Physics 325 Lecture 17 - Physics 325 Lecture 26 In several...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 325 Lecture 26 In several of the examples that we have done in the past two lectures, we have come upon situations in which the derivative of the Lagrangian with respect to one of the generalized coordinates is zero 0 i L q = (26.1) Given the Lagrange equations 0 ii LdL qd tq = ∂∂ ± Equation (26.1) leads to the conservation law 0 i dL dt q = ± (26.2) or constant i L q = ± In Cartesian coordinates we have LT mx p xx = == ± ±± Thus 0 i L x = leads to conservation of linear momentum along the x i direction. Now, in Cartesian coordinates we also have that i LU F = =− which is the component of the force along the x i direction. In the situation where Equation (26.1) is satisfied, then, we have F i =0 and the component of the momentum along the x i direction is a constant in time. This is good. For the puck in a bowl problem from the last lecture, we found that the Lagrangian had no explicit φ dependence ( 0 L ∂∂= ), and this led to conservation of the z -component of angular momentum, as expected from the vanishing of the z- component of the net torque on the system. Inspired by these discoveries, we define the generalized momentum, p j , associated with the generalized coordinate q j as j i L p q ± and we note that p j is conserved whenever 0 j Lq ∂= . In a conservative system the potential energy is independent of the velocity. Therefore
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
0 and ii UL qq i T q ∂∂ == ±± ± We now defined the generalized force . If the system is conservative we have FU =−∂ x , so ij ji j j xx UU FQ qx q q = ∑∑ where we have defined the generalized force as i i i i x U Q F =− = (26.3) Note that if q i is a length, then the partial derivative is dimensionless and Q i is a force. If q i is an angle, then the partial derivative has the dimensions of length ( x i ) and Q i is a torque (force x length). Let’s see. In polar coordinates we have 12 12 xy yx xc o s x s i n Q =F F F F sin cos Q F F where 0 0 z xy xr yr xx xy ry r x k FF φ φφ ττ ∂∂ ∂∂ +=+ ∂∂∂∂ = = =−= = GG Systems with Holonomic Constraints We consider a system of n particles ( 3n coordinates) with m equations of constraint: ( ) , 0 1,. .., kj gqt k m (26.4)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

Physics 325 Lecture 17 - Physics 325 Lecture 26 In several...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online