# Chapter: Circular Motion 🔄

When a body moves in a circular path, its motion is called **circular motion**. In such motion, the body's **velocity** continuously changes due to the change in direction at every point, even if its **speed** remains constant.

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## **Key Concepts in Circular Motion**

### **Angular Displacement ($\theta$)**
Angular displacement is the angle swept by the radius vector of a body in a circular path. It is denoted by $\theta$.
* If a body moves from point A to B along a circular path of radius 'r', covering a linear distance (arc length) 's', the angular displacement is given by:
    $$ \theta = \frac{\text{Arc}}{\text{Radius}} = \frac{s}{r} $$
* This gives the relationship between linear and angular displacement:
    $$ s = r\theta $$
* **Units**: The unit of angular displacement is the **Radian (Rad)** in the MKS system and the **degree** in the CGS system.

### **Angular Velocity ($\omega$)**
Angular velocity is the rate of change of angular displacement. It is denoted by $\omega$ (omega).
* **Formula**:
    $$ \omega = \frac{d\theta}{dt} $$
* **Unit**: Radian per second (Rad/sec).

### **Angular Acceleration ($\alpha$)**
Angular acceleration is the rate of change of angular velocity. It is denoted by $\alpha$.
* **Formula**:
    $$ \alpha = \frac{d\omega}{dt} $$
* **Unit**: Radian per second squared (Rad/sec²).

### **Time Period (T) and Frequency (f)**
* **Time Period (T)**: The time taken by a body to complete one full rotation or revolution.
* **Frequency (f)**: The number of revolutions completed in one second. Its unit is **Hertz (Hz)** or per second ($s^{-1}$).
* **Relationship**: Frequency is the reciprocal of the time period.
    For one complete revolution, the angular distance covered is $2\pi$ radians in time T. Therefore:
    $$ \omega = \frac{2\pi}{T} $$
    Since $f = 1/T$, we can also write:
    $$ \omega = 2\pi f $$

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## **Relationship Between Linear and Angular Velocity**

We can derive the relationship between linear velocity (v) and angular velocity ($\omega$).
* We start with the formula for linear displacement:
    $s = r\theta$
* Differentiating both sides with respect to time (t):
    $\frac{ds}{dt} = \frac{d(r\theta)}{dt}$
* Since 'r' is a constant radius:
    $\frac{ds}{dt} = r \frac{d\theta}{dt}$
* We know that linear velocity $v = \frac{ds}{dt}$ and angular velocity $\omega = \frac{d\theta}{dt}$.
* Therefore, the relationship is:
    $$ v = r\omega $$

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## **Centripetal Acceleration ($a_c$)**

When a body moves in a circular orbit at a constant speed, its velocity changes direction at every point. This change in velocity means there is an acceleration. This acceleration is called **centripetal acceleration**, and it is always directed **towards the center** of the circular orbit.

### **Expression for Centripetal Acceleration**

**Derivation:**
Consider a body moving at a constant speed 'v' in a circular path of radius 'r'. Let its velocity be $V_A$ at point A and $V_B$ at point B.
* The change in velocity, $\Delta V$, is found by considering the vector components. For a small angular displacement $\theta$, the change in velocity is directed towards the center.
    $\Delta V = V_A \sin(\theta/2) + V_B \sin(\theta/2)$
* Since the speed is constant, $|V_A| = |V_B| = v$.
    $\Delta V = 2v \sin(\theta/2)$
* The time taken to move from A to B is $\Delta t = \frac{\text{Arc AB}}{v}$.
* From the definition of an angle, Arc AB = $r\theta$. So, $\Delta t = \frac{r\theta}{v}$.
* The acceleration is $a = \frac{\Delta V}{\Delta t}$:
    $a = \frac{2v \sin(\theta/2)}{r\theta/v} = \frac{2v^2 \sin(\theta/2)}{r\theta}$
* We can rewrite this as:
    $a = \frac{v^2}{r} \frac{\sin(\theta/2)}{\theta/2}$
* For a very small time interval ($\Delta t \to 0$), the angle $\theta$ also approaches zero. We use the small angle approximation limit:
    $\lim_{\theta \to 0} \frac{\sin(\theta/2)}{\theta/2} = 1$
* This leaves us with the final expression for centripetal acceleration:
    $$ a = \frac{v^2}{r} $$

