Question

1. (pendulum and peg) a pendulum of length \\(l\\) is held with its string horizontal, and then released. the string runs into a peg a distance \\(d\\) below the pivot, as shown in the figure below. what is the smallest value of \\(d\\) for which the string remains taut at all times?

Explanation:

Identify physical conditions for circular motion

To keep the string taut at all times, the pendulum bob must successfully complete a full circle of radius \(r = L - d\) around the peg. The critical point where the string is most likely to go slack is at the very top of this smaller circular path.

Determine critical velocity at the top

At the top of the loop, gravity and tension both pull downward. For the string to remain taut, the tension \(T\) must be greater than or equal to zero:

$$T = \frac{m v_{\text{top}}^2}{r} - mg \ge 0$$

This gives the minimum velocity required at the top:

$$v_{\text{top}}^2 \ge gr = g(L - d)$$

Apply conservation of mechanical energy

Let the lowest point of the swing be the reference height \(y = 0\).

• The initial release height is \(y_i = L\).
• The height at the top of the smaller loop is \(y_f = (L - d) + r = 2(L - d)\).

Using conservation of energy between release and the top of the loop:

$$E_i = E_f$$

$$mgL = mgy_f + \frac{1}{2}mv_{\text{top}}^2$$

$$mgL = 2mg(L - d) + \frac{1}{2}mv_{\text{top}}^2$$

Solve for the minimum distance d

Substitute the minimum velocity condition \(v_{\text{top}}^2 = g(L - d)\) into the energy equation:

$$gL = 2g(L - d) + \frac{1}{2}g(L - d)$$

Divide both sides by \(g\):

$$L = \frac{5}{2}(L - d)$$

$$\frac{2}{5}L = L - d$$

$$d = \frac{3}{5}L$$

Answer:

\(d = \frac{3}{5}L\)