Dynamics Applying Newton’s Laws Circular Motion Lana Sheridan De Anza College April 27, 2015 Overview • Circular motion and force • Centripetal force • Examples • Non-uniform circular motion 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 Serway & Jewett. 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 6= 0 1 Figures from Serway & Jewett. 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 6= 0 n’s Laws Any object moving in a circular (or curved) path must be experiencing a force. nsion)We call this the centripetal force. S SJOHPGDPOTUBOUMFOHUI wirled in a circle PSDFBDUJOHPOBQMBOFU the Sun in a perfectly apter 13) FBDUJOHPOBDIBSHFE 1 Figures from Serway & Jewett. !F S ac r S v Uniform Circular Motion For an object moving in a uniform circle, a = ac = This gives the expression for centripetal force: F = ma so, Fc = mv 2 r v2 r . Centripetal Force Something must provide this force: 6.1 Extending the Particle in S A force Fr , directed toward the center of the circle, keeps the puck moving in its circular path. S Fr r m When the string breaks, the puck moves in the direction tangent to the circle. r S Fr Figure 6.1 It could be tension An overhead view of a puck moving in a circular path in a inhorizontal a rope.plane. Figure 6.2 The string holding th puck in its circular path breaks. S ction of the velocity vector will be smaller in a given time Centripetal Force red tension in the string is smaller. As a result, the string r radius. Something must provide this force: the Car? AM S fs a curve as shown efficient of static maximum speed a large circle so a It could be friction. we model the car on. The car is not um in the vertical S n Centripetal Force Consider the example of a string constraining the motion a puck:the Parti 6.1 of Extending S A force Fr , directed toward the center of the circle, keeps the puck moving in its circular path. S Fr r m When the string breaks, the puck moves in the direction tangent to the circle. r S 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 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. Centripetal Force Question. What will the puck do if the string breaks? (A) Fly radially outward. (B) Continue along the circle. 6.1 Extending the Particle in Uniform Circular Motion Model ← (C) Move tangentially to the circle. S A force Fr , directed toward the center of the circle, keeps the puck moving in its circular path. S Fr r m When the string breaks, the puck moves in the direction tangent to the circle. S S Fr Figure 6.1 An overhead view of a puck moving in a circular path in a horizontal plane. r v Figure 6.2 The string holding the puck in its circular path breaks. 15 Determine (a) the astronaut’s orbital speed and (b) the of the orbit. Findingperiod Centripetal Force Example odel tutorial available in WebAssign 3. In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.20 3 10 6 m/s. Find utorial available in Enhanced (a) the force acting on the electron as it revolves in a 210 BIO circular Page 169,in#orbit 4 of radius 0.529 3 10 m and (b) the deo solution available Q/C centripetal acceleration of the electron. blem WebAssign 4. A curve in a road forms part of a horizontal circle. As a car goes around Sit 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 0.480 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 Finding Centripetal Force Example Page 169, # 4 Fc = mv 2 r Finding Centripetal Force Example Page 169, # 4 Fc = m r mv 2 r = Fc v2 = 130 N (14.0m/s)2 = 0.6633 kg/m Finding Centripetal Force Example Page 169, # 4 Fc = m r mv 2 r = Fc v2 = 130 N (14.0m/s)2 = 0.6633 kg/m Fc = m 2 v r = (0.6633 kg/m)(18.0 m/s)2 = 215 N angle u. Therefore, if g is smaller, as it is on Mars, the speed v with which the roa er. Ferris Wheel Forces A Ferris wheel is a ride you tend to see at fairs and theme parks. e Ferris Wheel AM Top S v erris wheel as shown in a vertical circle of of 3.00 m/s. S nbot S nt R by the seat on the xpress your answer in mg. S v Figure 6.6a. Based S S mg mg on a Ferris wheel or Bottom c way, you would expect a b path. During Similarly, the you ride the speed, v , is constant. Figure 6.6 (Example 6.5) (a) A child rides on a Ferris whee e bottom of the path. Ferris Wheel Forces Quick Quiz 6.11 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, Serway & Jewett Ferris Wheel Forces Quick Quiz 6.11 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, Serway & Jewett Ferris Wheel Forces Quick Quiz 6.11 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, Serway & Jewett Ferris Wheel Forces Quick Quiz 6.11 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, Serway & Jewett 1 35.0 m 2 1 9.80 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.Sv 6.5). Similarly, 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 S nbot H ATchild I F ? moves Imagine this circle same of roadway were built on Mars in the future to connect different colony centers. S .