CELE Engineering Mechanics — Dynamics: KineticsStudy Notes

Newton's second law is the cornerstone of kinetics: the net force acting on a body is directly proportional to its mass and acceleration, expressed as:

ΣF = ma

where:

• ΣF = sum of all forces acting on the body (in Newtons, N)
• m = mass of the body (in kilograms, kg)
• a = acceleration of the body (in meters per second squared, m/s²)

Drawing the Free-Body Diagram (FBD) is the critical first step. Identify all forces—applied forces, weight (W = mg), normal force (N), friction (f), tension, and any others—and arrange them head-to-tail. Then sum forces along the direction of motion and perpendicular to it.

Sign Convention: Choose a positive direction along the motion. Forces acting in that direction are positive; opposing forces are negative. On an incline, common practice is to take the positive direction along the slope.

Friction Force: When a surface is involved, include friction:

• Kinetic friction (body sliding): f = μₖN, opposing the direction of motion
• Static friction (body stationary): f ≤ μₛN, up to a maximum

where μₖ and μₛ are the coefficients of kinetic and static friction, respectively.

D'Alembert's Principle: An elegant restatement treats a dynamics problem like statics by adding an inertia force –ma to the body. The equation becomes ΣF – ma = 0, or ΣF = 0 in the inertial reference frame. This pseudo-equilibrium approach is useful for solving coupled systems and for visualizing forces.

Application in Civil Engineering: When designing the braking system of an elevator or analyzing the forces on a derrick during load movement, Newton's second law directly gives the required force or the resulting acceleration.

The work–energy principle states that the total work done on a body equals its change in kinetic energy:

U₁→₂ = ΔKE = ½mv₂² – ½mv₁²

where:

• U₁→₂ = total work done from state 1 to state 2 (in Joules, J)
• m = mass (kg)
• v₁, v₂ = velocities at states 1 and 2 (m/s)

Calculating Work:

1. Work of a constant force: U = F·d·cos(θ)

   • F is the magnitude of the force
   • d is the displacement
   • θ is the angle between F and d
   • If force is parallel to displacement, U = Fd; if perpendicular, U = 0; if opposing, U is negative

2. Work of gravity: U = ±mgh

   • Positive when the body moves downward (gravity aids motion)
   • Negative when the body moves upward (gravity opposes motion)
   • h is the vertical height change

3. Work of a spring: U = ½k(x₁² – x₂²)

   • k is the spring constant
   • x₁, x₂ are initial and final spring extensions/compressions
   • Work is positive when the spring returns toward its natural length (releases energy)
   • Note: This differs from the stored potential energy ½kx², which is always positive

4. Work of friction: U_f = –μN·d

   • Always negative (friction opposes motion)
   • Depends on the normal force N and distance traveled d
   • Independent of path if the surface is uniform

When to Use Work–Energy:
This principle is ideal when the problem involves a distance and a final speed, but time is not given or needed. It elegantly sidesteps the need to find acceleration and works seamlessly with variable forces.

Energy Conservation:
When only conservative forces (gravity, springs) act—no friction or external work—total mechanical energy is conserved:

KE₁ + PE₁ = KE₂ + PE₂

where PE includes gravitational and elastic potential energy. This is extremely useful for problems involving motion down a slide, a pendulum, or a mass–spring system.

Application in Civil Engineering: When analyzing the descent of a load on a cable, the work–energy method quickly determines the speed at impact without tracking every moment. For vibrating structures or damped oscillators, work done by friction reveals energy dissipation.

The impulse–momentum principle relates the impulse (force × time) to the change in momentum:

ΣF·Δt = m·v₂ – m·v₁ = Δ(mv)

where:

• ΣF = sum of forces (N)
• Δt = time interval (s)
• m = mass (kg)
• v₁, v₂ = velocities before and after the interval (m/s)
• Δ(mv) = change in momentum (kg·m/s)

Impulse is the product of force and the time duration for which it acts. It represents the "cumulative effect" of a force over time. Momentum is the product of mass and velocity; it quantifies how much "motion" a body carries.

Linear Momentum Conservation:
If no external forces act on a system (or external forces cancel), the total momentum of the system is conserved:

Σ(m·v)_before = Σ(m·v)_after

This is the foundation for analyzing collisions and explosions.

When to Use Impulse–Momentum:
Use this principle when:

• The problem involves time and force (or time and velocity change)
• You need to find an average force over a time interval
• Analyzing collisions or impacts
• External forces are absent or cancel (conservation applies)

Sign Convention for Momentum:
Velocities must be treated as signed quantities along a chosen positive direction. A leftward velocity is negative if rightward is positive. This is where many students make sign errors, especially in collision problems.

