Motion in a plane and equilibrium of rigid bodies

Introduction: Welcome to the World of Motion in a Plane and Equilibrium of Rigid Bodies!

"I can handle calculations when an object moves along a single straight line, but as soon as it is launched diagonally or starts rotating, it suddenly feels so difficult..."
Have you ever felt that way? It might seem tricky at first, but don't worry!
There are really only two main themes to master in this chapter:
1. Motion in a Plane (Projectile Motion, etc.): Once you resolve it into horizontal and vertical components, it's actually nothing more than a combination of the linear motions you learned in Basic Physics.
2. Equilibrium of Rigid Bodies: We will master the "conditions for an object not to rotate" when its size cannot be ignored.
If you organize the fundamental principles one by one, you'll be able to solve Common Test questions with surprising ease. Let's master it together step by step!

Section 1: Velocity and Acceleration in Curvilinear Motion

When an object moves in a plane (along a curved path), we use vectors (quantities having both magnitude and direction) to describe its position and motion accurately.

1. Vector Representation of Displacement, Velocity, and Acceleration

The golden rule for motion in a plane is to break it down into two perpendicular components: the \(x\)-axis (such as the horizontal direction) and the \(y\)-axis (such as the vertical direction).
- Displacement Vector \(\vec{r}\): Represents the change in position. In component form, it is written as \(\vec{r} = (x, y)\).
- Velocity Vector \(\vec{v}\): The displacement per unit time. In component form, it is expressed as \(\vec{v} = (v_x, v_y)\), and its magnitude (speed) \(v\) is given by the Pythagorean theorem: \(v = \sqrt{v_x^2 + v_y^2}\). The direction of the velocity vector is always tangent to the trajectory.
- Acceleration Vector \(\vec{a}\): The change in velocity per unit time. In component form, it is written as \(\vec{a} = (a_x, a_y)\), and its magnitude \(a\) is \(a = \sqrt{a_x^2 + a_y^2}\). The direction of acceleration is the same as the direction in which velocity changes (the direction of the net force acting on the object).

Key Point: The Strategy for Planar Motion is "Separating into Two Linear Motions"

Whenever you encounter motion in a plane, the first step is to treat the "motion in the \(x\)-direction" and the "motion in the \(y\)-direction" as completely independent. This is known as the independence of motions. The two motions do not interfere with each other at all.

Common Pitfall

A very common mistake when finding the magnitude of velocity is simply adding the components together as \(v = v_x + v_y\). Because velocity is a vector, you must use the Pythagorean theorem \(v = \sqrt{v_x^2 + v_y^2}\), just like finding the hypotenuse of a right-angled triangle.

Section 2: Projectile Motion (Horizontal and Oblique Projectiles)

The curved path of an object moving through the air under the influence of gravity alone is a parabola, and this type of motion is called projectile motion.
Since no force acts horizontally, the object undergoes uniform linear motion in the horizontal direction; vertically, it experiences a constant downward gravitational force, undergoing uniformly accelerated linear motion (acceleration \(g\)).

1. Horizontal Projection

This is the motion of an object projected horizontally with an initial velocity \(v_0\) from a cliff of height \(h\). Let horizontally to the right be the positive \(x\)-direction, and vertically downward be the positive \(y\)-direction.
- Horizontal Direction (\(x\)-direction): Since no external force acts, it remains in uniform linear motion.
Velocity: \(v_x = v_0\)
Position: \(x = v_0 t\)
- Vertical Direction (\(y\)-direction): Uniformly accelerated linear motion, identical to free fall from rest (initial velocity \(0\)).
Velocity: \(v_y = gt\)
Position: \(y = \frac{1}{2}gt^2\)
By eliminating time \(t\), we obtain the equation of the trajectory: \(y = \frac{g}{2v_0^2}x^2\), mathematically confirming that the path traces a parabola (a quadratic function).