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## **Centripetal and Centrifugal Force**

### **Centripetal Force ($F_c$)**
This is the force required to keep a body moving in a circular path at a constant speed. It is always directed **towards the center** of the circle.
* From Newton's second law, $F = ma$.
    $$ F_c = m \frac{v^2}{r} $$
* In terms of angular velocity (since $v=r\omega$):
    $F_c = m \frac{(r\omega)^2}{r} = m \frac{r^2\omega^2}{r}$
    $$ F_c = mr\omega^2 $$

### **Centrifugal Force**
This is a fictitious or apparent force that acts **outward** from the center of the circular path. Its magnitude is equal to the centripetal force but it acts in the opposite direction.
$$ \text{Centrifugal Force} = - \text{Centripetal Force} $$

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## **Applications of Centripetal Force**

### **1. Motion of a Car on a Level Curve**
For a car moving on a flat (unbanked) curved road, the necessary centripetal force is provided by the force of **static friction** between the tires and the road.
* $F_c = F_{\text{friction}}$
    $\frac{mv^2}{r} = \mu_s N = \mu_s mg$
* The maximum safe speed for the car without skidding is:
    $$ v = \sqrt{\mu_s rg} $$
    This shows the safe velocity depends on the radius of the curve and the coefficient of friction.

### **2. Banking of Roads**
Banking is the phenomenon of raising the outer edge of a curved road with respect to the inner edge. This is done to provide the necessary centripetal force from a component of the normal reaction force, reducing reliance on friction.
* The normal reaction force (R) is resolved into two components.
    * $R\cos\theta$ balances the weight of the car ($mg$).
    * $R\sin\theta$ provides the necessary centripetal force.
* Equations:
    1. $R\cos\theta = mg$
    2. $R\sin\theta = \frac{mv^2}{r}$
* Dividing equation (2) by (1):
    $\frac{R\sin\theta}{R\cos\theta} = \frac{mv^2/r}{mg} \implies \tan\theta = \frac{v^2}{rg}$
* The ideal speed for a given banking angle is:
    $$ v = \sqrt{rg \tan\theta} $$

### **3. Bending of a Cyclist**
A cyclist bends inward while taking a turn to provide the necessary centripetal force from a component of the normal reaction force from the ground. The derivation is identical to the banking of a road.
* The angle of bending $\theta$ with the vertical is given by:
    $$ \tan\theta = \frac{v^2}{rg} $$

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## **Motion in a Vertical Circle**

Consider a mass 'm' tied to a string and whirled in a vertical circle of radius 'r'. The tension in the string varies depending on the position.

* **At the Highest Point (A):**
    The centripetal force is the sum of tension and weight, as both act downwards.
    $T_A + mg = \frac{mv_A^2}{r} \implies T_A = \frac{mv_A^2}{r} - mg$
    This is the point of **minimum tension**.

* **At the Lowest Point (C):**
    The centripetal force is the difference between tension (upward) and weight (downward).
    $T_C - mg = \frac{mv_C^2}{r} \implies T_C = \frac{mv_C^2}{r} + mg$
    This is the point of **maximum tension**.

* **At the Middle Points (B and D):**
    The weight acts vertically downwards and does not contribute to the centripetal force, which is provided solely by tension.
    $T_B = \frac{mv_B^2}{r}$

### **Minimum Velocities in a Vertical Circle**

* **Minimum Velocity at the Highest Point (Critical Velocity):**
    The minimum velocity required to complete the circle occurs when the tension at the highest point is zero ($T_A=0$).
    $0 = \frac{mv_1^2}{r} - mg \implies \frac{mv_1^2}{r} = mg$
    $$ v_1 = \sqrt{rg} $$
    Below this speed, the string will go slack, and the object will not complete the circle.

