Notes on Force, Moment, Centre of Gravity, Uniform Circular Motion, and Centripetal/Centrifugal Forces

Moment and Equilibrium: Force, Torque, and Turning Effects

Force concept recap
- A force causes motion in a rigid body; for a rigid body, force can produce translational and/or rotational motion.
- Mathematical form (constant mass): $\mathbf{F} = \frac{d\mathbf{p}}{dt} \quad\text{where } \mathbf{p}=m\mathbf{v}.$
- For constant mass, $\mathbf{F}=m\mathbf{a}.$
- Units: SI unit of force is newton (N). Weight is a force: 1 kgf = g N with g ≈ 9.8 m s^{-2}. In metre-kilogram-second (MKS): $1\ \text{kgf} = g\ \text{N} \Rightarrow 1\ \text{kgf} = 9.8\ \text{N}.$ Also, in CGS: dyne and gram-force (gf).
Moment (turning effect) of a force (torque)
- When a force F is applied to a body pivoted about a point O, the turning effect (moment) about O is:
  $M = F \times \text{perpendicular distance from O to the line of action of F} = F \times OP.$
- Convention for sign: anticlockwise moment is positive; clockwise moment is negative.
- Maximum turning effect occurs when the line of action is farthest from the axis (maximum perpendicular distance).
- Moment of force is a vector quantity with direction along the axis of rotation: anticlockwise (positive) outward; clockwise (negative) inward.
- Units of moment: SI is N·m. In CGS, dyne·cm. Note: 1 N·m is not a joule; torque is a vector, while work/energy is scalar.
- Relationship of units:
  1\ \text{Nm} = 10^{7}\ \text{dyne·cm} ,\quad 1\ \text{kgf·m} = 9.8\ \text{Nm},\quad 1\ \text{gf·cm} = 980\ \text{dyne·cm}.
Clockwise and anticlockwise moments
- Moment magnitude: $M = F \times d,$
  where d is the perpendicular distance from the pivot to the line of action.
- If the effect is anticlockwise about the pivot, the moment is positive; if clockwise, the moment is negative.
Examples of turning effects (conceptual)
- Opening/shutting a door: push at the handle near the far end from hinges to maximize moment (small force, large moment arm).
- Turning a wheel with a handle: force application farther from the axis (larger lever arm) increases turning effect.
- Spanner (wrench): long handles produce large moments by applying small forces at the end.
- A pair of equal and opposite forces (a couple) can produce rotation without translation.
Couple (two equal and opposite forces)
- A single external force on a pivoted body cannot cause rotation by itself; rotation is produced by a pair of forces called a couple.
- A couple consists of two equal and opposite parallel forces not acting along the same line.
- For a bar with ends A and B and equal forces F at A and B, separated by a distance AB = d (the couple arm), the total moment (couple moment) is:
  $\text{Moment of couple} = F\times AB = F\times d.$
- In components:
  $\text{Moment about A} = F\times OA, \quad \text{Moment about B} = F\times OB, \quad \text{Total} = F\times OA + F\times OB = F\times AB.$
- The couple produces rotation (anticlockwise in the example) without any net resultant force.
Equilibrium of bodies
- A body is in equilibrium when it experiences no change in its state of rest or motion.
- Conditions for equilibrium (static or dynamic):
- Resultant force is zero: $\sum \mathbf{F} = 0.$
- Algebraic sum of moments about any point (often the pivot) is zero: $\sum M = 0.$
- Types of equilibrium:
- Static equilibrium: body at rest under several forces (e.g., a book on a table, a beam balanced horizontally).
- Dynamic equilibrium: body moves with constant velocity or constant angular velocity such that net force or net moment is zero (e.g., raindrop reaching Earth with terminal velocity; airplane in steady flight; a mass on a string undergoing circular motion with tension providing centripetal force).
Principle of moments
- If multiple forces act about a pivot, the sum of anticlockwise moments equals the sum of clockwise moments for equilibrium:
 $\sum M{anticlockwise} = \sum M{clockwise}.$
- The principle is used in physical balances (beam balances) and many measurement devices.
- Verification example (metre rule with weights): balance on a pivot by adjusting two masses W1 and W2 at distances OA1 and OB1 so that the net moment is zero:
 anticlockwise moment = clockwise moment → $W1 \times OA1 = W2 \times OB1.$
