Physics 2210 Fall smartphysics 20 Conservation of Angular Momentum 21 Simple Harmonic Motion 11/23/2015

Transcription

1 Physics 2210 Fall 2015 sartphysics 20 Conservation of Angular Moentu 21 Siple Haronic Motion 11/23/2015

2 Exa 4: sartphysics units Midter Exa 2: Day: Fri Dec. 04, 2015 Tie: regular class tie Section 01 12:55 1:45 p Section 10 02:00 02:50 p Location: FMAB 15 practice probles for Exa 4 posted on CANVAS and will be posted to Class Web Page Soon

3 Law of Conservation of Angular Moentu Definition (1) : Angular Moentu of a rotor (sign of L follows sign of ω) about P L Iω, L II Definition (2) : Angular Moentu of a particle, in linear otion, about P L r Here r is the lever-ar: perpendicular distance (of closest approach) fro the line-of-otion to the rotation axis Sign of L: is positive if particle isses P to the right Negative if it isses P to the left I P ω L II Total angular oentu of a syste (rotors and particles defined above) is the su (with sign: they are vectors!!!) of the angular oenta of all coponents. In the absence of external torque about a pivot (rotation axis) P, the total angular oentu (su of individual angular oenta) of a syste is conserved. v L P r Line-of- Motion

4 Exaple 19-1 (1/3) A sall puck of ass =100 g travels with initial speed v 0 =25/s to the right. It strikes and sticks to a rod (initially at rest) that is free only to rotate about the pivot P at one end, in the plane of the page. The rod has ass M=1.4 kg and length l=1.2. The point of ipact is at a distance of b=0.80 fro the pivot. Find the angular velocity ω of the stick after the collision. Solution: TOP VIEW OVER ICE In general, collisions are NOT elastic CANNOT assue energy is conserved We would norally say that total oentu (of the syste of the stick and puck) is conserved, but only if there are no external forces acting But the pivot provides WHATEVER EXTERNAL FORCE NECESSARY to prevent the stick fro detaching fro the rotation axis: Total Moentu is DEFINITELY NOT conserved Fortunately, the force of the pivot acts at the rotation axis: so it exerts NO (EXTERNAL) TORQUE abut P Total Angular Moentu of the syste (puck +stick) about P is CONSERVED v 0 b M P l

5 Exaple 19-1 (2/2) Puck =100 g, v 0 =25 /s, hits and sticks to rod: M=1.4 kg, l=1.2 at b=0.80. Find ω of the puck+rod after the collision. Solution by Conservation of Angular Moentu Angular Moentu (about P) is conserved Before: (stick at rest) L i = 0 r = 0 b After: rod and putty becoe joined, with total oent-of-inertia: I (rrr+pppp) = I rrr + b 2 And the syste now rotates at angular velocity ω L f = I (rrr+pppp) ω = Ml b 2 ω Total Angular Moentu is conserved: L f = L i, Ml b 2 ω = 0 b 0 b ω = Ml b = 0.10 kg 25 s kg kg = 2.72 rad/s Note the change in kinetic energy K i = 1 2 v 0 2 = kg 25 s 2 = J K f = 1 2 I (rrr+pppp)ω 2 = kg kg rad/s 2 = 2.72 J

6 Exaple 19-2 (1/4) A sall puck of ass =100 g travels with initial speed v 0 =25/s to the right. It strikes and sticks to a rod (initially at rest) that is free both to translate and rotate in the plane of the page. The rod has ass M=1.4 kg and length l=1.2. The point of ipact is at a distance of b=0.80 fro the pivot. Find the final linear and angular velocity. TOP VIEW OVER ICE v 0 b M l In general, collisions are NOT elastic CANNOT assue energy is conserved NO external force (in the x and y directions) Total Moentu is conserved NO (EXTERNAL) TORQUE abut the center-of-ass (CM) of puck+rod assebly Total Angular Moentu of the syste (puck+rod) about CM is CONSERVED How do you conserve both linear and angular oentu??? The linear oentu is that of a (fictional) point particle at the CM with total ass + M Angular oentu is conserved about the CM of the syste at oent of collision

