Question

QUESTION 5 [25 marks] Two masses mi and m are joined by an inextensible string of length I, as shown in Figure 2. The string passes over a massless pulley with frictionless bearings and radius R. The acceleration of gravity g points vertically downwards (a) 13 marks] Write down the Lagrangian, using the position of mass mi as the generalized coordinate m1 (b) 12 marks] Find the Lagrange equation of motion and solve it for 白m2 acceleration of mass mi Figure 2

Answer 1

(a) Total kinetic energy = 1/2 m_1 x^2 + 1/2 m_2 x^2 Total potential energy = -m_1 gx - m_2 g(l - x - pi R) So, Lagrangian, L = 1/2 m_1 x^2 + 1/2 m_2 x^2 + m_1 gx + m_2 g(l - x - pi R) (b) eq^n of motion can be found by, d/dt(dL.dx) - dL/dx = 0 rightarrow d/dt (m_1 x^. + m_2 x^.) - m_1 g + m_2 g = 0 rightarrow (m_1 + m_2) x^.. = m_1 g - m_2 g rightarrow x^.. = (m_1 - m_2)g/m_1 + m_2 This is the expression for acceleration of masses. Mas of Acceleration of both masses is same but direction is different if we take into account the rotational motion of pulley, K.E = 1/2 m_1 x^2 + 1/2 m_2 x^2 + 1/2 I x^2/R^2 with this, x^.. = (m_1 - m_2)g/m_1 + m_2 + I/R^2