A point A(x,y) moves uniformly round the circle x²+y² =

A point $A(x,y)$ moves uniformly round the circle $x^{2}+y^{2} = r^{2}$, completing one revolution in $T$ seconds, starting from $A_{0}(r,0)$ at $t = 0$. Find the velocity and acceleration of its projection onto $Ox$.

Step 1: The angular velocity

Uniform motion means the angle grows linearly, and one revolution of $2\pi$ takes $T$ seconds: $\omega = \frac{2\pi}{T}, \qquad \varphi(t) = \frac{2\pi t}{T}$.

Step 2: The law of motion of the projection

The starting point $(r,0)$ means $\varphi(0) = 0$, so

\[ x(t) = r\cos\frac{2\pi t}{T}. \]

Step 3: Velocity

\[ v_{x} = \frac{dx}{dt} = -r\omega\sin\omega t = -\frac{2\pi r}{T}\sin\frac{2\pi t}{T}. \]

Step 4: Acceleration

\[ w_{x} = \frac{d^{2}x}{dt^{2}} = -r\omega^{2}\cos\omega t = -\frac{4\pi^{2}r}{T^{2}}\cos\frac{2\pi t}{T} = -\frac{4\pi^{2}}{T^{2}}\,x. \]

The last form shows that the projection performs simple harmonic motion of amplitude $r$ and circular frequency $\dfrac{2\pi}{T}$.

Step 5: Verification

\[t = 0:\ x = r,\ v_{x} = 0,\ w_{x} = -\frac{4\pi^{2}r}{T^{2}} \ \text{(at rest at the extreme)} \;\;\checkmark\]

\[t = \frac{T}{4}:\ x = 0,\ \bigl|v_{x}\bigr| = \frac{2\pi r}{T},\ w_{x} = 0 \ \text{(fastest at the centre)} \;\;\checkmark\]

\[\text{The speed of } A \text{ itself is } r\omega = \frac{2\pi r}{T}, \text{ constant — the circumference divided by the period} \;\;\checkmark\]