Concept Card: Dynamics of Uniform Circular Motion & Banking of Roads
1. Centripetal Acceleration & Centripetal Force ($F_c$):
A particle of mass $m$ executing uniform circular motion of radius $r$ at speed $v$ (angular speed $\omega = \frac{v}{r}$) experiences a continuous inward radial acceleration:
$a_c = \frac{v^2}{r} = \omega^2 r = v\omega$
By Newton’s Second Law, the required net inward radial force is:
$$F_c = m a_c = \frac{m v^2}{r} = m\omega^2 r$$
- Physical Meaning: Centripetal force is not a new physical force of nature; it is simply the vector sum of real physical forces acting towards the center of curvature (e.g., tension, gravity, normal reaction, friction).
- Zero Work Done: Since $\vec{F}_c \perp d\vec{s}$ at every instant in UCM, $W = \int \vec{F}_c \cdot d\vec{s} = 0$. Centripetal force alters the direction of velocity without changing speed or kinetic energy.
2. Motion of a Vehicle on a Level (Unbanked) Curved Road:
On a flat, horizontal turn of radius $r$, the inward centripetal force is provided solely by static friction between tires and road:
$$f_s = \frac{m v^2}{r} \le \mu_s N = \mu_s mg \implies v \le \sqrt{\mu_s r g}$$
$$\mathbf{v_{\text{max}} = \sqrt{\mu_s r g}}$$
If speed exceeds $\sqrt{\mu_s r g}$, static friction is exceeded and the vehicle skids radially outward.
3. Banking of Roads & Railway Tracks:
Raising the outer edge of a curved road at angle $\theta$ utilizes the horizontal component of the normal contact force to provide centripetal acceleration.
- Frictionless (Design / Optimum) Speed ($v_0$):
At this speed, the turn is negotiated without relying on friction ($f = 0$):
$N\sin\theta = \frac{m v_0^2}{r}, \quad N\cos\theta = mg \implies \tan\theta = \frac{v_0^2}{r g}$
$v_0 = \sqrt{r g \tan\theta} \quad \text{and} \quad \theta = \arctan\left(\frac{v_0^2}{r g}\right)$
Superelevation of Rails (gauge $b$): $h \approx b\tan\theta = \frac{b v_0^2}{r g}$. - Rough Banked Road – Maximum Safe Speed ($v_{\text{max}}$):
Friction acts downward along the banked incline to resist outward skidding:
$v_{\text{max}} = \sqrt{r g \left(\frac{\tan\theta + \mu_s}{1 – \mu_s\tan\theta}\right)}$
(If $\mu_s \ge \cot\theta$, $v_{\text{max}} \to \infty$; the car will never skid outward). - Rough Banked Road – Minimum Safe Speed ($v_{\text{min}}$):
Friction acts upward along the banked incline to resist inward sliding:
$v_{\text{min}} = \sqrt{r g \left(\frac{\tan\theta – \mu_s}{1 + \mu_s\tan\theta}\right)} \quad (\text{for } \tan\theta > \mu_s)$
(If $\tan\theta \le \mu_s$, $v_{\text{min}} = 0$; the vehicle can rest parked on the bank without slipping down).
4. The Conical Pendulum:
A bob of mass $m$ suspended by a string of length $L$ revolving in a horizontal circle of radius $r = L\sin\theta$:
- String tension: $T = \frac{mg}{\cos\theta} = m\omega^2 L$
- Angular frequency: $\omega = \sqrt{\frac{g}{L\cos\theta}} = \sqrt{\frac{g}{h}}$ (where $h = L\cos\theta$ is vertical depth).
- Time period: $\tau = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{h}{g}}$
Solved Examples
Example 1 (Maximum Safe Speed on a Level Curve & Friction Modulation):
An unbanked circular highway turn has a radius $r = 100.0\text{ m}$. The coefficient of static friction on dry pavement is $\mu_s = 0.64$. Taking $g = 10\text{ m/s}^2$:
(a) Find the maximum safe speed $v_{\text{max}}$ in $\text{m/s}$ and $\text{km/h}$.