Practical examples and worked problems (selected)
- Example 1: A force 10 N applied at 30 cm from pivot. Moment: M = F r = 10\,\text{N} \times 0.3\,\text{m} = 3\,\text{N·m}.
- Example 2: If a moment is 2 N·m for a force of 5 N, distance r: $2 = 5 \times r \Rightarrow r = 0.4\,\text{m}.$
- Example 3: Nut with a lever: 150 N applied with lever 0.4 m; required lever length with 50 N: solve for L from $150\times 0.4 = 50\times L \Rightarrow L = 1.2\ \text{m}.$
- Example 4: A 3 m wide door opened by 100 N at middle (1.5 m from hinges) → torque = 100 \times 1.5 = 150\ \text{N·m}. For least force, apply at far end from hinges (3 m from hinges) → $F' \times 3 = 150 \Rightarrow F' = 50\ \text{N}.$
- Example 5: A wheel with axle O; horizontal force F1 at A (OA = 2.5 cm) and vertical force F2 at B (OB' = 1.5 cm). In equilibrium, clockwise moment from F1 equals anticlockwise moment from F2: $F1 \times OA = F2 \times OO'.$
- Example 6: Two equal and opposite forces F1 and F2 separated by 2 m. A point X midway, distance to each is 1 m; total moment about X is 5\text{ N} \times 1\text{ m} + 5\text{ N} \times 1\text{ m} = 10\ \text{N·m} \text{ (clockwise)}.
- Example 7: Two forces 2 N at opposite ends of a 1 m rod pivoted at center: moments are 2\times 0.5 = 1.0 \text{ N·m} each, total 2.0 N·m (clockwise).
- Example 8: A uniform metre rule balanced on a knife edge at 60 cm with 10 g mass at one end: determine mass of the rule by balancing moments about the knife edge at 60 cm. If the rule mass is Mg and its center is at 50 cm, anticlockwise moment Mg×(60−50) must balance clockwise moment 10×(100−60). → Mg×10 = 10×40 → Mg = 40 g. Then mass of rule is obtained accordingly.
- Example 9–12: See-saw and balance problems with weights, distances, and lever arms; use the principle of moments to find where to place masses to achieve horizontal equilibrium. Common results include that heavier weights must be placed closer to the pivot to balance lighter weights further away, or vice versa depending on distances.
Centre of gravity (C.G.)
- Definition: The C.G. is the point at which the algebraic sum of moments of all parts (weights) of a body about that point is zero; equivalently, the total weight W acts at G for balancing purposes.
- Properties:
- The CG of a body depends on its shape and mass distribution; deformation can shift CG.
- The CG may lie outside the material (e.g., a ring, hollow sphere, or certain shaped objects).
- A body can be treated as a particle of weight W acting at its CG for many calculations.
- CG of regular shapes (typical positions):
- Rod: midpoint of the rod.
- Circular disc: geometric center.
- Solid or hollow sphere: geometric center.
- Solid or hollow cylinder: mid-point on the axis.
- Solid cone: at height h/4 from the base (along axis).
- Hollow cone: at height h/3 from the base (along axis).
- Circular ring: at the center of ring.
- Triangular lamina: at the intersection of medians.
- Parallelogram, rectangle lamina, square, or rhombus: at the intersection of diagonals.
- Centre of gravity of irregular lamina
- Method of balance with a plumb line: suspend through different holes a, b, c and draw vertical lines along the plumb line corresponding to each suspension. The intersection of these lines locates G.
- Centre of gravity and balance point
- A body can be balanced when supported at its centre of gravity (e.g., a metre rule balanced at 50 cm). A square lamina can balance on a nail at its balance point.
Practical notes and MCQ prompts
- The position of the CG of a uniform ball is at its geometrical center.
- The CG of a hollow cone of height h is at distance from its vertex: for a hollow cone, often 2h/3 from the vertex along the axis; for the solid cone, h/4 from the base (equivalently 3h/4 from the vertex). The exact phrasing depends on the type of cone.
- The CG for a ring is at its center; for a circular lamina, at center; for a triangular lamina, at medians’ intersection; for a rectangle, at diagonals’ intersection; for a cylinder, at the axis midpoint.
- Experimental: to locate CG of irregular lamina, suspend from holes and draw lines with a plumb line; CG is the intersection of lines.
Uniform circular motion
