Linear Kinetics

PEP 294: Lecture Notes

XII. Linear Kinetics

Newton’s Laws of Motion (pp. 61, 360-363)

- Related Problems: IP1, 2, AP1

1. 1st Law: Law of Inertia

Law of inertia: a body remains current state of motion unless acted on by an external force

Inertia:

- the tendency of an object to keep the current state of motion

- difficulty in changing the state of motion

Properties of inertia:

- static inertia vs. dynamic inertia

- proportional to mass of the object:

"The more massive an object, the more it tends to maintain its current state of motion."

2. 2nd Law: Law of Acceleration

Law of acceleration: a force applied to a body causes an acceleration

- acceleration is proportional to the force, inversely proportional to mass.

- direction of the acceleration = direction of the force

- if F = 0 => a = 0 (constant velocity) => law of inertia

3. 3rd Law: Law of Reaction

Law of reaction: for every action, there is an equal and opposite reaction

- when one body exerts a force on a second, the second body exerts a reaction force that is equal in magnitude and opposite in direction on the first

F₁₂ = -F₂₁

Examples:

1. Bullet vs. gun

2. Fist fighting

3. Collision in ice hockey (p. 363)

m₁ = 90 kg, m₂ = 80 kg, F₁₂ = 450 N => F₂₁

Ground Reaction Force (GRF) (pp. 363-365):

- reaction supplied by the ground

- an important external force for human motion

- Examples:

1. Jogging (p. 364)

2. Jumping -- development of vertical force in sacrificing the horizontal velocity using the ground reaction force (p. 364)

Law of Gravitation (pp. 363-366)

Law of gravitation:

- all bodies are attracted to one another.

- gravitational force is proportional to their masses and inversely proportional to the square of distance between them

Weight:

- gravity acting on a body from the Earth.

where, g = 9.81 m/s²

- direction: downward

Mass vs. weight:

Friction (pp. 366-373)

- Related Problems: IP3, 4, 5, AP2, 3

Friction: force acting at the area of contact between two surfaces

- magnitude: proportional to the friction coefficient and the normal reaction force

F_f = m N

(where, m = friction coefficient, N = normal reaction force)

- direction: opposite that of motion or motion tendency

Sliding vs. rolling:

- sliding: due to relative motion of the surfaces

- rolling: due to deformation of the surfaces

m_sliding >> m_rolling

Static vs. kinetic friction:

- max. static friction: max. force required to initiate a motion

- kinetic friction: force required to maintain the motion

m_s > m_k

- Examples: m_s = 0.18, m_k = 0.15, W_sled = 200 N, W_boy = 250 N => F

Force to initiate motion:

F_f = m_s N = (0.18)(200+250) = (0.18)(450) = 81 N

Force to maintain motion:

F_f = m_k N = (0.15)(450) = 67.5 N

Friction vs. motion: if F > F_f => motion occurs

F_net = F - F_f

F_net = ma    =>   F - F_f = ma

- Example -- in pulling a box: F = (100, 50) N, W = 500 N, m = 0.2   =>   Can you move the box?

N = W - F_v = 500 - 50 = 450 N

F_f = (0.2)(450) = 90 N

Since F_h (100 N) > F_f (90 N)   =>   “move”

Lubrication: reduces sliding friction -- grease, synovial fluid

Biomechanical roles of friction:

- resistance vs. source of propulsion

- minimize friction as a resistance

- maximize friction as a source of propulsion

Momentum and Impulse (pp. 373-379)

- Related problems: IP6, 7 & AP6

1. Momentum

Momentum: amount of motion

- momentum = (mass)(velocity)

M = m v

- important in giving or receiving impact, collision, etc.

- vector

- unit: kg·m/s

Example -- Two hockey players (p.376)

m₁ = 90 kg, v₁ = 6 m/s, m₂ = 80 kg, v₂ = -7 m/s => M’s

M₁ = (90)(6) = 540 kg·m/s

M₂ = (80)(-7) = -560 kg·m/s

Principle of conservation of momentum:

- if F = 0 => total momentum of the system remains constant

- Example: from the previous example (p.376):

After impact. both travel together as a unit:

M_before = M₁ + M₂ = 540 + (-560) = -20 kg·m/s

M_after = (m₁ + m₂)(v_after) = ( 90 + 80 )(v_after) = (170)(v_after)

M_before = M_after => (170)(v_after) = -20

v_after = -20 / 170 = -0.12 m/s

Therefore, player 2 pushes player 1 at 0.12 m/s.