2. Oblique Projection

This is the motion of an object launched diagonally from the ground with an initial velocity \(v_0\) at an elevation angle \(\theta\) above the horizontal. Let horizontally to the right be the positive \(x\)-direction, and vertically upward be the positive \(y\)-direction.
First, resolve the initial velocity into components:
Horizontal initial velocity: \(v_{0x} = v_0 \cos\theta\)
Vertical initial velocity: \(v_{0y} = v_0 \sin\theta\)

- Horizontal Direction (\(x\)-direction): Uniform linear motion
Velocity: \(v_x = v_0 \cos\theta\)
Position: \(x = (v_0 \cos\theta)t\)
- Vertical Direction (\(y\)-direction): Vertical projection upward (acceleration \(-g\))
Velocity: \(v_y = v_0 \sin\theta - gt\)
Position: \(y = (v_0 \sin\theta)t - \frac{1}{2}gt^2\)

Crucial Checkpoints for Oblique Projection

- Time to reach maximum height \(t_1\): The point where the vertical velocity becomes instantaneously \(0\). From \(v_y = 0\), we get \(t_1 = \frac{v_0 \sin\theta}{g}\). Note carefully that the horizontal velocity \(v_x = v_0 \cos\theta\) is NOT zero at this point!
- Time to return to the ground (time of flight) \(t_2\): Due to symmetry, this is exactly twice the time to the highest point (\(t_2 = 2t_1 = \frac{2v_0 \sin\theta}{g}\)). Alternatively, it can be found by solving \(y = 0\).
- Horizontal Range \(L\): The horizontal distance traveled during the time of flight \(t_2\): \(L = v_x \times t_2 = (v_0 \cos\theta) \times \frac{2v_0 \sin\theta}{g} = \frac{2v_0^2 \sin\theta \cos\theta}{g}\). Using the trigonometric double-angle formula, this can be written neatly as \(L = \frac{v_0^2 \sin 2\theta}{g}\).

Quick Trivia: How to Launch the Farthest?

In the horizontal range formula \(L = \frac{v_0^2 \sin 2\theta}{g}\), for a fixed initial speed \(v_0\), the distance is maximized when \(\sin 2\theta\) reaches its maximum value of \(1\). This means launching at an angle of \(2\theta = 90^\circ \implies \theta = 45^\circ\) will send the projectile the farthest!

3. Falling Motion with Air Resistance

In the real world, the presence of air means an object experiences an air resistance (drag) force opposite to its direction of motion, proportional to its speed (or the square of its speed).
When an object falls freely, it initially accelerates under gravity alone; however, as it speeds up, the upward drag force increases.
Eventually, a state of balance is reached where "Gravity = Air Resistance Force." At this point, the net force becomes \(0\), and the object stops accelerating.
This constant final speed is called the terminal velocity. On the Common Test, you might be asked qualitative questions about how speed changes over time (a graph leveling off toward a constant value), so make sure you have a clear mental picture of this process!

Section 3: Equilibrium of Rigid Bodies

In Basic Physics, we treated objects as "point masses" whose physical size could be ignored. In reality, objects have dimensions, and applying a force at different positions can cause them to rotate. An ideal object that has a definite size and does not deform when forces are applied is called a rigid body.

1. Moment of a Force (Torque)

The measure of the tendency of a force to cause an object to rotate about a specific axis is called the moment of a force (or torque).
Let \(l\) be the perpendicular distance from the axis of rotation (pivot point) to the line of action of the force (the lever arm), and \(F\) be the magnitude of the applied force. The magnitude of the moment of force \(M\) is defined as:
\(M = F l\)
The SI unit is the newton-meter (symbol: \(\text{N}\cdot\text{m}\)).
If \(r\) is the distance from the pivot point to the point of application of the force, and \(\theta\) is the angle between the position vector and the force vector, the lever arm is \(l = r \sin\theta\), so it can also be calculated as \(M = F r \sin\theta\).