(b) If a car of mass $m = 1200\text{ kg}$ rounds this curve at $v = 15.0\text{ m/s}$, find the frictional force exerted by the road.
(c) What is the maximum safe speed if rain reduces static friction to $\mu_s’ = 0.16$?
Solution:
(a) $v_{\text{max}} = \sqrt{\mu_s r g} = \sqrt{0.64 \times 100.0 \times 10} = \sqrt{640} = 8\sqrt{10}\text{ m/s} \approx 25.30\text{ m/s} = 91.07\text{ km/h}$.
(b) Required centripetal friction: $f_s = \frac{m v^2}{r} = \frac{1200 \times 15.0^2}{100.0} = 12 \times 225 = 2700\text{ N}$.
(Limiting friction is $f_L = \mu_s mg = 7680\text{ N} > 2700\text{ N}$, so the car turns safely).
(c) Wet pavement: $v_{\text{max}}’ = \sqrt{0.16 \times 100.0 \times 10} = \sqrt{160} = 4\sqrt{10}\text{ m/s} \approx 12.65\text{ m/s} = 45.54\text{ km/h}$.
Example 2 (Optimum Banking & Railway Superelevation):
A circular railway curve of radius $r = 400.0\text{ m}$ on a broad-gauge track ($b = 1.67\text{ m}$) is designed for operating train speeds of $72.0\text{ km/h}$ ($20.0\text{ m/s}$). Taking $g = 9.8\text{ m/s}^2$:
(a) Find the optimum banking angle $\theta$ for zero lateral wheel flange thrust.
(b) Calculate the required elevation $h$ of the outer rail.
(c) Calculate the normal force exerted by the rails on an engine car of mass $m = 40,000\text{ kg}$ navigating the curve at this designed speed.
Solution:
(a) $\tan\theta = \frac{v_0^2}{r g} = \frac{20.0^2}{400.0 \times 9.8} = \frac{400.0}{3920.0} \approx 0.10204 \implies \theta \approx 5.82^\circ$.
(b) $h = b\tan\theta = (1.67\text{ m})(0.10204) \approx 0.1704\text{ m} = 17.04\text{ cm}$.
(c) Normal force: $N = \frac{mg}{\cos\theta} = mg\sqrt{1 + \tan^2\theta} = (40,000 \times 9.8)\sqrt{1 + 0.10204^2} \approx 394,035\text{ N}$.
Example 3 (Safe Speed Window on a Rough Banked Track):
A test track curve of radius $r = 50.0\text{ m}$ is banked at $\theta = 37^\circ$ ($\tan 37^\circ = 0.75$). The coefficient of static friction is $\mu_s = 0.25$. Taking $g = 10\text{ m/s}^2$:
(a) Find the optimum speed $v_0$.
(b) Find the maximum safe speed $v_{\text{max}}$.
(c) Find the minimum safe speed $v_{\text{min}}$.
Solution:
(a) $v_0 = \sqrt{r g \tan\theta} = \sqrt{50.0 \times 10 \times 0.75} = \sqrt{375} \approx 19.36\text{ m/s}$.
(b) $v_{\text{max}} = \sqrt{500 \left(\frac{0.75 + 0.25}{1 – 0.25(0.75)}\right)} = \sqrt{500 \left(\frac{1.0}{0.8125}\right)} = \sqrt{\frac{8000}{13}} \approx 24.81\text{ m/s}$.
(c) $v_{\text{min}} = \sqrt{500 \left(\frac{0.75 – 0.25}{1 + 0.25(0.75)}\right)} = \sqrt{500 \left(\frac{0.50}{1.1875}\right)} = \sqrt{\frac{4000}{19}} \approx 14.51\text{ m/s}$.
