 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2022/5/a/solution

h

be the displacement of the lower pulley and

z

that of the

4 m

mass. The two inextensible mechanical strings impose

x + y - 2 h = 0, z + h = 0, h = \frac{x + y}{2}, z = - \frac{x + y}{2} .

For ideal massless pulleys and mechanical strings, only the three masses contribute kinetic energy. Hence

K = \frac{1}{2} (4 m) (\frac{x ˙ + y ˙}{2})^{2} + \frac{1}{2} (5 m) \overset{x}{˙}^{2} + \frac{1}{2} (3 m) \overset{y}{˙}^{2} = m (3 \overset{x}{˙}^{2} + \overset{x}{˙} \overset{y}{˙} + 2 \overset{y}{˙}^{2}) .

U = 4 m g z + 5 m gx + 3 m g y = m g (3 x + y) .

L = K - U = m (3 \overset{x}{˙}^{2} + \overset{x}{˙} \overset{y}{˙} + 2 \overset{y}{˙}^{2}) - m g (3 x + y) .

 New to topics? Read the docs here!