + m 2 )a = m 2 g sin θ m 1 g a = (m 2 sin θ m 1 )g m 1 + m 2 m 1 (m 2 sin θ m 1 )g m 1 + m 2 = T m 1 g T = m 1m 2 (sin θ + 1)g m 1 + m 2 Does this agree with what we had for the Atwood machine when θ = 90?

5 Overview Circular motion and force Centripetal force Examples Non-uniform circular motion

6 Circular Motion - Now with Force If an object moves in a uniform circle, its velocity must always be changing. It is accelerating. 1 Figures from erway & Jewett.

7 Circular Motion - Now with Force If an object moves in a uniform circle, its velocity must always be changing. It is accelerating. F = ma F 0 1 Figures from erway & Jewett.

8 Circular Motion - Now with Force If an object moves in a uniform circle, its velocity must always be changing. It is accelerating. F = ma F 0 s Laws Any object moving in a circular (or curved) path must be experiencing a force. sion) We call this the centripetal force. irled in a circle he un in a perfectly pter 13) 1 Figures from erway & Jewett. F r a c v

9 Uniform Circular Motion For an object moving in a uniform circle, a = a c = v 2 r. This gives the expression for centripetal force: so, F = ma F c = mv 2 r As a vector: F c = mv 2 ˆr r

10 Centripetal Force omething must provide this force: 6.1 Extending the Particle in A force F r, directed toward the center of the circle, keeps the puck moving in its circular path. F r r m When the string breaks, the puck moves in the direction tangent to the circle. r Fr It could be tension in a rope. Figure 6.1 An overhead view of a puck moving in a circular path in a horizontal plane. Figure 6.2 The string holding th puck in its circular path breaks.

11 tion of the velocity vector will be smaller in a given time Centripetal Force ed tension in the string is smaller. As a result, the string radius. omething must provide this force: he Car? AM curve as shown fficient of static aximum speed f s large circle so It could be friction. e model the car n. The car is not m in the vertical a n

12 Centripetal Force Consider the example of a string constraining the motion 6.1 of Extending a puck: the Parti A force F r, directed toward the center of the circle, keeps the puck moving in its circular path. F r r m When the string breaks, the puck moves in the direction tangent to the circle. r Fr Figure 6.1 An overhead view of a puck moving in a circular path in a horizontal plane. Figure 6.2 The string hold puck in its circular path brea

13 Centripetal Force Question. What will the puck do if the string breaks? (A) Fly radially outward. (B) Continue along the circle. (C) Move tangentially to the circle.

14 Determine (a) the astronaut s orbital speed and (b) the Findingperiod Centripetal of the orbit. Force Example del tutorial available in ebassign 3. In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed lem of the electron is approximately m/s. Find torial available (a) in Enhanced the force acting on the electron as it revolves in a Pagecircular 169, # orbit 4 of BIOradius m and (b) the eo solution available centripetal in acceleration Q/C of the electron. ebassign 4. A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total horizontal force on the driver has magnitude 130 N. What is the total horizontal force on the driver if the speed on the same curve is 18.0 m/s instead? 5. In a cyclotron (one type of particle accelerator), a deuteron (of mass 2.00 u) reaches a final speed of 10.0% of the speed of light while moving in a circular path of radius m. What magnitude of magnetic force is required to maintain the deuteron in a circular path? 6. A car initially traveling y a 7. A d 3 w r t 8. C W o t m

15 Finding Centripetal Force Example Page 169, # 4

16 ngle u. Therefore, if g is smaller, as it is on Mars, the speed v with which the roa r. Ferris Wheel Forces A Ferris wheel is a ride you tend to see at fairs and theme parks. Ferris Wheel rris wheel as shown n a vertical circle of f 3.00 m/s. y the seat on the press your answer in g. AM Top v R n bot n t v Figure 6.6a. Based on a Ferris wheel or ay, you would expect path. During imilarly, the you ride the speed, v, is constant. Figure 6.6 (Example 6.5) (a) A child rides on a Ferris whee bottom of the path. mg mg Bottom a b c

17 Ferris Wheel Forces Quick Quiz You are riding on a Ferris wheel that is rotating with constant speed. The car in which you are riding always maintains its correct upward orientation; it does not invert. (i) What is the direction of the normal force on you from the seat when you are at the top of the wheel? (A) upward (B) downward (C) impossible to determine 1 Page 153, erway & Jewett

