Simple Harmonic Motion and Uniform Circular Motion — Question 8

Question 8

Angular acceleration spoils the simple harmonic projection

A point stays on a circle of fixed radius $R$, but its polar angle $\theta(t)$ need not increase uniformly. Its horizontal coordinate is $x=R\cos\theta$.

1. Derive $\ddot x$ in terms of $\dot\theta$ and $\ddot\theta$.

2. At $\theta=\pi/2$, suppose $\ddot\theta=\alpha\ne0$. Can the projection obey $\ddot x=-\omega^2x$ about zero at that instant? Explain the role of tangential acceleration.

Question 8 - Solution

Angular acceleration spoils the simple harmonic projection

(a) Derivation and calculation

$$x=R\cos\theta,\qquad\dot x=-R\dot\theta\sin\theta.$$$$\boxed{\ddot x=-R\dot\theta^2\cos\theta-R\ddot\theta\sin\theta}.$$$$\mathbf a=-R\dot\theta^2\hat{\mathbf e}_r+R\ddot\theta\hat{\mathbf e}_{\theta}.$$The two terms are the horizontal components of radial and tangential acceleration.

(b) Result and interpretation

$$\theta=\pi/2,\quad\ddot\theta=\alpha\ne0:\qquad x=0,\quad\boxed{\ddot x=-R\alpha\ne0}.$$$$\ddot x=-\omega^2x\ \text{would require}\ \ddot x=0.$$Uniform angular speed sets $\ddot\theta=0$ and $\dot\theta^2=\omega^2$ constant, recovering the harmonic projection.

Physical check

A circular geometric path does not by itself specify the time dependence along the path.