Safe speed window: $14.51\text{ m/s} \le v \le 24.81\text{ m/s}$.
Example 4 (Conical Pendulum Dynamics & Horizontal Limit):
A bob of mass $m = 0.50\text{ kg}$ is attached to a string of length $L = 1.0\text{ m}$ and revolves in a horizontal circle at semi-vertical angle $\theta = 60^\circ$. Taking $g = 10\text{ m/s}^2$:
(a) Calculate string tension $T$.
(b) Find radius $r$ and orbital speed $v$.
(c) Find the period of revolution $\tau$.
(d) Explain why revolving at $\theta = 90^\circ$ is physically impossible.
Solution:
(a) $T = \frac{mg}{\cos(60^\circ)} = \frac{0.50 \times 10}{0.50} = 10.0\text{ N}$.
(b) $r = L\sin(60^\circ) = \frac{\sqrt{3}}{2}\text{ m} \approx 0.866\text{ m}$.
$T\sin(60^\circ) = \frac{m v^2}{r} \implies v^2 = \frac{10.0 \times 0.75}{0.50} = 15.0 \implies v = \sqrt{15.0} \approx 3.87\text{ m/s}$.
(c) $\tau = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{0.50}{10}} = \frac{\pi}{\sqrt{5}} \approx 1.405\text{ s}$.
(d) As $\theta \to 90^\circ$, $\cos\theta \to 0 \implies T = \frac{mg}{\cos\theta} \to \infty$ and $v \to \infty$. Infinite force and energy are impossible, so the string can never be strictly horizontal.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A car of mass $m$ rounds an unbanked circular curve of radius $r$ on a level road with coefficient of static friction $\mu$. The maximum safe speed is:
(A) $\sqrt{\mu r g}$
(B) $\sqrt{\frac{r g}{\mu}}$
(C) $\mu r g$
(D) $\sqrt{\frac{\mu g}{r}}$
Problem 2 (JEE Main – Single Correct):
The banking angle $\theta$ for a curved road of radius $r$ designed for vehicles traveling at speed $v$ without depending on friction is given by:
(A) $\sin\theta = \frac{v^2}{r g}$
(B) $\tan\theta = \frac{v^2}{r g}$
(C) $\cos\theta = \frac{v^2}{r g}$
(D) $\tan\theta = \frac{r g}{v^2}$
Problem 3 (JEE Main – Single Correct):
The work done by the centripetal force on a particle of mass $m$ revolving in a circular orbit of radius $r$ with constant speed $v$ during one complete revolution is:
(A) $2\pi r \left(\frac{m v^2}{r}\right)$
(B) $m v^2$
(C) $\frac{1}{2}m v^2$
(D) Zero
Problem 4 (JEE Main – Single Correct):
The period of revolution of a conical pendulum of string length $L$ making an angle $\theta$ with the vertical is:
(A) $2\pi\sqrt{\frac{L}{g}}$
(B) $2\pi\sqrt{\frac{L\cos\theta}{g}}$
(C) $2\pi\sqrt{\frac{L\sin\theta}{g}}$
(D) $2\pi\sqrt{\frac{L\tan\theta}{g}}$
Problem 5 (JEE Main – Single Correct):
A vehicle negotiates a frictionless banked turn of angle $\theta$ at its designed optimum speed $v_0$. The normal reaction $N$ exerted by the road on the vehicle is:
(A) $mg\cos\theta$
(B) $\frac{mg}{\cos\theta}$
(C) $mg$
(D) $mg\tan\theta$
Problem 6 (JEE Advanced – One or More Correct):
For a car on a rough road banked at angle $\theta$ with static friction coefficient $\mu_s$:
(A) If $\mu_s \ge \cot\theta$, the car will never skid outward regardless of how high its speed is.
(B) If $\tan\theta \le \mu_s$, the car can remain parked at rest on the banked road without sliding down.
