Pendulums — Question 10

Question 10

A string must remain in tension for the pendulum path to apply

A point bob of mass $m$ on a massless string of length $\ell$ is released from rest at angle $\theta_A$ from downward vertical. Ignore drag. Use exact energy conservation, not the small-angle approximation.

1. Derive speed and string tension at an intermediate angle $\theta$ while the string is taut.

2. Find the release-angle range that keeps tension nonnegative on the swing from release to bottom. Explain why release from rest above the horizontal cannot follow the same circular path immediately.

Question 10 - Solution

A string must remain in tension for the pendulum path to apply

(a) Derivation and calculation

Energy from release at rest gives$$\tfrac12mv^2+mg\ell(1-\cos\theta)=mg\ell(1-\cos\theta_A).$$$$\boxed{v^2=2g\ell(\cos\theta-\cos\theta_A)}.$$The inward radial force equation is$$\mathcal T-mg\cos\theta=mv^2/\ell\Rightarrow\boxed{\mathcal T=mg(3\cos\theta-2\cos\theta_A)}.$$

(b) Result and interpretation

For a swing from rest below the horizontal, the smallest tension occurs at release:$$\mathcal T_{\min}=mg\cos\theta_A.$$$$\mathcal T_{\min}\ge0\iff\boxed{0\le\theta_A\le\pi/2}.$$Strict positive tension requires $\theta_A<\pi/2$. Above the horizontal the circular-path equation demands negative initial tension, so the string goes slack and the initial motion is free fall.

Physical check

At the bottom, $\mathcal T=mg(3-2\cos\theta_A)\ge mg$, consistent with the required inward centripetal acceleration.