- Definition: Motion of a particle in a circle at constant speed v. The path is circular; thus velocity vector changes direction, though speed is constant.
- Velocity direction: tangent to the path at any instant; for a circular path AB, BC, CD, DA, velocity directions rotate by 90° at each quarter turn.
- Acceleration: centripetal (towards the centre) with magnitude a_c = v^2 / r. The direction of acceleration is toward the centre at every instant.
- Key distinction: Uniform circular motion has constant speed but changing velocity; hence it is accelerated motion (unlike uniform linear motion, which has zero acceleration).
- Centripetal force is the net force causing this acceleration toward the centre; examples include gravitational force for planetary motion, tension in a string for a whirled stone, electrostatic force for electrons in atoms, etc.
Centripetal and centrifugal forces
- Centripetal force: directed toward the center of the circular path; it is the real force causing centripetal acceleration and circular motion.
- Centrifugal force: a fictitious (pseudo) force acting outward away from the centre, observed in a rotating frame of reference. It balances the centripetal force in that rotating frame, explaining stationary appearance for an observer on the rotating frame.
- Notation example: in a ball tied to a string on a merry-go-round, from an outside observer the string tension provides the centripetal force; from a rider on the merry-go-round, the ball appears stationary but a centrifugal force balances the tension, allowing the ball to appear at rest in that frame.
- Fundamental point: Centrifugal force is not an action-reaction pair to centripetal force; rather, it is a fictitious force arising in a non-inertial reference frame.
Exercises and key results (highlights)
- Moment of a force about a point O: M = F × d; anticlockwise positive, clockwise negative.
- For a pair of equal and opposite forces separated by a distance d (a couple): moment = F × d.
- Equilibrium conditions: ∑F = 0 and ∑M = 0 about the pivot.
- In a see-saw or beam balance, weights and distances must satisfy moments balance to be horizontal.
- For a rod with two forces at its ends, the sum of moments about its center is the total turning effect; if the forces are opposite and equal and applied at equal distances from the center, the net moment is the sum of each end’s moment about the center.
- Typical numerical relations you’ll encounter:
- Moment about a point: $M = F r$ where r is the perpendicular distance.
- For a door: maximum torque occurs when force is applied at the far end from hinges.
- If a weight W1 at distance OA1 and weight W2 at distance OB1 balance: $W1 \times OA1 = W2 \times OB1.$
- For a body pivoted at mid-point with equal and opposite forces at ends separated by d: total torque = $F \times d.$
- In a see-saw with weights on opposite sides, the balance condition: anticlockwise moment = clockwise moment about the pivot.
Quick reference formulas (LaTeX)
- Moment of a force about pivot: $M = F \times d$
- Sign convention: anticlockwise positive, clockwise negative.
- Moment of a couple: $M_{ ext{couple}} = F \times d = F \times AB.$
- Equilibrium conditions:
- $\sum \mathbf{F} = 0$
- $\sum M = 0$
- Centripetal acceleration: $a_c = \frac{v^2}{r}$
- Centripetal force (towards centre): $Fc = m ac = m \frac{v^2}{r}$
- Centrifugal force (fictitious, outward): $F_{ ext{cen}} = m \frac{v^2}{r}$ (in a rotating frame, direction outward; not an actual real force).
Connections to practical physics and real-world relevance
- Torque concepts explain how doors open with small forces if applied away from hinges, how spanners work, and how to design mechanical balances.
- Centre of gravity helps explain balance, stability in structures, and how irregular shapes balance on supports or suspensions.
- Uniform circular motion and centripetal force underpin celestial mechanics (planets, moons), atomic physics, and everyday demonstrations (ball on a string, merry-go-round).
- The idea of centrifugal force clarifies why observers in rotating frames perceive outward force even though the only real force is toward the centre (centripetal force). This distinction is essential in rotating machinery, engineering design, and physics interpretation.

