320 gives a there is no force in the r direction so

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Unformatted text preview: according + F, d ∂η m ¨ = m g − dV · ∂η f 2 = −2sin θ = 2 − , ∂x 1dη f 2 2 +∂ ∂η m g F. =⇒ m2 ¨2 == cos + = 2θ and ∂y 1 2 1+f . (318)(325) ¨ ¨ The condition 1 +becomes So Eq. (324) 2 = L implies 1 = − 2 . Multiplying the first equation of motion by m2 , the second by m1 , and then adding gives x2 = −2m1 m2 g/(m1 + m2 ). The mf ˙F F= negative sign means that the force points against the. direction of increasing η . In(326) 1+f 2 other words, it points up on both masses. 2 But x coordinates ˙ 6.37. Cartesian = v cos θ = v/ 1 + f , so we finally have 2+ 2 The first time derivative gives (xx + y y )/ xmf y 2 = 0. Taking another derivative ˙ ˙ v F= . (327) gives (1 + f 2 )3/2 2 2 2 2 (x + y )(x + y + xx + y y ) − (xx + y y )2 = 0. ˙ ˙ ¨ ¨ ˙ ˙ (319) Note: for the special case of the bottom point on a circle of radius R, we have f = 0. Multiplying this out, and then using x2 + y 2 = R, gives And you can show that f = 1/R. So the above result reduc...
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