Notes on Turning Forces, Equilibrium, Centre of Gravity, and Uniform Circular Motion

Translational and Rotational Motions

A rigid body can experience two kinds of motion when acted upon by a force: translational (linear) motion, where the body moves as a whole in a straight line in the direction of the resultant force, and rotational motion, where the body rotates about a pivot or axis. Examples include a ball pushed along a floor exhibiting translational motion, and a door, which rotates about its hinges when a force is applied at the handle. In general, a force can cause either type of motion, or both, depending on constraints and points of application.

Moment (Turning Effect) of a Force (Torque)

The turning effect of a force about an axis is called the moment of force or torque. The moment of a force is equal to the product of the force magnitude and the perpendicular distance from the axis of rotation to the line of action of the force:
$M = F \times d$
where

F is the force applied, and
d is the perpendicular distance from the axis of rotation to the line of action of the force.
The SI unit of moment is newton metre (N·m). In centimetre–gram–second (CGS) units the moment is dyne·cm, and in kilogram–force units it is kgf·m. The relationships between common units include:
1\ \text{N·m} = 10^7 \ \text{dyne·cm}, \quad 1\ \text{kgf·m} = 9.8\ \text{N·m}, \quad 1\ \text{gf·cm} = 980\ \text{dyne·cm}.
The moment is a vector quantity. Conventionally, anticlockwise moments about the axis are taken as positive, and clockwise moments as negative. The direction of the moment vector is along the axis of rotation: anticlockwise moments point outward from the rotation axis, while clockwise moments point inward.

Factors affecting the turning of a body

The turning effect depends on two factors:

the magnitude of the force, and
the perpendicular distance from the line of action to the axis of rotation. The larger either factor (or both), the greater the turning effect.

Maximum turning effect

For a given force, the turning effect is maximized when the line of action is as far as possible from the axis of rotation, i.e., when the perpendicular distance is maximum.

Common Examples of Moment of Force

Opening or closing a door: apply a force normal to the door at the handle, which is at the maximum distance from the hinges.
A hand flour grinder: the weight at the rim and a handle near the rim maximize the moment.
Turning a steering wheel: force applied tangentially on the rim; the sense of rotation depends on where along the rim the force is applied.
Turning a wrench (spanner): a long handle increases the moment for a given applied force to loosen or tighten a nut.
In a bicycle, the pedal force acts with a larger perpendicular distance on a larger gear to turn the wheel.

Coupling of Forces (Couples)

A single force applied to a pivoted body does not cause rotation by itself; rotation is produced by a pair of equal and opposite forces that are not collinear, called a couple. The forces produce equal and opposite moments about the pivot, so their resultant force is zero, but their moments do not cancel if they are separated; the rotation direction depends on the orientation of the forces.

A couple consists of two equal, opposite, parallel forces whose lines of action are separated by a perpendicular distance known as the couple arm, d. The moment of a couple is
$M_{\text{couple}} = F \times d.$
Examples of couples include opening a door with a force at the handle and an opposing reaction at the hinge, turning a wrench, turning taps, keys in locks, steering wheels, bicycle pedals, etc.

Equilibrium of Bodies

Equilibrium occurs when a number of forces acting on a body produce no net change in its state of rest or motion. There are two kinds of equilibrium:

Static equilibrium: the body remains at rest under the action of several forces.
Dynamic (or mechanical) equilibrium: the body continues in a state of uniform motion (translational or rotational) under the action of several forces.

Two conditions must be satisfied for equilibrium:
1) The resultant force acting on the body must be zero (net force = 0).
2) The algebraic sum of the moments of all the forces about any fixed axis through the point of rotation must be zero (sum of anticlockwise moments = sum of clockwise moments).

Principle of Moments

When several forces act on a pivoted body, they cause moments about the pivot. The resultant moment is the algebraic sum of the moments of the individual forces. If this sum is zero about the axis of rotation, the body is in equilibrium:
$\sum M = 0 \quad\text{(anticlockwise positives, clockwise negatives)}.$
A physical balance (beam balance) operates on this principle.