7 Exaple 19-2 (2/4) Puck of ass =100 g, v 0 =25/s, strikes and sticks to a rod: M=1.4 kg, l=1.2, at b=0.80. Find v f and ω f b r cc l/2 y ω f v f M l x The puck and rod fors a new syste whose CM is offset fro the center of the rod toward the poi t at which the puck is stuck r cc = 1 + M l b + M 2 = 0.1kg kg kg kg The CM of the puck+rod assebly will travel in a straight line Before the collision:p ii = v 0, P iy = 0. = After the collision the CM of the assebly ust be traveling in the +x direction at v f. And the assebly is rotating about the CM at angular velocity ω f

8 Exaple 19-2 (3/4) Puck of ass =100 g, v 0 =25/s, strikes and sticks to a rod: M=1.4 kg, l=1.2, at b=0.80. Find v f and ω f r cc = Before the collision: P ii = v 0, P iy = 0. After the collision P fx = + M v f, P fy = 0. Moentu is conserved + M v f = v 0 v f = v 0 0.1kg 25 = s + M 0.1kg + 1.4kg = 1.67 s Angular oentu about CM before collision: L i = +v 0 r = +v 0 b r cc = 0.1kg 25 s = 0.466kg 2 s After the collision: L f = Iω f Where I is the oent-of-inertia about the CM of the puck+rod assebly, I = I rrr + I pppp By Parallel Axes Theore: I rrr = 1 12 Ml2 + M r cc l 2 = kg kg = kg 2 2

9 Exaple 19-2 (4/4) Puck of ass =100 g, v 0 =25/s, strikes and sticks to a rod: M=1.4 kg, l=1.2, at b=0.80. Find v f and ω f r cc = 0.613, L i = +v 0 b r cc = 0.466kg 2 s After the collision: L f = Iω f, I = I rrr + I pppp By Parallel Axes Theore: I rrr = 1 12 Ml2 + M r cc l 2 2 = kg 2 And the puck is a point ass at a distance of b r cc fro the CM I pppp = b r cc 2 = b r cc 2 = 0.1kg = kg 2 So the total oent of inertia of the puck+rod assebly about its own CM is I = I rrr + I pppp = kg kg 2 = kg 2 Angular oentu about the CM of the rod+puck assebly is conserved Iω f = L i ω f = L i 0.466kg 2 s = = 2.72 rad/s I kg 2 Total final kinetic energy: K f = M v f Iω f 2 = 2.72 J

10 The archetype siple haronic oscillator b is a positive length L = L 0 b x = b F = +kkı L = L 0 x = 0 F = 0 Equilibriu Position L = L 0 + b x = +b F = kkı In each case we have the usual relationship between force (exerted by the spring on the block) and the position (displaceent fro equilibriu) of the block: F = kkı Applying Newton s 2 nd Law then we have (noting that a x = d 2 x dd 2 ): d 2 x dd 2 = k x This is an equation that involves the 2 nd derivative of position, x, with respect to tie, t

11 The archetype siple haronic oscillator The solution to the equation is equivalent to integrating twice: in other words, it will have TWO integration constants that are to be deterined by the particular conditions But first: let s guess at the solution we can try x 1 = c 1 sin ωω : dx 1 dd = ωc d 2 x 1 1 cos ωω dd 2 = d dx 1 = d dd dd dd ωc 1 cos ωω = ω ωc 1 cos ωω = ω 2 c 1 cos ωω But the equation requires: d 2 x 1 dd 2 = k x 1 ω 2 c 1 cos ωω = k c 1 cos ωω This eans x 1 = c 1 sin ωω can be a solution provided we satisfy the condition: Unit check: L = L 0 + x ω 2 = k ω = k k = N kg = kg s kg = s 2 = s 1 This is what we need: because the arguent of a sine/cosine function needs to be unitless. x F = kkı

