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Q.
When a car is moving along a circle on a level road the centripetal force is provided by $f$, where $f$ denotes as:
Laws of Motion
Solution:
Centripetal force $f \leq \mu_{s} N =\frac{ mv ^{2}}{ r }$ used in moving a car on a circular road. When car moves with speed $v$
centripetal acceleration required $=\frac{v^{2}}{r}$.
$a=\frac{ v [(-\sin \theta \hat{i})+(\cos \theta \hat{j})]-\hat{v}}{(r \theta / v)}$
$a=\frac{v^{2}}{r}\left(-\frac{\sin \theta}{\theta} \hat{i}-\frac{2 \sin ^{2} \frac{\theta}{2}}{\theta} \hat{j}\right) $
for $ \theta \rightarrow 0 \text { a } =\frac{v^{2}}{r}(-\hat{i})$
So, acceleration is $\frac{ v ^{2}}{ r }$ towards centre at $P$ so friction of ground provides centripetal force
$\mu_{s}=$ coefficient of static friction