Verification of the Principle of Moments (example)

Suspend a metre rule horizontally from a fixed support at O. Attach two spring balances A and B on either side with weights W1 and W2 at distances OA = l1 and OB = l2. When in balance, clockwise moment equals anticlockwise moment:
$W1 \, l1 = W2 \, l2.$

Some typical problems and solutions

1) A body is pivoted at a point. A force of 10 N is applied at a distance of 0.30 m from the pivot. The moment about the pivot is
M = F d = 10 \times 0.30 = 3\ \text{N·m}.
2) The moment of a force of 5 N about a point P is 2 N·m. The distance from the point to the line of action is
$d = \frac{M}{F} = \frac{2}{5} = 0.4\ \text{m}.$
3) A mechanic opens a nut by applying a force of 150 N with a wrench of length 40 cm (0.40 m). The moment needed is
M = 150 \times 0.40 = 60\ \text{N·m}.
If the force is reduced to 50 N, the required length L is obtained from
$50 L = 60 \Rightarrow L = 1.2\ \text{m}.$
4) For a door, the torque required is achieved by applying the force at the farthest point from the hinges, giving the least force needed.

Centre of Gravity (CG)

The centre of gravity of a body is the point G where the algebraic sum of moments of all the particle weights about G is zero. The entire weight W of the body can be considered to act at G. In other words, a body can be treated as a point mass W located at its centre of gravity.

Key points:

The location of CG depends on the distribution of mass; deformation can change CG.
The CG need not lie within the material of the body (e.g., a ring has CG at its centre, which is empty in the middle).
For regular objects, CG positions are known:
- Rod: midpoint
- Circular disc: geometric centre
- Sphere: geometric centre
- Cylinder: middle of the axis
- Solid cone: at a height h/4 from the base along the axis (solid cone)
- Hollow cone: at a height h/3 from the base along the axis
- Circular ring: centre of the ring
- Triangular lamina: intersection of medians
- Parallelogram, rectangle, square lamina: intersection of diagonals

Centre of gravity and balance points:

A solid body can be balanced by supporting it at its centre of gravity. For a uniform metre rule, the CG is at 50 cm; it can balance on a knife edge if it is exactly under the 50 cm mark.
When a body is freely suspended from a point, it comes to rest with its CG vertically below the suspension point. This method is used to locate the CG of irregular laminae using a plumb line: suspend from a hole, draw a line along the plumb line, repeat from other holes, and the intersection gives G.

Practical questions (conceptual)

Can CG lie outside the body? Yes (e.g., ring; hollow objects).
Determining CG experimentally uses a plumb line and multiple suspensions to find the common intersection point of the lines drawn by gravity.

CG for some irregular and regular objects

See diagrams showing CG positions for regular shapes: rod, disc, ring, triangle, square, rectangle, parallelogram, cylinder.

Centre of Gravity: Determination by Balance and Plumb-Line Method

To determine the CG of an irregular lamina, suspend it from different points (holes a, b, c) and let a plumb line hang. Draw lines along the plumb lines ad, be, cf. The intersection is the CG G. This provides a practical method to locate CG for irregular shapes.

Uniform Circular Motion

Uniform circular motion occurs when a particle moves with constant speed in a circle. The velocity is tangent to the path, and its direction changes continuously, producing acceleration even though speed is constant. If the time for one complete revolution is T, then the velocity remains constant in magnitude but varies in direction, implying acceleration with magnitude a_c but changing direction.

The velocity direction at a point on the circle is along the tangent to the path, changing as the particle moves around the circle.
The motion is an accelerated motion (centripetal acceleration directed toward the circle's centre).

Centripetal and Centrifugal Forces

In circular motion, a centripetal force is required to keep the body moving along the circular path; this force points toward the centre. The centripetal acceleration is ac = v^2/r and the centripetal force is Fc = m a_c = m v^2 / r = m ω^2 r (where ω is angular speed). The direction of this force is toward the circle’s centre.
Centrifugal force is a fictitious force observed in a rotating reference frame. It acts away from the centre, opposite to the centripetal force, and has the same magnitude in the rotating frame. In a rotating frame, a ball on a string appears to experience a centrifugal force that balances the tension, making the ball appear stationary relative to the rotating observer. The true physical force at play is the centripetal force provided by the tension in the string.

