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Physics 151: Lecture 5, Pg 1 Announcements: l Physics Learning Resource Center Open, –> room P207-C çOpen 9am - 5 pm Monday - Friday çHours also listed.

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Presentation on theme: "Physics 151: Lecture 5, Pg 1 Announcements: l Physics Learning Resource Center Open, –> room P207-C çOpen 9am - 5 pm Monday - Friday çHours also listed."— Presentation transcript:

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2 Physics 151: Lecture 5, Pg 1 Announcements: l Physics Learning Resource Center Open, –> room P207-C çOpen 9am - 5 pm Monday - Friday çHours also listed on www.phys.uconn.eduwww.phys.uconn.edu l NOTE: l Homework #1 (due this Fri. 9/8 by 5:00 pm EST on WebAssign) l Homework #2 (due next Fri. 9/15 by 5.00 pm)

3 Physics 151: Lecture 5, Pg 2 çProjectile motion çUniform circular motion çtext sections 4.1-4.5 Physics 151: Lecture 5 Today’s Agenda (Chapter #4)

4 Physics 151: Lecture 5, Pg 3 ACT -1 Question from past Exam-1 Which statement is true. 1.Initial speed of ball B must be greater than that of ball A. 2.Ball A is in the air for a longer time than ball B. 3.Ball B is in the air for a longer time than ball A. 4.Ball B has a greater acceleration than ball A. 5.None of above. Two balls, projected at different times so they don’t collide, have trajectories A and B, as shown.

5 Physics 151: Lecture 5, Pg 4 Review : Kinematics in 3D (2D) l Motion of objects in 3-D under constant acceleration (problem in projectile motion) :  y x ->independence of x- and y- components : y (t) x (t) a y = -g a x =0 yoyoyoyo xoxoxoxo t o = 0 t vxvx vyvy vovovovo v oy v ox y = y 0 + v 0y t - 1 / 2 g t 2 v y = v 0y - g t x = v x t v x = v 0x x-components (a=0) y-components (a=-g) Animation (v) Animation (a)

6 Physics 151: Lecture 5, Pg 5 Review : Kinematics in 3D (2D) (cont.)  y x y (t) x (t) a y = -g a x =0 yoyoyoyo xoxoxoxo t o = 0 t vxvx vyvy vovovovo v oy v ox l How many parameter determine projectile motion ? at t o = 0: x o, y o, v ox, v oy at t = t : x(t), y(t), v x, v y 9 variables plus a=g { we have 4 equations ! } 4 independent variables !

7 Physics 151: Lecture 5, Pg 6 Projectile Motion Typical Questions, 1.  x: How far will it go ? 2.  y: How high will it be at some distance ? 3. t: How long until it hits ? 4.  : At what angle should I start ? 5. v 0 : How fast must I start ? v0v0  g

8 Physics 151: Lecture 5, Pg 7 Projectile Motion Useful Things to Know If projectile begins and ends at same height, maximum distance is achieved for  = 45°. (prove it) x distance is same for angles about 45° if everything else remains the same. Time in flight depends on y equation if no barriers other than the earth interrupt the flight path. v0v0  g

9 Physics 151: Lecture 5, Pg 8 Projectile Motion / Example Problem v0v0  g UConn football team wants to complete a 45m pass (about 50 yards). Our qb can throw the ball at 30 m/s. At what angle must he throw the ball to get it there ? SOLUTION:  = 15° or 75°which gets there first ?

10 Physics 151: Lecture 5, Pg 9 Problem 3 (correlated motion of 2 objects in 3-D) l Suppose a projectile is aimed at a target at rest somewhere above the ground as shown in Fig. below. At the same time that the projectile leaves the cannon the target falls toward ground. t = t 1  y x v0v0v0v0 t = 0 TARGET PROJECTILE ( A ) MISS ( B ) HIT ( C ) CAN’T TELL Would the projectile now miss or hit the target ?

11 Physics 151: Lecture 5, Pg 10 Review ( displacement, velocity, acceleration ) Velocity : Acceleration : v a = dv / dt l 3-D Kinematics : vector equations: rrvrar r = r(t)v = dr / dta = d 2 r / dt 2 r v = dr / dt r v av =  r /  t v a av =  v /  t vv2vv2 vvvv -v -v 1 vv2vv2 y x path vv1vv1

12 Physics 151: Lecture 5, Pg 11 General 3-D motion with non-zero acceleration: l Uniform Circular Motion is one specific case of this :a v path t aa a = 0 because either or both: -> change in magnitude of v -> change in direction of v a = 0 a a a a = + Animation

13 Physics 151: Lecture 5, Pg 12 Uniform Circular Motion l What does it mean ? l How do we describe it ? l What can we learn about it ? See text: 4-4

14 Physics 151: Lecture 5, Pg 13 What is Uniform Circular Motion (UCM) ? l Motion in a circle with: çConstant Radius R v ç Constant Speed v = |v| çacceleration ? R v x y (x,y) See text: 4-4 = 0a a = const.a

15 Physics 151: Lecture 5, Pg 14 How can we describe UCM? l In general, one coordinate system is as good as any other: ç Cartesian: » (x,y) [position] » (v x,v y ) [velocity] ç Polar: » (R,  )[position] » (v R,  ) [velocity] l In UCM: ç R is constant (hence v R = 0).   (angular velocity) is constant. ç Polar coordinates are a natural way to describe UCM! R v x y (x,y)  See text: 4-4

16 Physics 151: Lecture 5, Pg 15 Polar Coordinates: l The arc length s (distance along the circumference) is related to the angle in a simple way: s = R , where  is the angular displacement.  units of  are called radians. l For one complete revolution: 2  R = R  c   c = 2   has period 2 . R v x y (x,y) s  See text: 4-4 1 revolution = 2  radians

17 Physics 151: Lecture 5, Pg 16 Polar Coordinates... l In Cartesian co-ordinates we say velocity dx/dt = v. ç x = vt In polar coordinates, angular velocity d  /dt = .   =  t   has units of radians/second. l Displacement s = vt. but s = R  = R  t, so: R v x y s  t v =  R

18 Physics 151: Lecture 5, Pg 17 Period and Frequency l Recall that 1 revolution = 2  radians çfrequency (f) = revolutions / second (a)  angular velocity (  ) = radians / second (b) l By combining (a) and (b)   = 2  f l Realize that: çperiod (T) = seconds / revolution  So T = 1 / f = 2  /  R v s   = 2  / T = 2  f v R s

19 Physics 151: Lecture 5, Pg 18 Lecture 5, ACT 2 Uniform Circular Motion l A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

20 Physics 151: Lecture 5, Pg 19 Acceleration in UCM: l This is called l This is called Centripetal Acceleration. l Now let’s calculate the magnitude: vv1vv1 vv2vv2 vvvv vv2vv2 vv1vv1 R RRRR Similar triangles: But  R = v  t for small  t So:

21 Physics 151: Lecture 5, Pg 20 Lecture 5, ACT 2 Uniform Circular Motion l A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ? (a) 20 m (b) 200 m (c) 2000 m (d) 20,000 m

22 Physics 151: Lecture 5, Pg 21 Recap for today: l Recap of Projectile Motion l Introduce Uniform Circular Motion l Reading assignment for Monday çReread Circular Motion: Ch 4.4-4, pp.91-95 çRead Relative Velocity: Ch 4.5, pp.95-99 Homework #1 (due this Fri. 9/8 by 5:00 pm EST on WebAssign) Problems from Chapter 1 and 2 l Homework #2 (due next Fri. 9/15 by 5.00 pm) Problems from Chapter 3 and 4


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