18 Ferris Wheel Forces Quick Quiz You are riding on a Ferris wheel that is rotating with constant speed. The car in which you are riding always maintains its correct upward orientation; it does not invert. (ii) From the same choices, what is the direction of the net force on you when you are at the top of the wheel? (A) upward (B) downward (C) impossible to determine 1 Page 153, erway & Jewett

19 m m/s 2 dius r banked at a fixed angle u. Therefore, if g is smaller, as it is on Mars, the speed v with which the roadway s traveled that the speed ly is also smaller. v is proportional to the square root of g for a roadway Finalize Equation (3) shows that the banking angle is independent of the mass of the vehicle negotiating the curve. If a ore, g is the smaller, it is speed vacceleration with which thetherefore, roadway carif rounds curve at aas speed lesson thanmars, 13.4 m/s,the the centripetal decreases. the normal force, which is unchanged, is sufficient to cause two accelerations: the lower centripetal acceleration and an acceleration of the car down the inclined roadway. Consequently, an additional friction force parallel to the roadway and upward is needed AM Top 6.5 to keepriding the carthe fromferris slidingwheel down the bank (to the left in Fig.v 6.5). imilarly, a driver attempting to negotiate the curve top net at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig. 6.5). mass m rides on a Ferris wheel as shown nbot H ATchild I F? moves Imagine this circle same of roadway were built on Mars in the future to connect different colony centers..6a. W The in a that vertical ntop it be speed traveled them/s. same speed? m atcould a constant of at 3.00 Ferris Wheel Assume the speed, v, is constant. n < mg : F points down R el mine Answer theam forcethe exerted by the seat ontop theforce on Mars would mean that the car is not pressed as tightly to the roadway. The reduced gravitational bottom of the ride. Express your answer inv component of the normal force toward the center of the circle. This smaller reduced normal force results in a smaller e weight of the child, component wouldmg. not be sufficient to provide the centripetal acceleration associated with the original speed. The censhown the speed v. tripetal acceleration must be reduced, which can be done by reducing nbot N Mathematically, notice that Equation (3) shows that the speed v is proportional to the square root of g for a roadway rcleofof fixed radius r banked at a fixed angle u. Therefore, if g is smaller, as it is on Mars, the speed vnwith which the roadway v top ize Look carefully at Figure 6.6a. Based mg mg can be safely traveled is also smaller. nces you may have had on a Ferris wheel or Bottom c r small hills on a roadway, you would expect R a b he at the top of the path. imilarly, you hter Figure 6.6 (Example 6.5) (a) A child rides on a Ferris wheel. ct to feel of the path. net bot wer in heavier6.5at the bottom (b) The forces at the bottom of the path. AMacting on the child Top Example e bottom of the path and Riding the top,the the Ferris nor- Wheel top of the path. (c) The forces acting on the child at the v avitational forces on the child act in opposite childsum of mass m rides on a Ferris TheAvector of these two forces gives awheel as shown nbot inmagnitude Figure 6.6a. The child a vertical circle of path at a constant speed. To yield net force vecnstant that keeps themoves child in moving in a circular radius 10.0 m at a constant speed of 3.00 m/s. e same magnitude, the normal force at the bottom must be greater than that at the top. R continued (A) Determine the force exerted by the seat on the v child at the bottom of the ride. Express your answer in Based terms of the weight of the child, mg. mg mg n heel or Bottom OLUTION expect a Conceptualize Look carefully at Figure 6.6a. Based y, you on experiences you may have had on a Ferris wheel or Figure 6.6 b > mg : F points up ntop c v Bottom mg (Example 6.5) (a) A child rides on a Ferris wheel. mg

20 A Banked Turn m/s m/s m harp turns in roads are often banked inwards to assist cars in making the turn: the centripetal force comes from the normal force, not friction. for the dry road. Figure 6.5 (Example 6.4) A car moves into the page and is roundn x y in Example 6.3 in such a way he curve without skidding. In negotiate the curve even when ed, which means that the roade opening photograph for this o be 13.4 m/s (30.0 mi/h) and the curve be banked? n u n y and Example 6.3 is that the 5 shows the banked roadway, the left of the figure. Notice rticipates in causing the car s u F g as a particle in equilibrium in