12 The archetype siple haronic oscillator L = L 0 + x We can also try x 2 = c 2 cos ωω : dx 2 dd = ωc d 2 x 2 2 sin ωω dd 2 = ω 2 c 2 cos ωω = d dd dx 2 dd = d dd ωc 2 sin ωω = ω ωc 2 cos ωω But the equation requires: d 2 x 2 dd 2 = k x 2 ω 2 c 2 cos ωω = k c 2 cos ωω This eans x 2 = c 2 cos ωω can ALSO be a solution provided we satisfy the SAME condition: ω 2 = k ω = k NOTE: The equation involves only the function x(t), and its 2 nd derivative x d 2 x dd 2, BUT NOT its powers (like x 2, or x 2 ). We say that the equation is LINEAR This eans any linear cobination of the two solutions is also a solution x(t) = c 1 sin ωω + c 2 cos ωω is also a solution for ω = k Since this for involves TWO integration constants, c 1 and c 2, it is a GENERAL solution Another general solution (sartphysics is inconsistent) has the for x t = A cos ωω + φ x F = kkı

13 The archetype siple haronic oscillator k, x(t) = c 1 sin ωω + c 2 cos ωω and x t = A cos ωω + φ are equivalent : For ω = Trigonoetric identity So we can apply the second to sin u + v = sin u cos v + sin v cos u cos u + v = cos u cos v sin v sin u A cos ωω + φ = A cos ωω cos φ A sin φ sin ωω Now recalling that sine is an odd function, and cosine an even function A cos ωω + φ = A sin φ sin ωω + A cos φ cos ωω In other words to ake the two fors be equivalent, we identify Or conversely L = L 0 + x c 1 = A sin φ, c 2 = A cos φ A = c c 2 2, φ = tan 1 c 1 c 2 x F = kkı

14 A cos ωω φ x YES I know that the pre-lecture used the for A sin ωω φ, and the picture they gave is a better representation for A cos ωω φ But the A ccc ωω + φ is ore conventional: and the ain points fro sartphysics text uses this for instead (see next slide)

15 Unit of angular frequency ω: s 1 or rad/s*** To coplete one period of the otion, the arguent to the cosine/sine ust advance by 2π i.e. the period, T, is given by ωt = 2π T = 2π/ω (unit of period: seconds) The frequency (nuber of coplete cycles per unit tie) is given by f = 1 T = ω/2π Unit of frequency: s 1 heeee (Hz) The axiu displaceent fro equilibriu is A, known as the aplitude (unit: eters) Maxiu speed: v = ωω, axiu acceleration: a = ω 2 A

16 x t = A cos ωω + φ can be represented graphically as the x-coponent of a particle undergoing unifor circular otion with angular velocity ω, and starting at angle φ.

17 Poll A ass on a spring oves with siple haronic otion as shown. Where is the acceleration of the ass ost positive? A. x = -A B. x = 0 C. x = +A

18 Poll In the two cases shown the ass and the spring are identical but the aplitude of the siple haronic otion is twice as big in Case 2 as in Case 1. How are the axiu velocities in the two cases related? A. V ax,2 = V ax,1 B. V ax,2 = 2 V ax,1 C. V ax,2 = 4 V ax,1

19 Deo spring does not really copress 1. Hang vertically fro pivot 2. With 100 g ass attached: the spring stretches 10 c fro relaxed length 3. With 200 g attached, stretches 20 c fro relaxed length Calculate Spring Constant: k F kg 9.8 s2 = L kg 9.8 s2 = If we just hang a 500 g ass, and take configuration (2) to be equlibriu = 9.8 N Then for displaceent x about the equilibriu configuration of (2) we have d2 x dd 2 = kk x = A sin ωω φ ω = k = 9.8 N 0.50 kg = 4.43 s 1 He period is given by T = 2π ω = 2π = 1.42 s 4.43s 1 Deo/Exaple 21.1 (1) 10 c 100 g (2) x 10 c 100 g 100 g (3) 20 c

