Question

Derive the formula for centripetal acceleration of a particle which is moving with a constant speed on a circular path.

Answer 1

Consider the distance travelled by the body over a tome $\Delta t$

Arc length $S=V\Delta t$

also $S=r\Delta \theta$

$\therefore r\Delta \theta =V\Delta t\frac{\Delta \theta }{\Delta t}=\frac{V}{r}\Rightarrow \frac{d\theta }{dt}=\frac{V}{r}$ angular velocity

Now from the second diagram:

$\sin \left(\Delta \theta /2\right)=\frac{\Delta V}{2V}\Delta V=2V\sin \left(\Delta \theta /2\right)\frac{\Delta V}{\Delta \theta }=\frac{2V\sin \left(\Delta \theta /2\right)}{\Delta \theta }\Rightarrow \frac{\Delta V}{\Delta \theta }=\frac{V\sin \left(\Delta \theta /2\right)}{\left(\Delta \theta /2\right)}\Rightarrow \frac{dV}{d\theta }={lim}_{\Delta \theta ⟶0}\frac{V\sin \left(\Delta \theta /2\right)}{\left(\Delta \theta /2\right)}=V$

$\Rightarrow \frac{dV}{d\theta }=\frac{dv}{d\theta }\frac{d\theta }{dt}=V\times \frac{V}{r}=\frac{V^2}{r}$