* **Minimum Velocity at the Lowest Point:**
    Using the principle of conservation of energy between the highest and lowest points (a height difference of $h=2r$):
    Gain in KE = Loss in PE
    $\frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = mg(2r)$
    Substituting $v_1^2 = rg$:
    $v_2^2 - v_1^2 = 4gr \implies v_2^2 - rg = 4gr$
    $v_2^2 = 5gr$
    $$ v_2 = \sqrt{5rg} $$

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## **Conical Pendulum**

When a suspended body is whirled in a horizontal circle, it forms a conical pendulum.

* A mass 'm' is attached to a string of length 'L' and revolves in a horizontal circle of radius 'r'.
* The tension 'T' in the string is resolved into two components:
    1. $T\cos\theta$ balances the weight ($mg$).
    2. $T\sin\theta$ provides the necessary centripetal force.
* Equations:
    1. $T\cos\theta = mg$
    2. $T\sin\theta = \frac{mv^2}{r}$
* Dividing (2) by (1) gives the velocity:
    $\tan\theta = \frac{v^2}{rg} \implies v = \sqrt{rg \tan\theta}$

* **Time Period of Oscillation (t):**
    $t = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2\pi r}{v}$
    $t = \frac{2\pi r}{\sqrt{rg \tan\theta}} = 2\pi \sqrt{\frac{r^2}{rg \tan\theta}} = 2\pi \sqrt{\frac{r}{g \tan\theta}}$
* From the geometry of the triangle, $\tan\theta = r/h$ (where h is the vertical depth).
    $t = 2\pi \sqrt{\frac{r}{g(r/h)}} = 2\pi \sqrt{\frac{h}{g}}$
* Also, $h = L\cos\theta$. Therefore, the time period is:
    $$ t = 2\pi \sqrt{\frac{L\cos\theta}{g}} $$

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## **Short Answer Questions**

**1) Why is it more difficult to whirl a stone tied with a longer string than a shorter string?**
The required centripetal force is given by $F = m\omega^2r$. For the same angular velocity ($\omega$), a longer string (larger 'r') requires a greater centripetal force to maintain the circular motion, making it more difficult to whirl.

**2) What is the working principle of a spin dryer in a washing machine?**
It works on the principle of **centrifugal force**. As the drum spins at high speed, the water particles are forced outward through the holes in the drum, separating them from the clothes and drying them faster.

**3) What is the significance of banking a curved road?**
When a road is unbanked, friction provides the necessary centripetal force, causing wear and tear on tires. By banking the road, the horizontal component of the normal reaction force provides the required centripetal force. This makes driving safer, reduces tire wear, and allows for higher speeds on curves.

**4) Why don't electrons collapse into the positively charged nucleus?**
The electrostatic attraction between the nucleus and the electron provides the necessary centripetal force for the electron to move in its orbit. This force is always perpendicular to the electron's velocity. Since no work is done on the electron, it continues its circular motion without losing energy and does not collapse into the nucleus.

**5) Why are passengers thrown away from the center of a curved path when a bus takes a turn?**
This is due to **inertia**. The passengers' bodies tend to continue moving in a straight line (Newton's First Law). When the bus turns, the passengers feel an outward push (centrifugal force) because their bodies resist the change in direction, throwing them away from the center of the curve.

**6) What is the difference between a simple and conical pendulum?**

| Feature | Simple Pendulum | Conical Pendulum |
| :--- | :--- | :--- |
| **Motion** | Moves in a to-and-fro (back and forth) motion. | Moves in a horizontal circle. |
| **Principle** | Obeys the principle of Simple Harmonic Motion (SHM).| Does not obey the principle of SHM. |
| **Acceleration** | Directed towards the mean position. | Directed towards the center of the circle. |
| **Time Period** | $t = 2\pi\sqrt{L/g}$ | $t = 2\pi\sqrt{(L\cos\theta)/g}$ |