Circular Motion – Concept Booster | Class 11 Physics CBSE

First, convert speed to m/s: $v = 18 \times \frac{5}{18} = 5 \text{ m/s}$.
The maximum safe speed on an unbanked road is given by $v_{\text{max}} = \sqrt{\mu_s r g}$.
$v_{\text{max}} = \sqrt{0.1 \times 3 \times 9.8} = \sqrt{2.94} \approx 1.71 \text{ m/s}$.
Since the cyclist’s speed ($5 \text{ m/s}$) is much greater than the maximum safe speed ($1.71 \text{ m/s}$), the required centripetal force exceeds the available friction, and the cyclist will skid outwards and slip.

1. Draw the FBD of the car. Normal force $N$ acts perpendicular to the banked surface (angle $\theta$). Weight $mg$ acts downwards. Limiting friction $f_s$ acts down the slope to prevent the car from skidding outwards.
2. Resolve forces vertically: $N \cos\theta = mg + f_s \sin\theta$.
Since $f_s = \mu_s N$, we get $N(\cos\theta – \mu_s \sin\theta) = mg$. (Equation 1)
3. Resolve forces horizontally (towards the center): $N \sin\theta + f_s \cos\theta = \frac{mv_{\text{max}}^2}{r}$.
Substituting $f_s = \mu_s N$: $N(\sin\theta + \mu_s \cos\theta) = \frac{mv_{\text{max}}^2}{r}$. (Equation 2)
4. Divide Eq 2 by Eq 1: $\frac{\sin\theta + \mu_s \cos\theta}{\cos\theta – \mu_s \sin\theta} = \frac{v_{\text{max}}^2}{rg}$.
Divide numerator and denominator by $\cos\theta$ to get: $v_{\text{max}} = \sqrt{rg \left(\frac{\tan\theta + \mu_s}{1 – \mu_s \tan\theta}\right)}$.

Let $v_B$ be velocity at bottom and $v_T$ be velocity at top.
Tension at bottom (Max): $T_{\text{max}} = \frac{mv_B^2}{L} + mg$
Tension at top (Min): $T_{\text{min}} = \frac{mv_T^2}{L} – mg$
By conservation of energy: $\frac{1}{2}mv_B^2 = \frac{1}{2}mv_T^2 + mg(2L) \implies v_B^2 = v_T^2 + 4gL$.
Given $T_{\text{max}} / T_{\text{min}} = 4 \implies T_{\text{max}} = 4 T_{\text{min}}$.
$\frac{m(v_T^2 + 4gL)}{L} + mg = 4 \left( \frac{mv_T^2}{L} – mg \right)$
Divide by $m$ and multiply by $L$: $(v_T^2 + 4gL) + gL = 4v_T^2 – 4gL$
$v_T^2 + 5gL = 4v_T^2 – 4gL \implies 3v_T^2 = 9gL \implies v_T^2 = 3gL$.
Substitute values: $v_T^2 = 3 \times 10 \times (\frac{10}{3}) = 100 \implies v_T = 10 \text{ m/s}$.