HW 6 Solutions

HW 6 Solutions - Problem 13.152 The bar rotates in the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 13.152 The bar rotates in the J'—_'l' plane with constant angular velocity (no = 12 radfs. The radial com- . ,. 1 . . ponent of acceleration of the collar C [111 1nfs' I is given as a function of the radial position in meters by a, = —3r. When :' = 1 1n. the radial component of velocity of C is l-‘r = 2 III-“s. Determine the velocity of C in terms of polar coordinates when r = 1.5 in. Strategy: Use the chain rule to write the first term in the radial component of the acceleration as d2? (for tie, dr' do, — ——n d£2=F—FE— dr r Solution: We have air as 3 S dfz (tr—JF—r a") _—t ra-s jr. air (it? 3 a 2 1 _ c m2 = E — ['3 radt's‘} r : {[12 rad-“5] — [S rad-“s ];u' : ([13e rad!s')r Using the supplied strategy we can solve for the radial velocity dflr tin,- m: = a? = (135 md-‘szjr u, q 1.5 fl. f 1min, : [136 rad-‘3‘)f rdr 2 n_.I's 1 m 2 I 2 2 1' {2 mist . [1.5 In] [1 1'11] Tr — T =t13+s rad-‘5‘) (T — 2 Solving we find. or = 13.2 rn.."s_ “re also have v.9 : r? = [1.5 Injljli radish = 1.3 this. Thus r = (13.22,. +1393J1'nfsi Problem 13.154 The hydraulic actuator mores the pin P upward with constant velocity v=2j unis). Determine the acceleration of the pin in terms of polar coordinates and the angular acceleration of the slotted bar when E? = 3.3". Solution: From Problem 13.15?- V = Ej rafs. constant i" = 2L}. seceiisecfl tan 315': + 3 see: {99‘ fl' ' I = ' = _" i 5 " _ E a I‘ 2 fl=_‘IED 3 we Ilé.lr'1d." 9:[ 2sct9tnt9]u9,'l ‘1'; = 35 sect? 5‘ 2 :0 j? = 2 m-"s 5‘ = Boil-1 rad-9's: 6‘ = or I. = D .r- H I-J u. GI CI IJ Cb '15. Problem 13.168 A private pilot wishes to fly from a f -0 — cit}: P to a city Q that is 200 km directly north of city P. The airplane will fly with an airspeed of EQO Kin-'11. At the altitude at which the aiiplane will be flying. there is — an east wind (that is. the Wind's direction is first) with a speed of 50 kIl'JflL What direction should the pilot point the airplane to fly directly from city P to city Q? How long will the trip take? w E 203 Km — 50 km'm Solution: Assume an angle 6'. measured ccv.‘ front. the east. “Plane-"Grode = Y?!“ s_.-'.4!.I' + Y.4!.".-"G.“J|m|'z_ “Pl'firrsl.-'_.1|'v = (EQEI knu'hit'cosai + sin 9]] p v.-il'”_."Cv”'D:Hld‘ 2 —[5D kill-5111i “Harri-Erasers = [[29I] cos 9 — SDJi + [290 sin Ejj] knu‘h we want the airplane to travel due north therefore 5a 2900059 — so = o :9 a = cos—1(fi): SI].U'.-‘° Thus the heading is [30° — SUD? : 993° east of north The ground. speed is new U = (29D kmth sint'SEI.1°} = 285.6 krn'h The time is . ‘fln km =o.?oo h = 42.0 min 235.6 kmt'h Problem 13.184 In the cam-follower mechanism. the slotted har rotates with a constant angular velocity at = 12 radis. and the radial position of the follower A is determined by the profile of the stationary cam. The slotted bar is pinned a distance it = 0.2 m to the left of the center of the circular cam. The follower moves in a circular path of 0.42 in radius. Determine the velocity of the follower when IE? 2 40° {a} in terms of polar coordinates. and [bi in terms of cartesian coordinates. Solution: (3] The first step is to get an equation for the path of the follower in terms of the angle 6. This can be most easily done by referring to the diagram at the right. Using the law of cosines. we can write R2 2ft: + r1 — "hr cosfi'. This can be reuritten as r: — 3hr cost? + [W1 — R3) = Cl. We need to find the components of the velocity. These are u, = F and 1.59 = Hi. ‘We can differentiate the relation derived from the law of oosines to get F. Carrying out this differentiation. we get 2”" — Zfit‘cosfi + Marti sinti 2 III. Solving for .l". we get jl _ magma _ thcosfl —.l":l‘ Recalling that a:- 2 ti' and substituting in the numerical values. i.e.. R = 0.42 m. k :02 In. an :12 rad-“s. and i9 = 40°. we get r =ll553 m. y, = —2.13 nus, and 1:9 =fi.64 this The transformation to earteaian coordinates can be derived from er = cos Si + ain tij. and e3 2 — ain 9i + cos i9j. Substituting these into it = Lifer + 1:999. we get i: = tur cost?| — v.9 sinSJi + (v, sine? + v9 cosSJj. Substituting in the numbers. 1: = —S.9|Ji + 3.?lj (nits) ...
View Full Document

This note was uploaded on 04/29/2008 for the course MAE 212 taught by Professor Mr. during the Spring '08 term at ASU.

Page1 / 3

HW 6 Solutions - Problem 13.152 The bar rotates in the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online