Problem 4. Consider a pendulum attached to a vertical spring, as shown in the at

Question

Problem 4. Consider a pendulum attached to a vertical spring, as shown in the attached fig parts are constrained to move in the x -y plane. The spring has natural length a and spring constant k. The pendulum rod has length a. At the point where the pendulum bar is joined to the spring, there is a frictionless pivot with mass mi. The pendulum bob has mass m2. Gravity is downwards. Let ri and r2 be the position vectors of the pivot and the pendulum bob, respectively. ure. All Use y and 0 as generalized coordinates. Finding ri is simple: it's ryj I. Express the position vector r2 as a function of the generalized coordinates y and and the standard unit vectors i and j 2. Find the system's total kinetic energy T and potential energy V as functions of y, e, and their time derivatives. 3. Find equations of motion for y and e 4. Repeat part 3 using Mathematica and FullSimplify Remember: as messy as this problem gets, it would be much worse without Lagrangian mechanics!

Explanation / Answer

a) vector r2 is y(j) + b*cos(theta) (j) + b*sin(theta) (i)

can get it by resoving b vector and using parallelogram law of vector addition.

b) T = (0.5)*(m1)*((y')^2) + 0.5*(m2)*((b*(theta') + y')^2) (where a' is derivative of a with respect to time)

and V = 0.5*k*(y^2) - m1*g*y - m2*g*(y + b*cos(theta))

c) now your lagrangian is L = T-V

and substitute it in the euler-lagrangian equation

for y :

m1*(y'')+m2*(y'')+b*m2*(theta") = k*y + (m1+m2)*g ----->1

for theta:

b*m2*(y") + (b^2)*m2*(theta") = -m2*g*b*sin(theta) ------->2

now multiply eqn 1 with "b" and subtract 2 from it

m1*(y") = ky + (m1 +m2*(1 + sin(theeta))g

from the above eqn get y" and substitute in either 1 or 2 to get an eqn in only theeta"

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