Equating order frequencies we have found set to

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Unformatted text preview: es to F = mv 2 /R, as it should. R2 (xx + y y ) + (xy − y x)2 = 0. ¨ ¨ ˙ ˙ (320) 6.39. Bead on stick, using F = ma Using mx = F (x/R) and my = −mg + F (y/R) to eliminate the x and y in Eq. ¨ ¨ ¨ ¨ (320) gives (a) There is no force in the r direction, so the first of Eqs. (3.51) gives (using ˙ θ = ω ) r = rω 2 . Multiplying through by r and integrating yields ¨ ˙ Fy Fx R2 x +y −g + (xy − y x)2 = 0 ˙ ˙ 12 1 1 1 mR mR rr dt = ω 2 ¨˙ rr dt =⇒ ˙ r = r2 ω 2 + C =⇒ ˙ mr2 − mr2 ω 2 = E, ˙ 2 2 2 2 y m =⇒ F = mg − 3 (xy − y x)2 . ˙ ˙ (321)(328) R R where E ≡ mC is a constant of integration. ˙ ˙ Finally, using x = R sin θ =⇒ x = Rθ cos θ, and y = R cos θ =⇒ y = −Rθ sin θ, ˙ ˙ ¨ ˙ (b) Since θ θ −0, the2second of Eqs. (3.51) gives (using θ = ω ) Fθ = 2mω r. So the ˙ ˙ gives F = mg cos = mRθ , as desired. work done on the bead is θ W= t (2mω r)r(ω dt) = mr2 ω 2 ˙ F θ ( r dθ ) = θ0 r . (329) r0 t0 Since W = ∆K = K − K0 , the kinetic energy is 2 K = K0 + W = K0 + mr2 ω 2 − mr0 ω 2 ≡ mr2 ω 2 +...
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This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.

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