20 The point of solving an equation like d 2 x dd 2 = k x, Initial Value Proble L = L 0 + x x t = A cos ωω + φ Is to be able to predict the otion of the block for all ties. The general solution does not actually do that unless we deterine the value of A and φ for a given case -- i.e. find the particular solution. To do so we need to have TWO given conditions (2 eqns 2 unknowns). For exaple: We are given initial conditions x 0 = x 0 and x (0) v 0 = v 0. To Find A and φ, we first write down expressions for x 0 and v 0 (or whatever is given) fro the general solution x 0 = A cos ω 0 + φ = A cos φ v t = dd dd = d A cos ωω + φ = ωa sin ωω + φ, dd v 0 = ωa sin ω 0 + φ = ωω sin φ And then we set these expressions to the given values: A cos φ = x 0, ωa sin φ = v 0 And then we have A = x v 0 ω 2, φ = tan 1 v 0 ωx 0 x F = kkı

21 Solution (a): (b) (c) General solution is ω = k 807 kg = s kg T = 2π ω = 2π rad s f = 1 T = s = rad s = s = s = Hz x t = A cos ωω + φ We are given (released at t = 0) : x 0 = x 0 = 0.078, and x 0 = v 0 = v 0 = 0 Using the general for: Exaple 21.2 (1/3) (30p12wh) A 4.6-kg object on a frictionless horizontal surface is attached to one end of a horizontal spring that has a force constant k = 807 N/. The spring is stretched 7.8 c fro equilibriu and released. (a) What is the frequency of the otion? (b) What is the period of the otion? (c) What is the aplitude of the otion? (e) What is the axiu acceleration of the otion? (f) When does the object first reach its equilibriu position? x 0 = A cos ω 0 + φ = A cos φ

22 (c) continued: x t = A cos ωω + φ v t = dd dd = ωa sin ωω + φ, v 0 = ωa sin φ Now we put in the data: Exaple 21.2 (2/3) (30p12wh) A 4.6-kg object on a frictionless horizontal surface is attached to one end of a horizontal spring that has a force constant k = 807 N/. The spring is stretched 7.8 c fro equilibriu and released. (a) What is the frequency of the otion? (b) What is the period of the otion? (c) What is the aplitude of the otion? (e) What is the axiu acceleration of the otion? (f) When does the object first reach its equilibriu position? x 0 = A cos φ = v 0 = ωa sin φ = 0 The only possible solution for ωa sin φ = 0 is sin φ = 0, of which there is an infinite nuber of equivalent solutions, but we choose here φ = 0 (principal branch), and thus A cos φ = A = (e) The acceleration is given by a t = dv dd = ω 2 A cos ωω + φ So the axiu a t occurs when cos ωω + φ = 1 a = ω 2 A = rad s = s 2 (f) The particular solution of the oscillator here is x t = cos rad s t + 0 We will use the iplicit for x = A cos ωt for the oent.

23 (f) continued: x = A cos ωt For the oscillator first to return to equilibriu: we require x = 0 In this case it eans finding the sallest positive solution for t in A cos ωω = 0 Which gives us Exaple 21.2 (3/3) (30p12wh) A 4.6-kg object on a frictionless horizontal surface is attached to one end of a horizontal spring that has a force constant k = 807 N/. The spring is stretched 7.8 c fro equilibriu and released. (a) What is the frequency of the otion? (b) What is the period of the otion? (c) What is the aplitude of the otion? (e) What is the axiu acceleration of the otion? (f) When does the object first reach its equilibriu position? ωω = π 2 i.e. t = π 2ω = π rad s = 0.12 s