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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online