Application in Civil Engineering: When designing a crash barrier, the impulse–momentum principle determines the required stopping time (or distance) to safely reduce a vehicle's momentum to zero. In blast analysis, the principle helps quantify the force and duration of impact from explosions.

When two bodies collide, linear momentum is always conserved (in the absence of external impulses), but kinetic energy may not be. The coefficient of restitution (e) quantifies the relative motion after impact and distinguishes between elastic, partially elastic, and perfectly plastic collisions.

Definition of Coefficient of Restitution:

e = (v₂ – v₁) / (u₁ – u₂)

where:

• v₁, v₂ = velocities of bodies 1 and 2 after collision (m/s)
• u₁, u₂ = velocities of bodies 1 and 2 before collision (m/s)
• The numerator is the relative velocity of separation (how fast they move apart)
• The denominator is the relative velocity of approach (how fast they come together)
• All velocities are signed along the line of impact

Range of e:

• e = 1: Perfectly elastic collision; kinetic energy is conserved; bodies bounce apart at the same relative speed they approached
• 0 < e < 1: Partially elastic collision; some energy is lost to deformation, sound, heat
• e = 0: Perfectly plastic (or inelastic) collision; bodies stick together and move as one; maximum energy loss

Critical Point: The sign convention is paramount. If body 1 moves right at +5 m/s and body 2 moves left at –3 m/s before collision, the approach velocity is 5 – (–3) = 8 m/s. After collision, if body 1 moves left at –1 m/s and body 2 moves right at +2 m/s, the separation velocity is 2 – (–1) = 3 m/s. Then e = 3/8 = 0.375.

Solving Collision Problems:

1. Draw a diagram showing initial velocities with signs.
2. Apply momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
3. Apply the restitution equation: e = (v₂ – v₁)/(u₁ – u₂)
4. Solve the two equations simultaneously for v₁ and v₂.

Application in Civil Engineering: In impact analysis of vehicles against barriers, the coefficient of restitution determines the likelihood of a second collision (rebound). In seismic design, the energy absorption during impact affects structural safety.

Energy Loss in Collisions:
The kinetic energy before and after can be compared. The loss is:
ΔKE = KE_before – KE_after = ½m₁u₁² + ½m₂u₂² – ½m₁v₁² – ½m₂v₂²

For a perfectly elastic collision, ΔKE = 0. For perfectly plastic, the loss is maximum.

When a body moves along a curved path, acceleration is no longer confined to a single direction. Newton's second law must be applied in tangential and normal (radial) components.

Tangential and Normal Components:

For a body on a curve:

• Tangential acceleration (a_t): Along the path; changes the speed. a_t = dv/dt
• Normal acceleration (a_n): Perpendicular to the path (toward the center of curvature); changes the direction. a_n = v²/r

where v is the instantaneous speed and r is the radius of curvature.

Newton's Second Law in Component Form:

ΣF_t = m·a_t (along the path)

ΣF_n = m·a_n = m·(v²/r) (toward the center of curvature)

Physical Interpretation:
The tangential forces speed up or slow down the body along its path. The normal forces provide the centripetal acceleration—the "push" toward the center that keeps the body from flying off in a straight line. On a horizontal curve, this might be the friction between tires and road or the banking angle of a road.

Circular Motion at Constant Speed:
If v is constant (uniform circular motion), a_t = 0, and only a_n = v²/r remains. The required centripetal force is:

F_c = m·v²/r

This force must be supplied by tension in a string, normal force from a banked surface, or friction.

Banking of Roads:
On a banked curve (incline angle θ), the normal force N has components that provide centripetal force and support the weight. At the ideal banking angle (no friction needed):

tan(θ) = v²/(r·g)

where v is the safe speed, r is the radius, and g is gravity. If the actual speed is less than this ideal speed, friction acts up the slope; if greater, friction acts down.

Looping and Vertical Circular Motion:
At the top of a vertical loop of radius r, the centripetal force points downward (toward the center). Both weight and normal force contribute:

m·g + N = m·v²/r

For the body to maintain contact with the track, N ≥ 0, so:

v_min = √(g·r) at the top

Below this speed, the body loses contact and falls.

Application in Civil Engineering: Road design requires calculating the minimum radius for a given speed and friction. Roller coasters and ramps must ensure adequate centripetal force without excessive G-forces. Bridge vibrations under wind or seismic loading involve curved motion of the deck.

