Pendulums — Question 5

Question 5

A pendulum in an accelerating cart swings about a tilted direction

A simple pendulum of length $\ell=1.0\,\mathrm{m}$ is suspended inside a cart accelerating horizontally rightward at $a=6.0\,\mathrm{m/s^2}$. Take downward gravity $g=8.0\,\mathrm{m/s^2}$. Analyze small swings relative to the cart after defining its equilibrium.

1. Find the equilibrium tilt from vertical and its direction.

2. Find the period about that equilibrium and compare with the stationary-cart period. Why is replacing $g$ by $g-a$ incorrect?

Question 5 - Solution

A pendulum in an accelerating cart swings about a tilted direction

(a) Derivation and calculation

$$\mathbf g_{\mathrm{eff}}=-a\,\hat{\mathbf x}-g\,\hat{\mathbf y},\quad g_{\mathrm{eff}}=\sqrt{g^2+a^2}=\sqrt{8^2+6^2}=10\,\mathrm{m/s^2}.$$$$\boxed{\tan\theta_e=a/g=3/4},\quad\boxed{\theta_e=36.87^\circ\ \text{left of vertical}}.$$$$\ddot q+\frac{g_{\mathrm{eff}}}\ell q\simeq0.$$

(b) Result and interpretation

$$\boxed{T=2\pi\sqrt{\ell/g_{\mathrm{eff}}}=2\pi\sqrt{1/10}=1.98692\,\mathrm s}.$$$$T_{\mathrm{stationary}}=2\pi\sqrt{1/8}=2.22144\,\mathrm s,\quad\frac{T}{T_{\mathrm{stationary}}}=\sqrt{8/10}=.89443.$$The inertial and gravitational accelerations are perpendicular vectors; their magnitudes cannot be subtracted.

Physical check

The equilibrium leans opposite the support acceleration, and the larger effective field shortens the period.