A point M moves along the circle x²+y² = a² with constant

A point $M$ moves along the circle $x^{2}+y^{2} = a^{2}$ with constant angular velocity $\omega$, starting at $(a,0)$ when $t = 0$. Find the law of motion of its projection $M_{1}$ onto $Ox$, together with the velocity and acceleration of $M_{1}$: at the initial moment, at the passage through the origin, and their maximum values.

Step 1: The law of motion

Constant angular velocity means the polar angle is $\varphi = \omega t$, so $M = \bigl(a\cos\omega t,\ a\sin\omega t\bigr)$ and the projection onto $Ox$ has abscissa

\[ x(t) = a\cos\omega t. \tag{1} \]

Step 2: Velocity and acceleration

\[ v = \dot{x} = -a\omega\sin\omega t, \qquad w = \ddot{x} = -a\omega^{2}\cos\omega t = -\omega^{2}x. \tag{2} \]

The last equality identifies the motion as simple harmonic: the acceleration is proportional to the displacement and directed against it.

Step 3: At the initial moment

At $t = 0$: $\cos0 = 1,\ \sin0 = 0$, so $x = a$ and

\[ v(0) = 0, \qquad w(0) = -a\omega^{2}. \]

The projection is momentarily at rest at the extreme right, with the largest possible acceleration, directed towards the origin.

Step 4: At the passage through the origin

Here $x = 0$, so $\cos\omega t = 0$ and $\bigl|\sin\omega t\bigr| = 1$: \[\bigl|v\bigr| = a\omega, \qquad w = -\omega^{2}\cdot0 = 0.\]

The acceleration vanishes exactly where the speed is greatest — the two extremes never coincide.

Step 5: Maximum values

\[ \bigl|v\bigr| = a\omega\bigl|\sin\omega t\bigr| \le a\omega, \qquad \bigl|w\bigr| = a\omega^{2}\bigl|\cos\omega t\bigr| \le a\omega^{2}. \]

\[x = a\cos\omega t, \qquad v_{\max} = a\omega \ \text{(at } x = 0), \qquad w_{\max} = a\omega^{2} \ \text{(at } x = \pm a)\]

Step 6: Summary

\[\begin{array}{c|c|c|c} \text{moment} & x & v & w \\ \hline t = 0 & a & 0 & -a\omega^{2} \\ \text{through } O & 0 & \mp a\omega & 0 \\ \text{far extreme} & -a & 0 & +a\omega^{2} \\ \hline \text{maximum } \bigl|\cdot\bigr| & a & a\omega & a\omega^{2} \end{array}\]

\[\text{Energy identity: } v^{2}+\omega^{2}x^{2} = a^{2}\omega^{2}\sin^{2}\omega t+a^{2}\omega^{2}\cos^{2}\omega t = a^{2}\omega^{2} \;\;\checkmark\]

\[\text{It confirms both extremes at once: } \bigl|v\bigr| = a\omega \iff x = 0, \text{ and } v = 0 \iff \bigl|x\bigr| = a \;\;\checkmark\]