Centre of Gravity (CG)

Key concept
- CG is the point where the resultant weight W of a body acts and where the algebraic sum of moments of all particle weights about that point is zero.
- For many geometries, the CG coincides with familiar geometric centers or intersection points:
- Rod: midpoint.
- Circular disc: geometric center.
- Solid sphere / hollow sphere: geometric center.
- Cylinder: midpoint along axis.
- Cone (solid): h/4 from the base; Cone (hollow): h/3 from the base.
- Ring: center of ring.
- Triangle lamina: intersection of medians.
- Parallelogram / rectangle lamina / square / rhombus: intersection of diagonals.
Practical implications
- A body can be balanced when supported at its centre of gravity.
- A freely suspended body comes to rest with its CG directly below the suspension point.
Experimental method
- To locate CG of an irregular lamina, suspend it from holes a, b, c near the edge using a plumb line; for each suspension, draw the vertical line along the plumb line, and the intersection of these lines is the CG.
Notes from exercises
- The CG position depends on mass distribution; it may lie outside the material (e.g., ring, hollow sphere).
- Examples of CG placement in common shapes help in solving balance problems and in designing stable structures.

Uniform Circular Motion (UCM)

Definition and key features
- UCM: motion with constant speed around a circle; velocity changes direction, hence acceleration is present even though speed is constant.
- Direction of velocity at any instant is tangent to the circular path.
- The velocity vector completes a full rotation after one period T, with quarter-turns in time T/4 each.
- Acceleration is centripetal (toward the center) and its magnitude is $a_c = \frac{v^2}{r}.$
Distinctions from linear motion
- Uniform linear motion: speed and velocity are constant, acceleration is zero.
- Uniform circular motion: speed is constant, velocity is not; acceleration exists due to changing direction.
Centripetal vs centrifugal (revisited)
- Centripetal force is the real force causing the inward acceleration toward the circle's centre.
- Centrifugal force is a fictitious force observed in a rotating frame; it acts outward and is not an actual interaction on the body in an inertial frame.

Centripetal and Centrifugal Force: Details and Examples

Centripetal force examples
- Electron around nucleus (electrostatic attraction acts as centripetal force).
- Planet around the sun (gravity provides centripetal force).
- Moon around Earth (gravity provides centripetal force).
- Stone whirled on a string (tension provides centripetal force).
Centrifugal force interpretation
- In a rotating frame, an observer perceives outward force (centrifugal) balancing inward centripetal force to explain a stationary viewpoint on the rotating frame.
- Important caution: Centrifugal force is not a real force; it is a fictitious force arising from non-inertial reference frames.

Quick Exercise and Problem-Solving Notes (Conceptual Highlights)

Conditions for equilibrium (restate):
- Resultant force equals zero: $\sum \mathbf{F} = 0.$
- Sum of moments about the pivot equals zero: $\sum M = 0.$
When does a body rotate? How to change rotation direction?
- Rotation occurs when net moment about the pivot is nonzero; direction can be changed by applying force closer to or farther from the pivot or by changing the force direction to switch the sense of turning.
How to obtain greater moment from a given force? Increase the perpendicular distance (lever arm) or apply larger force.
How to reduce moment? Move the line of action closer to the pivot or apply forces to produce opposite moments.
How to determine the effect of two forces at two different points? Compute their moments about a chosen axis; if they are equal and opposite and produce rotation in the same sense, they form a couple; otherwise, net translation and/or rotation depends on the resultant force and net moment.

Summary of Key Formulas (LaTeX)

Moment of a force about a point: $M = F \times d$
Moment of a couple: $M_{\text{couple}} = F \times d$
Equilibrium conditions: $\sum \mathbf{F} = 0, \quad \sum M = 0$
Centripetal acceleration: $a_c = \frac{v^2}{r}$
Centripetal force: $Fc = m ac = m \frac{v^2}{r}$
Centrifugal force (fictitious): $F_{\text{cen}} = m\frac{v^2}{r}$ (in rotating frame, outward direction)
Units (reference):
- 1 N·m (torque), 1 N = 1 kg m s^{-2}, 1 kgf = 9.8 N, 1 kgf·m = 9.8 N·m, 1 gf·cm = 980 dyne·cm, 1 Nm = 10^7 dyne·cm.

Notes on Real-World Applications and Conceptual Clarity

Torque and lever arms explain why it is easier to turn heavier doors at their far ends and why longer wrenches reduce the required force.
Centre of gravity considerations are crucial for the stability of vehicles, aircraft, structures, and even everyday objects like rulers and flat laminae.
The distinction between centripetal and centrifugal forces is essential to avoid misinterpreting rotating systems; centrifugal force appears only in non-inertial reference frames.

Quick Reference: Common Exam-Type Scenarios

Determine the moment of a force given F and distance r: M = F r.
Determine the required lever length to achieve a desired moment: L = M / F.
Find balance position using the principle of moments: choose a pivot and equate anticlockwise and clockwise moments about that pivot.
Locate the centre of gravity for common shapes using the standard reference points (midpoints, centers, intersections of medians or diagonals).
Distinguish between real centripetal force and fictitious centrifugal force in rotating frames; apply the correct frame of reference when solving a problem.

If you’d like, I can tailor these notes to a specific section or convert this into a printable cheat-sheet with more worked examples and additional practice problems.