Cyclist on Curved Paths – Detailed Solution and Interactive

Cyclist on Curved Paths – Detailed Solution and Interactive Quiz

Cyclist on Curved Paths

Statement I: A cyclist is moving on an unbanked road with a speed of \(6\,\text{km/h}\) and takes a sharp circular turn of radius \(2\,\text{m}\) without reducing the speed. The static friction coefficient is \(0.2\). The cyclist will not slip and can pass the curve. (Use \(g = 9.8\,\text{m/s}^2\).)

Statement II: If the same road is banked at an angle of \(45^\circ\), the cyclist can cross the same curve of \(2\,\text{m}\) radius with a speed of \(12\,\text{km/h}\) without slipping. (Use \(g = 9.8\,\text{m/s}^2\).)

Choose the option containing the correct statements.

(1) Statement I is incorrect and Statement II is correct
(2) Both Statement I and Statement II are true
(3) Statement I is correct and Statement II is incorrect
(4) Both Statement I and Statement II are false

Detailed Step‑by‑Step Explanation

Analysis of Statement I (Unbanked Road):

Step 1: Convert the given speed: \[ 6\,\text{km/h} = \frac{6}{3.6} \approx 1.67\,\text{m/s}. \]

Step 2: Required centripetal force for circular motion: \[ F_c = \frac{mv^2}{r} \propto v^2. \] (The mass \(m\) cancels out in comparison with friction capacity.)

Step 3: Maximum static friction available on a level (unbanked) road: \[ f_{\text{max}} = \mu mg. \] For a typical block of mass, \(mg\) is large compared to the tiny required centripetal force at such a low speed.
With \(v=1.67\,\text{m/s}\) and \(r=2\,\text{m}\): \[ \text{Required } F_c \propto \frac{(1.67)^2}{2} \approx \frac{2.7889}{2} \approx 1.39\,[\text{(units per mass)}]. \] Whereas, \[ f_{\text{max}} \propto 0.2 \times 9.8 \approx 1.96\,[\text{(units per mass)}]. \] Since \(1.39 < 1.96\), the static friction is sufficient. Hence, Statement I is true.

Analysis of Statement II (Banked Road):

Step 1: Convert the speed: \[ 12\,\text{km/h} = \frac{12}{3.6} \approx 3.33\,\text{m/s}. \]

Step 2: For a frictionless banked curve, the ideal speed is given by: \[ v_0 = \sqrt{g\,R\,\tan\theta}. \] Here, with \(R=2\,\text{m}\), \(\theta = 45^\circ\) (so \(\tan 45^\circ = 1\)) and \(g=9.8\,\text{m/s}^2\): \[ v_0 = \sqrt{9.8 \times 2 \times 1} = \sqrt{19.6} \approx 4.43\,\text{m/s}. \] This frictionless ideal speed (approximately \(4.43\,\text{m/s}\) or \(16\,\text{km/h}\)) is higher than \(3.33\,\text{m/s}\).

Step 3: At a speed lower than the ideal frictionless speed, the net force provided by the banking alone is not sufficient to give the required centripetal acceleration. The cyclist would tend to slide down the bank. Thus, if no extra frictional force is provided, the cyclist would slip; Statement II is false.

Final Conclusion: Statement I is correct and Statement II is incorrect.

Hence, the correct option is (3) Statement I is correct and Statement II is incorrect.

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