Dynamics

Dynamics is the study of forces and motion. Unlike statics where objects are at rest, dynamics deals with acceleration and changing motion.

Kinematics vs Kinetics

Kinematics: Describing Motion

Linear Motion (1D)

Key Variables:

• Position: x (meters)
• Velocity: v = dx/dt (m/s)
• Acceleration: a = dv/dt = d²x/dt² (m/s²)

Kinematic Equations (Constant Acceleration)

v = v₀ + at                    (velocity)
x = x₀ + v₀t + ½at²            (position)
v² = v₀² + 2a(x - x₀)          (velocity-position)
x = x₀ + ½(v₀ + v)t            (average velocity)

Where:
- v₀ = initial velocity
- v = final velocity
- a = acceleration (constant)
- t = time
- x₀ = initial position
- x = final position

Example: Braking Car

A car traveling at 30 m/s brakes with constant deceleration of 5 m/s². How far does it travel before stopping?

Given:
v₀ = 30 m/s
v = 0 m/s (stops)
a = -5 m/s² (deceleration is negative)

Using: v² = v₀² + 2a(x - x₀)
0² = 30² + 2(-5)(x - 0)
0 = 900 - 10x
x = 90 m

The car travels 90 meters before stopping.

Angular Motion (Rotation)

Key Variables:

• Angular position: θ (radians or degrees)
• Angular velocity: ω = dθ/dt (rad/s)
• Angular acceleration: α = dω/dt (rad/s²)

Relationships to Linear Motion:

s = rθ         (arc length)
v = rω         (tangential velocity)
a_t = rα       (tangential acceleration)

Where r = radius

Rotational Kinematic Equations:

ω = ω₀ + αt
θ = θ₀ + ω₀t + ½αt²
ω² = ω₀² + 2α(θ - θ₀)

Same form as linear equations, just with angular variables!

Example: Spinning Wheel

A wheel starts from rest and accelerates at 2 rad/s² for 5 seconds. What is its final angular velocity?

ω₀ = 0 rad/s
α = 2 rad/s²
t = 5 s

ω = ω₀ + αt = 0 + 2(5) = 10 rad/s

Newton's Second Law

The foundation of kinetics:

ΣF = ma

Where:
- ΣF = sum of all forces (N)
- m = mass (kg)
- a = acceleration (m/s²)

In component form:

ΣFx = max
ΣFy = may
ΣFz = maz

Example: Pulling a Sled

A 20 kg sled is pulled with a force of 100 N at 30° above horizontal on frictionless ice. What is the acceleration?

        F = 100 N
       /
      / 30°
    [====]  m = 20 kg
    --------- (ice, no friction)

Components:
Fx = F cos(30°) = 100 × 0.866 = 86.6 N
Fy = F sin(30°) = 100 × 0.5 = 50 N

Horizontal acceleration:
ΣFx = max
86.6 = 20 × ax
ax = 4.33 m/s²

Example: Elevator Acceleration

You stand on a scale in an elevator (your mass = 70 kg). What does the scale read when:

1. Elevator at rest
2. Accelerating upward at 2 m/s²
3. Accelerating downward at 2 m/s²

Forces on you:
- Weight: W = mg = 70 × 9.81 = 686.7 N (downward)
- Normal force: N (upward, from scale)

ΣF = ma

1. At rest (a = 0):
   N - W = 0
   N = 686.7 N (scale reads 70 kg)

2. Accelerating up (a = +2 m/s²):
   N - W = ma
   N = W + ma = 686.7 + 70(2) = 826.7 N (feels heavier!)
   Scale reads: 826.7/9.81 = 84.3 kg

3. Accelerating down (a = -2 m/s²):
   N - W = ma
   N = W + ma = 686.7 + 70(-2) = 546.7 N (feels lighter!)
   Scale reads: 546.7/9.81 = 55.7 kg

Work and Energy

Work

Work is energy transferred by a force acting over a distance.

W = F × d × cos(θ)

Where:
- W = work (Joules, J)
- F = force magnitude (N)
- d = distance moved (m)
- θ = angle between force and displacement

Special cases:

• If force is in direction of motion: W = F × d
• If force is perpendicular to motion: W = 0
• If force opposes motion: W is negative

Kinetic Energy

Energy of motion:

KE = ½mv²

Where:
- m = mass (kg)
- v = velocity (m/s)

Potential Energy

Energy due to position:

PE = mgh

Where:
- m = mass (kg)
- g = 9.81 m/s²
- h = height above reference (m)

Work-Energy Theorem

W_net = ΔKE = KE_final - KE_initial

Or:
W_net = ½mv² - ½mv₀²

Net work done on an object equals its change in kinetic energy.

Example: Sliding Down a Ramp

A 5 kg box slides down a frictionless ramp from height h = 3 m. What is its velocity at the bottom?

