NEET-XII-Physics
07: Circular Motion
- #20A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate
dvdt=a. The friction coefficient between the road and the tyre is μ. Find the speed at which the car will skid.Ans : Let v be the speed of the car.
Since the motion is non-uniform, the acceleration has both radial (ar) and tangential (at) components.
`` {a}_{r}=\frac{{v}^{2}}{R}``
`` {a}_{t}=\frac{dv}{dt}=a``
`` \,\mathrm{\,Resultant \,}\,\mathrm{\,magnitude \,}=\sqrt{{\left(\frac{{v}^{2}}{R}\right)}^{2}+{a}^{2}}``
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From free body diagram, we have:
`` mN=m\sqrt{{\left(\frac{{v}^{2}}{R}\right)}^{2}+{a}^{2}}``
`` \Rightarrow \mu mg=m\sqrt{{\left(\frac{{v}^{2}}{R}\right)}^{2}+{a}^{2}}``
`` \Rightarrow {\mu }^{2}{g}^{2}=\frac{{v}^{4}}{{R}^{2}}+{a}^{2}``
`` \Rightarrow \frac{{v}^{4}}{{R}^{2}}=({\,\mathrm{\,\mu \,}}^{2}{g}^{2}-{a}^{2})``
`` \Rightarrow {v}^{4}=({\,\mathrm{\,\mu \,}}^{2}{g}^{2}-{a}^{2}){R}^{2}``
`` \Rightarrow v=\left[\right({\,\mathrm{\,\mu \,}}^{2}{\,\mathrm{\,g \,}}^{2}-{\,\mathrm{\,a \,}}^{2}){R}^{2}{]}^{1/4}``
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