21 18.00 m/s2 2 A Banked Turn.80 m/s m for the dry road. A turn has a radius r. What should the angle θ be so that a car traveling at speed v can turn the corner without relying on friction? n x y in Example 6.3 in such a way he curve without skidding. In negotiate the curve even when ed, which means that the roade opening photograph for this o be 13.4 m/s (30.0 mi/h) and the curve be banked? n u n y and Example 6.3 is that the 5 shows the banked roadway, the left of the figure. Notice rticipates in causing the car s u F g as a particle in equilibrium in lar motion in the horizontal Figure 6.5 (Example 6.4) A car moves into the page and is rounding a curve on a road banked at an angle u to the horizontal. When

22 18.00 m/s2 2 A Banked Turn.80 m/s m for the dry road. A turn has a radius r. What should the angle θ be so that a car traveling at speed v can turn the corner without relying on friction? n x y in Example 6.3 in such a way he curve without skidding. In negotiate the curve even when ed, which means that the roade opening photograph for this o be 13.4 m/s (30.0 mi/h) and the curve be banked? n u n y and Example 6.3 is that the 5 shows the banked roadway, the left of the figure. Notice rticipates in causing the car s u F g as a particle in equilibrium in lar motion in the horizontal Figure 6.5 (Example 6.4) A car moves into the page and is rounding a curve on a road banked at an angle u to the horizontal. When Hint: consider what the net force vector must be in this case.

23 A Banked Turn n x uch a way dding. In ven when the roadh for this i/h) and? n u n y that the roadway, e. Notice the car s u F g ibrium in orizontal accelerathe preever, the Figure 6.5 (Example 6.4) A car moves into the page and is rounding a curve on a road banked at an angle u to the horizontal. When friction is neglected, the force that causes the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the normal force.

24 A Banked Turn.187 n x h a way ing. In n when e roadfor this h) and n u n y hat the adway, Notice he car s rium in izontal ccelerahe prever, the u F g Figure 6.5 (Example 6.4) A car moves into the page and is rounding a curve on a road banked at an angle u to the horizontal. When friction is neglected, the force that causes the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the normal force.

25 Non-Uniform Circular Motion If the speed is changing the net force will not be pointed into the center of the circle. The net force exerted on the particle is the vector sum of the radial force and the tangential force. Figure 6.7 When the net f ticle moving in a circular pa component o F t, the particle F r F F t

26 at points,, and. (b) uppose the stant tangential acceleration as it moves resenting the force on the bead at points Non-Uniform Circular Motion Figure 6.8 (Quick Example Quiz 6.2) A bead slides along a curved wire. Consider a ball of mass m swinging in a vertical circle, radius R. What is the tension T when the string makes an angle θ with the n the Ball vertical AMif the speed at that instant is v? end of a cord of length bout a fixed point O as tangential acceleration at any instant when the kes an angle u with the v top R T top mg he sphere in Figure 6.9 sociated with Example th. Unlike the child in phere is not uniform in ng the path, a tangenthe gravitational force ticle under a net force and article in uniform circues discussed in this sec- mg cos u u mg T mg sin u Figure 6.9 (Example 6.6) The forces acting on a sphere of mass m connected to a cord of length R and u O T bot mg v bot

27 Non-Uniform Circular Motion Example

28 le moving in a uniform magnetic field (Chapter 29) One More Example: Conical pendulum us in the Bohr model of the hydrogen atom (Chapter 42) A small ball of mass m is suspended from a string of length L. The ball revolves with constant speed v in a horizontal circle of radius r as shown in the figure. Find an expression for v in terms of the geometry in the figure. L. The ball revolves shown in Figure 6.3. system is known as a ometry in Figure 6.3. T L u T cos u u a and convince yourin a horizontal circle. tically. Therefore, we It experiences a cendeled as a particle in mg 1 erway & Jewett, page 152. a r mg Figure 6.3 (Example 6.1) (a) A b T sin u conical pendulum. The path of the ball is a horizontal circle. (b) The

29 One More Example: Conical pendulum

30 ummary Uniform circular motion with forces Centripetal force Banked turns Non-uniform circular motion (Uncollected) Homework erway & Jewett, Ch 6, onward from page 169. Questions: ection Qs 1, 5, 7, 9, 15, 17