Using energy conservation:
PE_top = KE_bottom

mgh = ½mv²
gh = ½v²
v = √(2gh) = √(2 × 9.81 × 3) = √58.86 = 7.67 m/s

Power

Power is the rate of doing work:

P = W/t = F × v

Where:
- P = power (Watts, W)
- W = work (J)
- t = time (s)
- F = force (N)
- v = velocity (m/s)

Example: A motor lifts a 100 kg mass at constant speed of 2 m/s. What power is required?

Force needed = Weight = mg = 100 × 9.81 = 981 N
P = F × v = 981 × 2 = 1962 W ≈ 2 kW

Momentum and Impulse

Linear Momentum

p = mv

Where:
- p = momentum (kg⋅m/s)
- m = mass (kg)
- v = velocity (m/s)

Impulse

Change in momentum:

J = Δp = FΔt

Where:
- J = impulse (N⋅s)
- F = average force
- Δt = time interval

Impulse-Momentum Theorem:

FΔt = mΔv = m(v_f - v_i)

Example: Baseball Bat

A 0.15 kg baseball approaches at 40 m/s and leaves the bat at 50 m/s in the opposite direction. Contact time is 0.001 s. What is the average force?

Δv = v_f - v_i = 50 - (-40) = 90 m/s  (change direction!)

F = mΔv / Δt = 0.15 × 90 / 0.001 = 13,500 N

That's about 3000 lbf — huge force over short time!

Conservation of Momentum

In a closed system (no external forces):

Σp_before = Σp_after
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Example: Collision

Cart A (2 kg, 5 m/s) collides with cart B (3 kg, at rest). After collision, they stick together. What is their final velocity?

Before: p = m_A v_A + m_B v_B = 2(5) + 3(0) = 10 kg⋅m/s
After: p = (m_A + m_B)v_f = 5 v_f

Conservation: 10 = 5 v_f
v_f = 2 m/s

Rotational Dynamics

Moment of Inertia

Resistance to rotational acceleration (like mass for rotation):

I = Σ m_i r_i²

Where:
- I = moment of inertia (kg⋅m²)
- r_i = distance from rotation axis

Common Shapes:

Rotational Equation of Motion

ΣM = Iα

Where:
- M = moment/torque (N⋅m)
- I = moment of inertia (kg⋅m²)
- α = angular acceleration (rad/s²)

This is the rotational equivalent of F = ma!

Rotational Kinetic Energy

KE_rotational = ½Iω²

Where:
- I = moment of inertia
- ω = angular velocity (rad/s)

Example: Rolling Cylinder

A solid cylinder (m = 5 kg, r = 0.1 m) rolls down a ramp from height h = 2 m. What is its velocity at the bottom?

Total energy = Translational KE + Rotational KE

mgh = ½mv² + ½Iω²

For solid cylinder: I = ½mr²
For rolling without slipping: v = ωr, so ω = v/r

mgh = ½mv² + ½(½mr²)(v/r)²
mgh = ½mv² + ¼mv²
gh = ¾v²

v = √(4gh/3) = √(4 × 9.81 × 2 / 3) = √26.16 = 5.11 m/s

Notice this is slower than sliding (7.67 m/s) because some energy goes into rotation!

Practice Problems

Problem 1: Acceleration

A 1000 kg car accelerates from 0 to 25 m/s in 8 seconds. What force is required (assuming no friction)?

<details> <summary>Solution</summary>

a = Δv/t = 25/8 = 3.125 m/s²
F = ma = 1000 × 3.125 = 3,125 N

</details>

Problem 2: Work and Energy

How much work is done lifting a 20 kg box 3 meters vertically?

<details> <summary>Solution</summary>

W = mgh = 20 × 9.81 × 3 = 588.6 J

</details>

Problem 3: Momentum

A 60 kg person jumps off a 40 kg skateboard at 3 m/s. What is the skateboard's recoil velocity?

<details> <summary>Solution</summary>

Initial: both at rest, p = 0
Final: p = m_person v_person + m_board v_board

0 = 60(3) + 40(v_board)
v_board = -180/40 = -4.5 m/s (opposite direction)

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Key Takeaways

✓ Kinematics: Describes motion without considering forces
✓ F = ma: Foundation of dynamics, relates forces to acceleration
✓ Work-Energy: W = Fd, kinetic and potential energy
✓ Power: Rate of doing work, P = Fv
✓ Momentum: p = mv, conserved in collisions
✓ Rotational: Same principles apply, use I and ω instead of m and v

Next Steps

With forces and motion mastered, time to study how materials respond:

• Chapter 04: Mechanics of Materials: stress, strain, and failure
• Practice more kinematics and dynamics problems
• Study impact and vibration (advanced dynamics)

Pro Tip: Draw FBDs even for dynamics problems! Show all forces and acceleration direction.