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 S 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 S 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 S rcleofof S 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 S S 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. Similarly, 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 S 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 S 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 S v child at the bottom of the ride. Express your answer in Based S S terms of the weight of the child, mg. mg mg n heel or Bottom SOLUTION 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 S ntop c S v S Bottom mg (Example 6.5) (a) A child rides on a Ferris wheel. S mg A Turn 1 8.00Banked m/s 2 2 9.80 m/s 2 1 35.0 m 2 2 5 0.187 Sharp turns in roads are often banked inwards to assist cars in for the dry road. making the turn: the centripetal force comes from the normal force, not friction. M n x ay in Example 6.3 in such a way the curve without skidding. In negotiate the curve even when ked, which means that the roadhe opening photograph for this to be 13.4 m/s (30.0 mi/h) and d the curve be banked? S n u ny ← e and Example 6.3 is that the .5 shows the banked roadway, o the left of the figure. Notice articipates in causing the car’s as a particle in equilibrium in u S Fg Figure 6.5 (Example 6.4) A car moves into the page and is round- 1 8.00 m/s 2 2 5 0.187 9.80 m/s2 2 1 35.0 m 2 A Banked Turn 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? M ay in Example 6.3 in such a way the curve without skidding. In negotiate the curve even when ked, which means that the roadhe opening photograph for this to be 13.4 m/s (30.0 mi/h) and d the curve be banked? e and Example 6.3 is that the .5 shows the banked roadway, o the left of the figure. Notice articipates in causing the car’s as a particle in equilibrium in ular motion in the horizontal nx S n u u ny S Fg 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 1 8.00 m/s 2 2 5 0.187 9.80 m/s2 2 1 35.0 m 2 A Banked Turn 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? M ay in Example 6.3 in such a way the curve without skidding. In negotiate the curve even when ked, which means that the roadhe opening photograph for this to be 13.4 m/s (30.0 mi/h) and d the curve be banked? e and Example 6.3 is that the .5 shows the banked roadway, o the left of the figure. Notice articipates in causing the car’s as a particle in equilibrium in Hint: consider what ular motion in the horizontal nx S n u u ny S Fg Figure 6.5 (Example 6.4) A car moves into the page and is round- the vector ing anet curveforce on a road bankedmust at an be in this case. angle u to the horizontal. When A Banked Turn 5 0.187 such a way idding. In even when t the roadph for this mi/h) and d? s that the roadway, re. Notice g the car’s librium in horizontal l acceleran the prewever, the nx S n u y -direction (vertical): ny Fy ,net = 0 ny − mg u S Fg 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. = 0 A Banked Turn 5 0.187 such a way idding. In even when t the roadph for this mi/h) and d? s that the roadway, re. Notice g the car’s librium in horizontal l acceleran the prewever, the nx S n u y -direction (vertical): ny Fy ,net = 0 ny − mg = 0 n cos θ = mg n = u S Fg 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. mg cos θ A Banked Turn 0.187 ch a way ding. In en when he roadh for this /h) and that the oadway, . Notice he car’s brium in rizontal ccelerathe preever, the x-direction (horizontal): nx S n u u ny S Fg 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. Fx,net = Fc mv 2 nx = r A Banked Turn 0.187 ch a way ding. In en when he roadh for this /h) and that the oadway, . Notice he car’s brium in rizontal ccelerathe preever, the x-direction (horizontal): nx S n u u ny S Fg 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. Fx,net = Fc mv 2 nx = r mv 2 n sin θ = r mg sin θ = cos θ tan θ = mv 2 r v2 ⇒ θ = tan−1 rg v2 rg 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. S !F S ! Fr S ! Ft Figure 6.7 When the net f ticle moving in a circular pa component o F t , the particle at points !, ", and #. (b) Suppose the nstant tangential acceleration as it moves presenting the force on the bead at points # Non-Uniform Circular Motion Example Figure 6.8 (Quick 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 AMif the speed at that instant is v ? on the Ball vertical e end of a cord of length about a fixed point O as tangential acceleration d at any instant when the akes an angle u with the S vtop S Ttop S mg R O the sphere in Figure 6.9 ssociated with Example ath. Unlike the child in sphere is not uniform in ong the path, a tangenm the gravitational force rticle under a net force and particle in uniform circuues discussed in this sec- S T mg cos u u u S Tbot S vbot mg sin u S mg S mg Figure 6.9 (Example 6.6) The forces acting on a sphere of mass m connected to a cord of length R and Non-Uniform Circular Motion Example In the radial direction: Fr ,net = mv 2 R Non-Uniform Circular Motion Example In the radial direction: Fr ,net = mv 2 R T − mg cos θ = mv 2 R T T mv 2 + mg cos θ R 2 v = mg + cos θ Rg = NBHOFUJDGPSDFBDUJOHPOBDIBSHFE icle moving in a uniform magnetic field (Chapter 29) One More Example: Conical pendulum MFDUSJDGPSDFBDUJOHPOBOFMFDUSPOJOPSCJUBSPVOEB eus 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. h L. The ball revolves shown in Figure 6.3. system is known as a eometry in Figure 6.3. L u S T cos u T u r T sin u 3a and convince yours in a horizontal circle. rtically. Therefore, we . It experiences a cenodeled as a particle in S S mg mg a b Figure 6.3 (Example 6.1) (a) A conical pendulum. The path of the 1 Serway & Jewett, ball pageis 152. a horizontal circle. (b) The One More Example: Conical pendulum x-direction: Fnet,x = mv 2 = T sin θ r (1) y-direction: Fnet,y = 0 = T cos θ − mg T cos θ = mg (2) One More Example: Conical pendulum x-direction: Fnet,x = mv 2 = T sin θ r (1) y-direction: Fnet,y = 0 = T cos θ − mg T cos θ = mg Divide (1) by (2): tan θ = v2 rg Replace r = L sin θ and rearrange: p v = Lg sin θ tan θ (2) Summary • Uniform circular motion with forces • Centripetal force • Banked turns • Non-uniform circular motion (Uncollected) Homework Serway & Jewett, • Ch 6, onward from page 169. Questions: Section Qs 1, 5, 7, 9, 15, 17

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