2. Impulse

Impulse: accumulated effect of force exerted on an object for a period of time

- impulse = (force)(time)

I = F t

- increase in F or t => increase in I

- vector

- equal to the change in momentum of the system

I = DM = M₂ - M₁ = m v₂ - m v₁

Example -- Toboggan race: F = 100 N, t = 7 s, m = 90 kg => v

F t = m v₂ - m v₁

(100)(7) = (90)(v₂) - (90)(0) => v₂ = 7.78 m/s

Strategies in sports:

- giving impulse: maximize (batting, throwing, etc.)

- receiving impulse: minimize (landing, receiving, etc.)

Work, Power & Energy (pp. 383-390)

- Related problems: IP9, 10, AP8 & 9

1. Work

Work: force applied against a resistance

- work = (force)(displacement)

W = Fd cos q

(where, F = force, d = displacement, q = angle between the two vectors)

- scalar quantity

- unit: J (Joule) = Nm

Examples:

1. Upward weightlifting: weight = 500 N, lift height = 0.5 m => W

F = W = 500 N (upward), d = 0.5 m (upward)

W = Fd cos q = (500)(0.5)(cos 0) = 250 Nm = 250 J

2. Downward weightlifting: weight = 500 N, lift height = 0.5 m => W

F = W = 500 N (upward), d = 0.5 m (downward)

W = (500)(0.5)(cos p) = (500)(0.5)(-1.000) = -250 J

Positive vs. negative work:

- determined by the directions of the vectors

- if q > p/2 rad (90°) => negative work

Muscle contraction types vs. work:

- concentric contraction: positive work

- eccentric contraction: negative work

- isometric contraction: no work

Example -- Mountain climbing:

W₁ = Fd₁(cos q₁), W₂ = Fd₂(cos q₂)

d₁(cos q₁) = d₂(cos q₂) = h => W₁ = W₂

2. Power

Power: rate of work

- power = (work) / (elapsed time)

- scalar quantity

- unit: W (Watt) = J/s

Example -- stair climbing (p. 385): 30 steps, 25 cm/step, weight = 580 N, Dt = 15 s => P
h = (30)(0.25) = 7.5 m,

P = (Fd·cos q) / Dt = (580)(7.5)(cos 0) / (15) = 4350 / 15 = 290.0 W

Power = (force)(velocity)

3. Mechanical Energy

Mechanical energy: capacity to do mechanical work

- unit: J (same to work)

Potential energy vs. kinetic energy:

- kinetic energy: energy due to motion of the object

KE =

- potential energy: energy due to the position (height) of the object

PE = mgh (where, g = 9.81 m/s²)

- Examples:

1. m = 2 kg, v = 3 m/s => KE

KE = (0.5)(2)(3)² = 9 J

2. m = 50 kg, h = 1 m => PE

PE = (50)(9.81)(1) = 490.5 J

- ME = KE + PE

Conservation of mechanical energy:

- when the gravity is the only external force acting on a system, ME remains constant

ME = KE + PE = const.

- Examples:

1. Diving

ME = KE + PE = const

Position1:

KE = 0, PE = ME

Position 2:

KE > 0, PE > 0, KE + PE = ME

Position 3:

PE = 0, KE = ME

2. Ball drop: h = 1.5 m, m = 2 kg => v_impact

ME_release = KE_release + PE_release = 0 + mgh

ME_impact = KE_impact + PE_impact

= + 0

ME_release = ME_impact

= mgh

(0.5)(2)(v_impact)² = (2)(9.81)(1.5)

(v_impact)² = 29.43

v_impact = 5.42 m/s

Work vs. energy:

- work causes changes in the mechanical energy of the system

W = DME = DKE + DPE

- Example:

1. Diving

Position 1 to 2:

W = DME = DPE = mgh

Position 2:

ME = PE = W (KE = 0)

Position 3:

ME = KE + PE = W (KE > 0, PE > 0)

Position 4:

ME = PE = W (PE = 0)

2. Ball catching (p. 390):

v = 40 m/s, m = 1.3 kg => W to stop the ball?

v₁ = 40, v₂ = 0

W = DKE = (0.5)(1.3)(0)² - (0.5)(1.3)(40)² = -1040 J

The catcher did negative work by 1040 J.