(C) At the design speed $v_0 = \sqrt{r g \tan\theta}$, the frictional force between tires and road is zero.
(D) If the car travels at $v > v_0$, static friction acts down the banked incline.
Problem 7 (JEE Advanced – One or More Correct):
For a conical pendulum of mass $m$, string length $L$, and semi-vertical angle $\theta$:
(A) The tension in the string is $T = \frac{mg}{\cos\theta}$.
(B) The tension in the string is $T = m\omega^2 L$.
(C) The total mechanical energy of the pendulum bob is conserved.
(D) The angular momentum of the bob about the suspension point remains constant in direction.
Problem 8 (JEE Advanced – One or More Correct):
A cyclist negotiates an unbanked circular curve of radius $r$ at speed $v$. To prevent toppling, the cyclist leans inward at an angle $\theta$ to the vertical. Which of the following is/are correct?
(A) The angle of inclination satisfies $\tan\theta = \frac{v^2}{r g}$.
(B) The torque about the center of mass due to the normal reaction and friction balances to zero.
(C) The friction force provides the centripetal acceleration: $f_s = \frac{m v^2}{r}$.
(D) Leaning inward decreases the required centripetal acceleration.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A car is driven around a flat circular track of radius $r = 45.0\text{ m}$. The coefficient of static friction between the tires and the track is $\mu_s = 0.50$. Taking $g = 10\text{ m/s}^2$, find the maximum speed in $\text{m/s}$ at which the car can travel without skidding.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A circular roadway of radius $r = 80.0\text{ m}$ is banked at an angle $\theta$ such that $\tan\theta = 0.50$. Taking $g = 10\text{ m/s}^2$, find the optimum design speed $v_0$ in $\text{m/s}$ at which a vehicle can turn with zero lateral friction.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (D) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C) | 9. 15 | 10. 20
Solution 1:
$f_s = \frac{mv^2}{r} \le \mu mg \implies v_{\text{max}} = \sqrt{\mu r g}$.
Correct Answer: (A)
Solution 2:
$N\sin\theta = \frac{mv^2}{r}, \quad N\cos\theta = mg \implies \tan\theta = \frac{v^2}{rg}$.
Correct Answer: (B)
Solution 3:
$\vec{F}_c \perp d\vec{s} \implies W = \int \vec{F}_c \cdot d\vec{s} = 0$.
Correct Answer: (D)
Solution 4:
$\tau = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$.
Correct Answer: (B)
Solution 5:
$N\cos\theta = mg \implies N = \frac{mg}{\cos\theta} > mg$.
Correct Answer: (B)
Solution 6:
All statements (A, B, C, D) are verified fundamental properties of banked turns with friction.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) $T\cos\theta = mg \implies T = \frac{mg}{\cos\theta}$ (True).
– (B) $T\sin\theta = m\omega^2(L\sin\theta) \implies T = m\omega^2 L$ (True).
– (C) Speed and height remain constant $\implies$ mechanical energy is conserved (True).
– (D) The angular momentum vector precesses around the vertical axis; its direction is not constant (False).
Correct Answer: (A, B, C)
Solution 8:
– (A) Torque balance about CM: $\tan\theta = \frac{v^2}{rg}$ (True).
– (B) Cyclist maintains rotational equilibrium about CM (True).
– (C) Friction provides centripetal force (True).
– (D) Centripetal acceleration is $a_c = v^2/r$, independent of lean angle (False).
Correct Answer: (A, B, C)
Solution 9:
$v_{\text{max}} = \sqrt{\mu_s r g} = \sqrt{0.50 \times 45.0 \times 10} = \sqrt{225} = 15.0\text{ m/s}$.
Correct Answer: 15
Solution 10:
$v_0 = \sqrt{r g \tan\theta} = \sqrt{80.0 \times 10 \times 0.50} = \sqrt{400} = 20.0\text{ m/s}$.
Correct Answer: 20