Physics 207: Lecture 27. Announcements

Transcription

1 Physcs 07: ecture 7 Announcements ake-up labs are ths week Fnal hwk assgned ths week, fnal quz next week Revew sesson on Thursday ay 9, :30 4:00pm, Here Today s Agenda Statcs recap Beam & Strngs» What f a strng breaks? Hnged Beams Angular omentum: Defntons & Dervatons What does t mean? Rotaton about a fxed axs = Iω Example: Two dsks Student on rotatng stool Exam 3 Results Average score ~ 54.4%; edan ~ 53% 40 Exam 3 Raw Percentage Dstrbuton 30 Count Percentage Score A 6 percentage pont curve wll be appled to ths exam Page

2 Statcs Revew: In general, we can use the two equatons F = 0 τ = 0 to solve any statcs problem. When choosng axes about whch to calculate torque, we can be clever and make the problem easy!! 3 ecture 7, Act Statcs In whch of the statc cases shown below s the tenson n the supportng wre bgger? In both cases s the same, and the blue strut s massless. (a) case (b) case (c) same T T / case case 4 Page

3 ecture 7, Act Soluton Consder the torque about the hnge between the strut & the wall: ( 30 ) 0 τ total = g T sn = due to gravty due to wre T g T = 30 0 T = g does not depend on length of massless beam! 5 ecture 7, Act Soluton The tenson s the same n both cases. T = g 30 0 T = g 30 0 / case case 6 Page 3

4 Revew: Beam and Strngs Prevously we solved for the tensons n the strngs n the followng problem: T T = g 3 = 3 g T T x cm / /4 But what f a strng breaks... g x y 7 Beam and Strngs... If the left strng breaks, what s the ntal acceleraton of the C? The beam wll rotate about the axs at A. Usng F = ma n the y drecton: g - T = a T Fgure out I about A: I = I C + d. Recall I C I = = + d x C g d x A y x 8 Page 4

5 Beam and Strngs... Fgure out τ about A: τ = gd Now use τ = Iα and a = αd gd = + d a d T x C x A d d ( ) a = g + d a y x 9 Hnged Beams: Consder a structure made from two beams, attached to each other and the wall wth hnges: φ = hnges 0 Page 5

6 Hnged Beams... y What we want to fnd: (A x,a y ) and (B x,b y ). A y x What we know: A x Any forces present at C wll act n pars, and wll therefore cancel f we consder the entre structure. C φ m g B y B x m g Hnged Beams... y Frst use F NET = ma n x and y drectons: x x A x + B x = 0 A x = -B x (a) A y y A y + B y = (m +m )g (b) A x That s two equatons, but we have four unknowns... φ m g B y B x m g Page 6

7 Hnged Beams... y Now use some torque relatonshps. Frst, consder the torque on the whole structure about an axs though the hnge at B. x m g + mg tan φ Ax = 0 g ( m + m ) = tan φ Ax ( m + m ) g = tan φ Ax (c) C φ m g m g A x tanφ B (No torques from forces at C) 3 Hnged Beams... y If we knew somethng about A y or B y we would be done! Do the smplest thng we can thnk of! Consder the torque on the bottom beam about an axs through C: x By mg = 0 mg By = (d) C B y m g 4 Page 7

8 Hnged Beams... So we have the followng equatons: (a) A x = -B x A y (b) A y + B y = (m +m )g A x m (c) ( m + m ) g = tan φ Ax φ B y (d) mg By = m B x 5 Hnged Beams... And we solve these to get: ( m + m ) g = tan φ Ax Ay = m + m g A y A x ( m + m ) g = tan φ Bx mg By = φ m B y m B x 6 Page 8

9 ore on Stablty: Consder a truck movng a refrgerator of mass. The C of the frdge s a heght h above the bed of the truck and the wh of the frdge s w. If the truck s on a horzontal road, what s the maxmum acceleraton a that the truck can have wthout tppng the frdge? (Assume the frdge does not slp). w C h a 7 Frdge... Suppose the truck s acceleraton a s such that the frdge s just startng to tp. In ths case the weght of the frdge s supported by a normal force actng only at the back corner. There must also be a frctonal force actng to keep the frdge acceleratng: w C a h F = a N = g 8 Page 9

10 Frdge... Snce the frdge s not rotatng, the sum of all the torques about an axs through the C must be zero. gw = ah w a = g w h C a h F = a N = g 9 Frdge... Snce the torque due to the normal force can t be any bgger, f we ncreased the acceleraton, the net torque would be non-zero, and the frdge would flp. Ths must be the maxmum allowable acceleraton! w a = g w h C a h F = a N = g 0 Page 0

11 ecture 7, Act Rotatons A grl s rdng on the outsde edge of a merry-go-round turnng wth constant ω. She holds a ball at rest n her hand and releases t. Vewed from above, whch of the paths shown below wll the ball follow after she lets t go? (a) (b) (c) (d) ω ecture 7, Act Soluton Just before release, the velocty of the ball s tangent to the crcle t s movng n. ω Page

12 ecture 7, Act Soluton After release t keeps gong n the same drecton snce there are no forces actng on t to change ths drecton. ω 3 p = mv Angular omentum: Defntons & Dervatons We have shown that for a system of partcles F EXT dp = omentum s conserved f F EXT = 0 What s the rotatonal verson of ths?? The rotatonal analogue of force F s torque τ = r F Defne the rotatonal analogue of momentum p to be angular momentum = r p 4 Page

13 Defntons & Dervatons... d Frst consder the rate of change of : = ( r p) d dr dp = p r + ( r p) ( v mv) = = 0 d So d dp = r (so what...?) 5 Defntons & Dervatons... d dp = r Recall that F EXT dp d = = r FEXT 3 τ EXT d Whch fnally gves us: τ EXT = dp Analogue of FEXT =!! 6 Page 3

14 What does t mean? d τ EXT = where = r p and τ EXT = r F EXT In the absence of external torques τ EXT d = = 0 Total angular momentum s conserved 7 Angular momentum of a rgd body about a fxed axs: Consder a rgd dstrbuton of pont partcles rotatng n the x-y plane around the z axs, as shown below. The total angular momentum around the orgn s the sum of the angular momenta of each partcle: = r p = m r v We see that s n the z drecton. Usng v = ω r, we get = m r ω kˆ = I ω = m r v kˆ Analogue of p = mv!! v (snce r and v are perpendcular) m r ω m 3 j r 3 v 3 v r m 8 Page 4