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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 ras 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‘ = Boil1 rad9's:
6‘ = or I. = D .r
H
IJ
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.“Jm'z_ “Pl'ﬁrrsl.'_.1'v = (EQEI knu'hit'cosai + sin 9]] p
v.il'”_."Cv”'D:Hld‘ 2 —[5D kill5111i “HarriErasers = [[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 camfollower 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.9Ji +
3.?lj (nits) ...
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 Spring '08
 mr.
 2J, Polar coordinate system, 1°, 2L, Marti sinti

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