Written answers preferred, complete all parts required. Consider a uniform solid

Question

Written answers preferred, complete all parts required.

Consider a uniform solid sphere of mass m and radius R on an incline of angle beta. An axle passes through a diameter of the sphere, and a light yoke is attached to the axle. In turn the yoke is attached to a light linear spring of stiffness k which is anchored to a fixed wall at the top of the incline as shown. The sphere is free to roll about this diameter up and down the incline without slipping, compressing and stretching the spring as it does so. (The yoke merely prevents twisting.) x is the displacement of the center of the sphere from the equilibrium position of the spring, while the angle phi measures the angle through which the sphere rotates from some convenient reference. [1] Write the Lagrangian for this system in terms of the single generalized coordinate x, by imposing the no-slip constraint kinematically at the outset. Remember that the sphere is rolling. [3] Determine the equation of motion, and solve it for the initial conditions that the sphere is released from rest when x = 0. [1] Write the Lagrangian in terms of the two generalized coordinates x and phi, and write the constraint in the form f(x, phi) = 0. [2] Determine the two modified equations of motion in terms of the Lagrange multiplier lambda. Interpret the physical meaning and origin of lambda in both equations (by, for example appealing to Newton's laws), i.e. What is lambda measuring physically in each equation?

Explanation / Answer

a) L = T-U
L = 1/2 m x'^2 + 1/2 I phi'^2 - 1/2 k x^2 + m g x sin beta

but I = 2/5 MR^2
no slip condition phi' = x'/R

so L = 1/2 mx'^2 + 1/5 m x'^2 - 1/2 kx^2 + m g x sin beta

L = 7/10 m x'^2 - 1/2 kx^2 + m g x sin theetata

dL/dx' = 7/5 m x'
d/dt DL/dx' = 7/5 m x''

dL/dx = - k x + M g sin beta

7/5 m x'' = - k x + M g sin beta

b)
x'(0)=0 and x(0) = 0

so x(t) = (2 g M sin(beta) sin^2((sqrt(5/7) sqrt(k) t)/(2 sqrt(m))))/k

c) now
L = 1/2 m x'^2 + 1/2 I phi'^2 - 1/2 k x^2 + m g x sin beta
and x - r phi = 0

2. so L + constraint = 1/2 m x'^2 + 1/5 m r^2 phi'^2 - 1/2 k x^2 + m g x sin beta - lambda( x- r phi)

dL/dx' = m x'
dL/dx = - kx + m g sin beta - lambda
so
m x'' = - kx + m g sin beta - lambda


dL/dphi' = 2/5 m r^2 phi'
dL/dphi = lambda r
so
2/5 mr^2 phi'' = lambda r

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