Note on Sign Convention for a_n:
The normal acceleration is always positive (magnitude v²/r), directed toward the center of curvature. In the force equation, ΣF_n equals this positive value.

The art of solving kinetics problems is knowing which principle to apply. Here is a decision guide:

Use Newton's Second Law (ΣF = ma) when:

• You need to find acceleration or an unknown force
• Forces are given and clearly defined
• The problem is straightforward: identify forces, sum them, solve for a
• Time and distance are not the focus
• Example: A block slides down a slope; find the acceleration

Use Work–Energy Principle (U = ΔKE) when:

• The unknown is a final speed and you know the distance traveled
• Work can be easily calculated (constant forces or springs)
• You want to avoid finding acceleration as an intermediate step
• Only distance and velocity matter; time is not involved
• Example: A ball rolls down a hill; find the speed at the bottom

Use Impulse–Momentum (ΣF·Δt = Δ(mv)) when:

• The unknown involves time and force or velocity
• You need an average force over a time interval
• Analyzing collisions or impacts
• External impulses cancel and momentum is conserved
• Example: A car accelerates for a certain time; find the final speed

Use Coefficient of Restitution when:

• Two bodies collide and you need post-collision velocities
• Momentum is always conserved; restitution provides a second equation
• Example: Two balls collide; find their velocities after impact

Use Curvilinear Kinetics when:

• Motion is on a curved path (circular, banked, looping)
• Decompose forces into tangential and normal components
• Apply ΣF_t = m·a_t and ΣF_n = m·v²/r separately
• Example: A car on a banked curve; find the safe speed

Efficiency Tip for Board Exams:
Quickly scan the problem. What is given? What is asked? If you see distance–to–speed without time, reach for work–energy. If you see time–and–force, reach for impulse–momentum. If you see collision, think momentum and restitution. This quick mental sorting saves minutes on the exam.

Even strong students make systematic errors on kinetics problems. Awareness prevents these costly mistakes.

Pitfall 1: Confusing Mass and Weight
Newton's second law uses mass (kg), not weight (N). A common error:

• Wrong: ΣF = W·a (using weight as if it were mass)
• Correct: ΣF = m·a (mass is the coefficient)
• Weight is a force: W = m·g, where g ≈ 9.81 m/s²

Pitfall 2: Forgetting to Include All Forces
Students often omit normal forces, weight, or friction from the FBD. Always ask:

• Is there a normal force? (if the body is in contact with a surface)
• Is there friction? (if the surface is rough)
• Is there tension or an applied force?
• Does the body experience its weight?

Pitfall 3: Sign Errors in Friction Work
Friction always opposes motion and always does negative work:

• Wrong: U_f = +μNd (friction aids the motion)
• Correct: U_f = –μNd (friction resists the motion)

Pitfall 4: Wrong Sign Convention in Collisions
A collision is a sign-convention nightmare. If body 1 moves right at +5 m/s and body 2 moves left, it must be written as –3 m/s (not as a separate positive speed). The relative approach velocity is then 5 – (–3) = 8 m/s, not 5 + 3. Many students add instead of subtract and get wrong answers. Write a diagram showing which direction is positive.

Pitfall 5: Applying the Wrong Principle

• Using ΣF = ma when work–energy is far quicker (and then making an arithmetic error)
• Using impulse–momentum for a static equilibrium problem
• Forgetting to apply momentum conservation in collisions

Pitfall 6: Ignoring the Centripetal Acceleration on Curves
On a horizontal curve, students forget that a_n = v²/r is always present. The normal force (or friction) must supply the centripetal force; if it doesn't, the body skids.

Pitfall 7: Misinterpreting the Coefficient of Restitution

• Wrong: e is the ratio of absolute speeds before and after
• Correct: e is the ratio of relative velocities of separation to approach

Pitfall 8: Energy NOT Conserved with Friction
When friction is present, mechanical energy is not conserved. Use the work–energy principle instead:
U_total = ΔKE, where U_total includes friction work (negative).

Exam Strategy:

1. Read carefully. Identify what is given and what is asked.
2. Draw the FBD. Label every force with magnitude and direction.
3. Choose the method. ΣF = ma? Work–energy? Impulse–momentum? Restitution?
4. Set up the equation(s). Be explicit about sign conventions and units (all SI).
5. Solve step-by-step. Show intermediate results (not all in one line).
6. Check the answer. Does the magnitude make sense? Are the units correct? Does the physics make sense?
7. If stuck, try a different principle or re-examine your FBD.

On the PRC exam, partial credit is often awarded for correct setup and method, even if the final number is wrong. A clear, organized solution is more valuable than a correct answer with no work shown.