Question

In circular motion, assuming ${\displaystyle \overline{v}=\overline{w}\times \overline{r}}$ , obtain an expression for the resultant acceleration of a particle in terms of tangential and radial component.

Answer 1

Acceleration of a particle,

${\displaystyle a=\underset{\delta t\to 0}{lim}\left(\frac{\delta v}{\delta t}\right)...\delta t\to 0;\delta t\ne 0}$

${\displaystyle \therefore a=\frac{dv}{dt}}$

${\displaystyle But,v=r\omega }$

${\displaystyle \therefore a=\frac{d}{dt}\left(r\omega \right)}$

${\displaystyle =r\frac{d\omega }{dt}+\omega \frac{dr}{dt}}$

${\displaystyle \because \text{r is constant}}$

${\displaystyle \frac{dr}{dt}=0}$

${\displaystyle \therefore a=r\frac{d\omega }{dt}}$

${\displaystyle \because \frac{d\omega }{dt}=\alpha }$

${\displaystyle \alpha =r\alpha }$

Given that:

${\displaystyle \overline{\text{v}}=\overline{\omega }\times \overline{\text{r}}}$

Differentiating w.r.t. time

${\displaystyle \frac{d\overline{v}}{dt}=\frac{d}{dt}(\overline{\omega }\times \overline{\text{r}})}$

${\displaystyle \frac{d\overline{v}}{dt}=\frac{d\overline{\omega }}{dt}\times \overline{r}+\overline{\omega }\times \frac{d\overline{r}}{dt}}$

${\displaystyle \frac{d\overline{v}}{dt}=\overline{\alpha }\times \overline{r}+\overline{\omega }\times \overline{v}}$

${\displaystyle \therefore \overline{a}={\overline{a}}_T+{\overline{a}}_r}$

Where,

a = Linear acceleration

a_T = Tangential component of linear acceleration

a_r = Radial component of linear acceleration