 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/10/a/solution

Let the pivot's prescribed horizontal position be

x (t)

θ

from the downward vertical. The bob's coordinates are

X = x + l sin θ

and

Y = - l cos θ

. Therefore

\dot{X}^{2} + \dot{Y}^{2} = \overset{x}{˙}^{2} + 2 l \overset{x}{˙} \dot{θ} cos θ + l^{2} \dot{θ}^{2} .

L = \frac{m}{2} (\overset{x}{˙}^{2} + 2 l \overset{x}{˙} \dot{θ} cos θ + l^{2} \dot{θ}^{2}) + m g l cos θ .

Here

x (t)

is a prescribed input, not yet a free coordinate. Terms depending only on that prescribed time function can be dropped without changing the

θ

equation. The moving block's own kinetic energy becomes relevant when its motion is free in part (d).

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