PHYSICS 44 MECHANICS Homework Assignment II SOLUTION

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1 July 21, 23 PHYSICS 44 MECHANICS Homewk Assignment II SOLUTION Problem 1 AcartofmassM is placed on rails and attached to a wall with the help of a massless spring with constant k (as shown in the Figure below); the spring is in its equilibrium state when the cart is atadistancex from the wall. A pendulum of mass m and length ` is attached to the cart (as shown). (a) Write the Lagrangian L(x; _x; µ; _ µ)fthecart-pendulum system, where x denotes the position of the cart (as measured from a suitable igin) and µ denotes the angular position of the pendulum. (b) From your Lagrangian, write the Euler-Lagrange equations f the generalized codinates x and µ. (a) Using the generalized codinates x (the displacement of the cart from the equilibrium point of the spring) and µ (the displacement of the pendulum from the vertical) shown in the Figure, the codinates of the cart are x M = x and y M =whilethecodinates of the pendulum are x m = x + ` sinµ and y m = ` cos µ and thus the squared velocities are v 2 M = _x2 M + _y2 M = _x2 v 2 m = _x2 m + _y2 m = (_x + ` _µ cosµ) 2 + (` _µ sin µ) 2 = _x 2 + `2 _ µ 2 +2` _x _ µ cos µ:

2 The expression f the kinetic energy of the cart-pendulum system is therefe K = (m + M) _x2 2 + m`2 _ µ m`_x _ µ cos µ: The potential energy U of the cart-pendulum system is broken into two parts: the gravitational potential energy mg` cosµ of the pendulum and the elastic potential energy kx 2 =2sted in thespring. The Lagrangian L = K U of the cart-pendulum system is therefe L(x; _x; µ; _ µ) = (m + M) _x2 2 + m`2 _ µ m` cosµ ³ _x _ µ + g k 2 x2 : (b) The Euler-Lagrange equation f x _x = (m + M) _x + m` _µ cosµ d = (m + M)Äx + m` ³Ä µ cos µ µ _ 2 sin µ = kx (m + M)Äx + m` ³Ä µ cos µ _ µ 2 sin µ + kx = The Euler-Lagrange equation f µ µ _ d µ = m` ³ ` _µ + _x cos µ = m` ³ ` ĵ + Äx cos µ _x _ µ sin µ = m` _x _ µ sin µ mg` sin µ ` ĵ + Äx cos µ + g sin µ =

3 Problem 2 Show that the two Lagrangians L(q; _q; t) and L (q; _q; t) = L(q; _q; t) + df(q;t) ; dt where F(q;t) isanarbitrary function of the generalized codinates q(t), yield the same Euler-Lagrange equations. Hence, two Lagrangians which di er only by an exact time derivative are said to be equivalent. We call L = L + df=dt the new Lagrangian and L the old Lagrangian. The Euler- Lagrange equations f the new Lagrangian are d _q i i ; where Let us begin _q _q i df (q;t) @t j : 1 _q j _q i ; so that d = d _q i _q i Next, we nd i j i 1 + X k _q i : A i _q j : Using the symmetry properties _q j = _q i and ; we easily verify d _q i = d _q i = ; and thus since L and L = L + df=dt lead to the same Euler-Lagrange equations, they are said to be equivalent.

4 Problem 3 An Atwood machine is composed of two masses m and M attached by means of a massless rope into which a massless spring (with constant k) is inserted (as shown in the Figure below). When the spring is in a relaxed state, the spring-rope length is `. (a) Find suitable generalized codinates to describe the motion of the two masses (allowing f elongation compression of the spring). (b) Using these generalized codinates, construct the Lagrangian and derive the appropriate Euler-Lagrange equations. (a) The Figure below shows a suitable set of generalized codinates where x denotes the distance of mass M from the top of the pulley, ` x denotes the distance of the equilibrium point of the spring from the top of the pulley, and y denotes the distance of mass m from the equilibrium point of the spring (i.e., its elongation).

5 (b) Using the generalized codinates (x; y), the codinate of mass M is x M = x while the codinate of mass m is x m = ` x + y and thus the squared velocities are v 2 M = _x2 and v 2 m = (_y _x)2 : The expression f the kinetic energy of the system is therefe K = (m + M) _x2 2 + m _y2 2 m _x _y: The potential energy U of the system is broken into two parts: the gravitational potential energy Mgx mg(y x)andtheelastic potential energy ky 2 =2sted in thespring. The Lagrangian L = K U of the system is therefe L(x; _x; y; _y) = (m + M) _x2 2 + m _y2 2 m _x _y +(M m)gx + mg y k 2 y2 : (c) The Euler-Lagrange equation f x _x = (m + M) _x m _y d = (m + M)Äx m Äy = (M m) g (m + M)Äx m Äy = (M m) g The Euler-Lagrange equation f y _y = m (_y _x) d = m (Äy Äx) = mg ky m (Äy Äx) +k y = m g