Physics 201, Lecture 10

Transcription

1 Physics 201, Lecture 10 Today s Topics n Circular Motion and Newton s Law (Sect. 6.1,6.2) n Centripetal Force in Uniform Circular Motion n Examples n n Motion in Accelerated Frame (sec. 6.3, conceptual understanding) Motion with Resistance (sec. 6.4, slides at the end, self reading) n Hope You ve Previewed Chapter 6 (and also Chapter 5)

2 Uniform Circular Motion and Newton s Law q Recall: Centripetal Acceleration v v v a c a c r a c a c v v = rω = 2πr/T, always in tangential direction a c = rω 2 = v 2 /r, always pointing to the center a c v q Now, per Newton s 2 nd law, there must be a net-force that is responsible for a c. F = ma c this F is called Centripetal Force (F c ) q F c can be in the form of tension, friction, gravitation, or combination of them

3 Example: Ball on a String in Horizontal Circular Motion. q Exercise: A ball attached on a string of length r is in uniform circular motion, if the ball is moving at a (constant) linear speed v, what is the tension T in the string? Solution: the only force in the horizontal plane is the tension (which serves as the centripetal force) T = ma c = m v 2 /r demo: what if the string is cut? view in horizontal plane (top view)

4 Quick Quiz: What If the Centripetal Force is Lost q For the above circular motion, how will the ball continue to fly if the string is cut off? Ø answer path #3.

5 Example/Demo: Conical Pendulum What is the period of the conical pendulum? y x Solution Draw FBD as shown y direction: ΣF y = Tcosθ-mg =0 à T= mg/cosθ T x directioon: ΣF x = Tsinθ = ma x =mv 2 /r trigonometry: r = Lsinθ à (g/cosθ) sinθ = v 2 /Lsinθ v=sqrt(lg sinθtanθ) Period T= 2πr/v = 2π sqrt(lcosθ/g)

6 Example: Car at a Turn (Level Road) q When a car is turning along a horizontal curve, the static friction between the tire and the road surface supplies the required centripetal force. f s = ma c = m v 2 /r r let µ s be the coefficient of static friction f s < µ s n = µ s mg (can you see n=mg?) à v 2 /r< µ s g v < µ gr s Quiz: why is static friction used here? answer: there is no relative motion in radial direction

7 Example: Car Turning on a Banked Curve q In cases of low friction road surface, (or when speed is high), road turns are designed to be banked. In such cases, normal force provides the required centripetal force. De-compose normal force n: F c = n x = nsinθ. Exercise: show r v = rg tanθ (see board)

8 Demo/Exercise: Roller Coaster q What is the minimum speed at the top of a roller coaster? A q at top point A: F c = mg + T top = mv 2 /R à v 2 = (mg + T top ) R/m > mgr/m =gr (note: T top >0) à v > gr

9 Roller Coaster Quiz q In this roller coaster design that the cart is rolling above the track, at top point B, the cart s speed can not be too high, can not be too low, no limit. Ø answer: at top point B: F c = mg - N top = mv 2 /R < mg

10 Non-Uniform Circular Motion q In a generic (non-uniform) circular motion, acceleration usually has both centripetal and tangential components a = a c + a t ΣF = ΣF r + ΣF t a c a t ΣF r = ma c, ΣF t = ma t Conceptual understanding only for this course

11 Example of non-uniform Circular Motion q Consider a mass in vertical circular motion with varying speed At any point, the centripetal force is provided by a combination of tension T and a component of gravitation mgcosθ F c = T- mgcosθ = ma c = mv 2 /R 2 v T = m( + R g cosθ ) Conceptual understanding only for this course

12 Quiz: Block in Accelerating Car a 0 q A block on the frictionless floor of an accelerating train. To a bystander on the ground, what is the blocks acceleration? 0, +a 0, -a 0, other Newton s 2 nd Law (in earth frame) : F = ma, F=0, a=0 To the observer standing inside the train, what is the block s acceleration? (standing=no relative motion) 0, +a 0, -a 0, other Newton s 2 nd Law (in train frame) : F=0, a = - a 0, F=ma?

13 Fictitious Force q Newton s 2 nd Law is valid only in the inertia reference frame i.e. IF a is measure in an inertia reference frame F real = ma q In an accelerating frame (a 0 ), the 2 nd is not valid. v To force the form of 2 nd law, one has to add an fictitious force F fictitious = -Ma 0 into the equation F = F real + F fictitious = ma

14 Motion In Accelerated Frame q Newton s 2 nd Law Applies only in inertial reference frame q One can derive the 2 nd Law in accelerated (non-inertial) reference frame: Reference Frame A: inertial, F = ma =m dv/dt Reference Frame B: Moving w.r.t to Frame A with a 0 =dv 0 /dt In Frame B: vʹ = v - v 0 aʹ = dvʹ /dt = dv/dt dv 0 /dt maʹ = F ma 0 =Fʹ è a fictitious force F fictitious =-ma 0 has to be introduced to, artificially, keep the same form of the 2 nd Law Conceptual Understanding only

15 Newton s 2 nd Law: Two Practical Approaches q First Principle: Newton s 2 nd Law Applies only in inertial reference frame q Approach 1: Working in inertia reference frame: F = m a Straightforward, but may need to do Galilean transformation. q Approach 2: Working in a non-inertia frame of acceleration a 0 Introduce a fictitious force F fictitious = - ma 0 Add the fictitious force to the real force: F = F + F fictitious è So we can, artificially, keep the same form of the 2 nd Law maʹ =F (= F ma 0 )

16 Quiz: Test Your Imagination Quiz: A lady is sitting on a rotating table watch a wood bock which is also fixed on the table. The distance between the lady and the block is r. To the lady s view, the motion of the block is: No motion, Circular motion with radius r, Circular motion with radius R, Motion in more complicated curve r R

17 Example: Fictitious Force In Circular Motion T=ma T= mv 2 /r Centrifugal Force F fictitious =-T = -ma (a=0, in rotating frame)

18 One more Example: Fictitious Force In Linear Motion Inertial Observer: y: ΣF y = Tcosθ - mg = 0 x: ΣF x = Tsinθ = mgtanθ = ma tanθ = a/g Observer In the Car (Noninertial) y: ΣF y =Tcosθ-mg =0 x: ΣF x =Tsinθ ma =0 F fictitious = - ma in x direction Study after class

19 Motion with Resistance Force q So far, we have considered on free fall for projectile motion Free Fall m dv/dt = mg Falling with Resistance R m dv/dt = mg R examples Model 1: R=-bv Model 2: R=1/2 DρA v 2

20 Terminate Speed q If we only care about the maximum speed the falling object can reach, the math is quite simple: At maximum speed: mg=r (quiz: why?) This maximum speed is called terminate speed (v T ) Eg. if R = bv v T = mg/b o r if R = 1/2DρAv 2 v T = 2mg DρA Self reading to know the meaning of quantities in the equation.

21 Some Terminal Speeds