The centrifugal force is not a real force acting on the ball in an inertial frame; it is a perceived force arising in a non-inertial (rotating) frame of reference.
If the string breaks, the ball moves tangent to the circle (as observed from the ground frame), while an observer on the rotating frame would see the ball moving radially outward from the centre as the centripetal force ceases to act.

Important Formulas and Concepts (Summary)

Moment of a force about an axis:
$M = F \times d,$
where d is the perpendicular distance from the axis to the line of action.
Moment sign convention: anticlockwise = positive, clockwise = negative.
Moment of a couple:
$M_{\text{couple}} = F \times d,$
where d is the perpendicular distance between the two equal and opposite forces (the couple arm).
Equilibrium conditions:
1) Net force = 0. 2) Sum of moments about the axis = 0.
Centre of Gravity ( CG ) definition: The point G where the algebraic sum of moments of all particle weights about G is zero; the total weight W acts at G.
Regular object CG positions:
- Rod: midpoint; Disc: geometric centre; Sphere: geometric centre; Cylinder: axis midpoint; Cone (solid): h/4 from base; Cone (hollow): h/3 from base; Ring: centre; Triangle lamina: intersection of medians; Rectangle/parallelogram/square: intersection of diagonals.
Uniform Circular Motion:
- Speed is constant; velocity changes in direction; acceleration exists.
- Centripetal acceleration: $a_c = \frac{v^2}{r} = \omega^2 r,$
- Centripetal force: $Fc = m ac = m \frac{v^2}{r} = m \omega^2 r.$
- Centrifugal force: fictitious force in a rotating frame, directed away from the centre, equal in magnitude to the centripetal force in the rotating frame.

Practice and Quick Problems (Representative Examples)

Example 1: A body pivoted at a point; F = 10 N applied at r = 0.30 m. Moment: M = F r = 10 \times 0.30 = 3\ \text{N·m}.
Example 2: Moment given as 2 N·m for F = 5 N; distance r = M/F = 2/5 = 0.4 m.
Example 3: Nut and wrench: for opening torque of 60 N·m, with F = 150 N and r = 0.4 m. If F is reduced to 50 N, length needed is $L = \frac{60}{50} = 1.2\ \text{m}.$
Example 4: A door of width 3 m opened by applying 100 N at the middle; maximum distance from hinge is 1.5 m. Torque needed: M = 100 \times 1.5 = 150\ \text{N·m}. The least force would be when applied at the farthest point from the hinges, i.e., at the door’s edge (3 m from hinge), with force $F'\times 3 = 150 \Rightarrow F' = 50\ \text{N}.$
Example 5: Two equal and opposite forces of 5 N at ends A and B of a rod separated by a distance d produce a couple with moment $M_{\text{couple}} = 5 \times d.$ If the rod length is AB = d, this is the total turning effect.

Short Q&A and MCQ Highlights

The moment of a force about a given axis depends on both the magnitude of the force and its perpendicular distance from the axis. (Answer: both factors)
A body acted upon by two unequal forces in opposite directions but not in the same line will have both rotational and translational motion.
The center of gravity of a body can be outside the material (e.g., a ring or hollow object).
The centrifugal force is not real in an inertial frame; it is a fictitious force observed in a rotating reference frame.

Key Takeaways for Exam Preparation

Always identify the pivot/axis and measure the perpendicular distance to the line of action of each force.
Use sign conventions: anticlockwise +, clockwise −.
For equilibrium problems, ensure both net force and net moment about the chosen pivot are zero.
Remember that the CG depends on shape and mass distribution; practice locating CG for common shapes and irregular laminae using balance and plumb-line methods.
Distinguish between centripetal force (real, toward the center) and centrifugal force (fictitious, away from the center in a rotating frame).