Question:
A car goes on a horizontal circular road of radius $\mathrm{R}$, the speed increasing at a constant rate $\mathrm{dv} / \mathrm{dt}=$ a. The friction dt coefficient between the road and the tyre is $\mu$. Find the speed at which the car will skid.
Solution:
From free body diagram
$\mathrm{ff}=\sqrt{\left(F_{t}^{2}+F_{c}^{2}\right)}$
$\mu \mathrm{N}=\sqrt{(m a)^{2}+\left(\frac{m v^{2}}{R}\right)^{2}}$
$(\mu \mathrm{mg})=\sqrt{(m a)^{2}+\left(\frac{m v^{2}}{R}\right)^{2}}$
Squaring and solving,
$V=\left[\left(\mu^{2} g^{2}-a^{2}\right) R^{2}\right]^{1 / 4}$
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