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Chapter 3 103

# Chapter 3 103 - CHAPTER 3 PROBLEMS PLUS 253 common...

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Unformatted text preview: CHAPTER 3 PROBLEMS PLUS 253 common y—intercept is \$8 + %. We want to ﬁnd the value of 1:0 for which the distance from (0. \$3 + l) to (\$0.153) equals 1. The square of the distance is (\$0 — 0)2 + [:83 ~ (mg + 9122953 + : 1 42 m0 2 i323. For these ) Another solution: Let the center of the circle be (07 a). Then the equation of the circle is x2 + (y — (1)2 2 values of 1:0, the y—intercept is 233 + % 2 2, so the center of the circle is at (0. mm Solving with the equation of the parabola, y 2 322. we get 51:2 + (x2 — (1)2 2 1 (2 x2 + m4 — 211272 + a2 2 1 42) m4 + (1 — 2a):c2 + a2 — 1 2 0. The parabola and the circle will be tangent to each other when this quadratic equation in \$2 has equal roots; that is, when the discriminant is 0. Thus. (1 — 2a)2 — 4(a2 ~ 1) 2 0 2) 1~4a+4a2—4a2+420 <2 4a25,soa2§.Thecenterofthecircleis 0,5. 4 4 8.11mM_hm f(\$)2f(a)\/5E+\/E:' 7“” WE‘VE H1 f—Va «mm/5 m—vCL {IT—a lim [f(w)—f(a),(ﬁ+ﬁ)] ﬂawmwwa)=f’<a>-<¢a+¢a>:2vaf'<a> We can assume without loss of generality that 6 2 0 at time t 2 0. so that 6 2 127rt rad. [The angular velocity of the wheel is 360 rpm 2 360 - (2n rad)/(60 s) 2 1271' rad/s] Then the position of A as a function of time is 2 (40 cos 6, 40 sin 6) 2 (40 cos 127rt, 40 sin 127rt). so sina 2 y 2 40 51116 Slug 1 ' 1.2m 120 Z 3 :Esmlm‘ d (a) Differentiating the expression for sina we get cos a d_a 2 127r cos 127rt— — 47r cos 6. When62—3,wehavesina2gsin62‘?socosa2i/1~()§221/ﬁand da 47rcos§ 27r 47r\/?_) _ : _ : _ : m 6.56 rad/s. dt c0801 411/12 V11 (b) By the Law ofCosines, [API2 2 |OAl2 + |0Pl2 — 2 |OA| lOPl c086 2> 1202 : 402 + [01312 — 2 - 40 |OPl c0s6 => |0Pl2 — (80 c086) |0P| — 12.800 : 0 :> 10131: §(80c0s0 j: M6400 c032 9 + 51,200) = 40 cos6 : 40 x/cos2 6 + 8 2 40(cos6 + V 8 + cos2 6) cm (since lOPl > 0) As a check. note that lOP| 2 160 cm when 6 2 O and |OPl 2 80 \/§ cm when 6— — tall: (c) By part (b). the x—coordinate of P is given by m 2 40(cos 6 + v8 + cos2 6 ) so dzL' d2: d6 . 2 cos 6 sin 6 cos 6 —:———40 Sln6 '12 ——480 ' ‘ dt d0 dt ( 2M8 + cos2 0) W ””160 + v/s + cos2 a) cm/S' In particular, dw/dt 2 O cm/s when 6 2 O and dzv/dt 2 —4807r cm/s when 6 2 g. ...
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