- Sign of the Moment: Choose whether counterclockwise or clockwise is positive, and keep it consistent throughout (usually, setting up the equation as "Sum of counterclockwise moments = Sum of clockwise moments" is the most convenient approach).

2. Two Conditions for Equilibrium of a Rigid Body

For a rigid body to remain at rest (undergoing neither translational nor rotational motion), the following two conditions must be satisfied simultaneously:
1. Translational Equilibrium (Balance of Forces): The net force must be zero (condition for no translation).
\(\sum \vec{F} = \vec{0}\) (Sum of horizontal forces \(= 0\), Sum of vertical forces \(= 0\))
2. Rotational Equilibrium (Balance of Moments): The sum of the moments of the forces about any arbitrary point must be zero (condition for no rotation).
\(\sum M = 0\) (Sum of counterclockwise moments = Sum of clockwise moments)

Key Point: You Can Choose ANY Point as the Axis of Rotation!

When a rigid body is in equilibrium, the reference point (pivot axis) for calculating moments can be chosen anywhere, whether inside or outside the body.
The ultimate trick to solving problems quickly is to choose the point where unknown forces are concentrated as your pivot. Any force passing through this pivot has a lever arm of \(0\), making its moment zero. This eliminates that unknown force from your torque equation, making calculations dramatically easier!

3. Center of Gravity

Gravity acts on every tiny particle comprising an object. The point of application of the resultant of all these tiny gravitational forces is called the center of gravity. When analyzing the equilibrium of a rigid body, the entire gravitational force acting on the body can be treated as acting at this single point.
- For two particles (masses \(m_1, m_2\) at positions \(x_1, x_2\)), the position of the center of gravity \(x_{\text{G}}\) divides the distance between them internally in the inverse ratio of their masses:
\(x_{\text{G}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\)
- For symmetrical objects with uniform density (such as a uniform rod, disk, or sphere), the center of gravity coincides with the geometric center.

The Golden 4-Step Method for Rigid Body Equilibrium Problems

1. Draw a large, clear diagram: Draw arrows for all forces acting on the rigid body (gravity, normal force, tension, friction, etc.). The key is to draw the exact "point of application" of each force.
2. Resolve forces and set up force equilibrium equations: Resolve forces into horizontal and vertical components, setting up equations: Forces right = Forces left, and Forces up = Forces down.
3. Choose a pivot and set up the moment equilibrium equation: Pick a pivot point through which multiple unknown forces pass, and write: "Sum of counterclockwise moments = Sum of clockwise moments".
4. Solve the system of equations: Calculate the unknown quantities.

Common Pitfall

Be careful not to confuse the "line of action" with the lever arm. The lever arm \(l\) is NOT the direct distance from the pivot to the point of application; it is the perpendicular distance from the pivot to the line of action (the line extending along the force arrow). When a force acts at an angle, either resolve the force into a component perpendicular to the lever or carefully draw the perpendicular line to the line of action.

Chapter Summary (Last-Minute Common Test Checklist)

- Curvilinear Motion: Displacement, velocity, and acceleration are vectors. Always analyze the \(x\) and \(y\) directions completely independently.
- Horizontal Projection: Uniform linear motion horizontally (\(v_x = v_0\)), free fall vertically (\(v_y = gt, y = \frac{1}{2}gt^2\)).
- Oblique Projection: Resolve initial velocity into \(v_0 \cos\theta\) and \(v_0 \sin\theta\). At maximum height, vertical velocity \(v_y = 0\) (horizontal velocity remains!).
- Air Resistance: As falling speed increases, drag increases until it balances gravity, reaching a constant "terminal velocity."
- Equilibrium of Rigid Bodies: Two conditions required: "Balance of forces (no translation)" + "Balance of moments about any point (no rotation)."
- Calculating Torque: \(M = F \times (\text{lever arm})\). Choosing the point with unknown forces as your pivot is the best shortcut!