Q.Derive the expression for centripetal acceleration.

Step 1. Setup. Let a particle in uniform circular motion (radius $r$, speed $v$ constant) be at position vector $\vec r_1$ with velocity $\vec v_1$, and a short time $\Delta t$ later at $\vec r_2$ with velocity $\vec v_2$, with $|\vec r_1|=|\vec r_2|=r$ and $|\vec v_1|=|\vec v_2|=v$.

Step 2. Similar triangles. Because $\vec v\perp\vec r$ at every instant on a circle, the small angle $\theta$ through which $\vec r$ turns in time $\Delta t$ is the same angle through which $\vec v$ turns. So the isosceles triangle formed by $\vec r_1,\vec r_2,\Delta\vec r$ is geometrically similar to the isosceles triangle formed by $\vec v_1,\vec v_2,\Delta\vec v$ (both have the same apex angle $\theta$ and equal 'legs').

Step 3. Ratio from similarity. Similar triangles give $\dfrac{|\Delta\vec v|}{v}=\dfrac{|\Delta\vec r|}{r}$, i.e. $|\Delta\vec v|=\dfrac{v}{r}|\Delta\vec r|$; and since $\Delta\vec v$, by this construction, points radially inward (toward the centre), we can write $\Delta\vec v=-\dfrac{v}{r}\Delta\vec r$. …

Use the similarity of the position-vector triangle and the velocity-vector triangle (both subtending the …

• Concluding there is no acceleration because speed is constant — acceleration exists because direction changes …