Dynamics

LessonAbout 1.5 h

Dynamics

Key Ideas

  • Study of motion and forces causing it
  • Covers translation, rotation, and curvilinear motion
  • Applies Newton’s laws, work-energy, and momentum

Dynamics is the branch of mechanics dealing with bodies in motion and the forces acting upon them.

1. Rectilinear Translation

Rectilinear translation is motion along a straight line path.

1.1 Uniform Motion

In uniform rectilinear motion, velocity remains constant and acceleration is zero.

Uniform Rectilinear Motion

\[ s = vt \]

Where: \(s\) = displacement, \(v\) = constant velocity, \(t\) = time

1.2 Variable Acceleration

In variable acceleration, velocity changes non-uniformly as a function of time, displacement, or velocity.

Variable Acceleration Relations

\[ v = \dfrac{ds}{dt} \]
\[ a = \dfrac{dv}{dt} \]
\[ v\,dv = a\,ds \]

Where: \(s\) = position, \(v\) = velocity, \(a\) = acceleration, \(t\) = time

EE Board October 1996

Example. A particle moves along a straight line such that its position is \(x(t) = 3.4t^3 - 5.4t\) meters. Determine its acceleration at \(t = 5\ \text{seconds}\).

Solution. Differentiate position twice with respect to time:

\[ v(t) = \dfrac{dx}{dt} = 10.2t^2 - 5.4 \]
\[ a(t) = \dfrac{dv}{dt} = 20.4t \]

Evaluate at \(t = 5\ \text{s}\):

\[ a(5) = 20.4(5) = 102\ \text{m/s}^2 \]

Answer: \(102\ \text{m/s}^2\)

1.3 Constant Acceleration

In uniformly accelerated motion, the rate of change of velocity is constant.

Constant Linear Acceleration Formulas

\[ v = v_0 + at \]
\[ s = v_0 t + \dfrac{1}{2}at^2 \]
\[ v^2 = v_0^2 + 2as \]

Where: \(v_0\) = initial velocity, \(v\) = final velocity, \(a\) = acceleration, \(s\) = displacement, \(t\) = time

ME Board April 1996

Example. Calculate the acceleration of an object that increases its speed from \(20\ \text{m/s}\) to \(40\ \text{m/s}\) in \(3\ \text{seconds}\).

Solution. Apply the constant acceleration equation:

\[ v = v_0 + at \]
\[ 40 = 20 + a(3) \]
\[ a = 6.67\ \text{m/s}^2 \]

Answer: \(6.67\ \text{m/s}^2\)

2. Rotation or Angular Motion

Rotation is the motion of a rigid body where all particles move in circular paths about a fixed axis.

2.1 Uniform Motion

In uniform angular motion, angular velocity remains constant and angular acceleration is zero.

Uniform Angular Motion

\[ \theta = \omega t \]

Where: \(\theta\) = angular displacement in radians, \(\omega\) = angular velocity, \(t\) = time

2.2 Uniform Acceleration

In uniformly accelerated rotation, the rate of change of angular velocity is constant.

Constant Angular Acceleration Formulas

\[ \omega = \omega_0 + \alpha t \]
\[ \theta = \omega_0 t + \dfrac{1}{2}\alpha t^2 \]
\[ \omega^2 = \omega_0^2 + 2\alpha \theta \]

Where: \(\omega_0\) = initial angular velocity, \(\omega\) = final angular velocity, \(\alpha\) = angular acceleration, \(\theta\) = angular displacement

Example. A \(15\ \text{cm}\) diameter flywheel accelerates uniformly from rest to \(500\ \text{rpm}\) in \(20\ \text{seconds}\). Find its angular acceleration.

Solution. Convert rotational speed to radians per second:

\[ \omega = 500\ \text{rpm} \times \dfrac{1\ \text{min}}{60\ \text{s}} \times \dfrac{2\pi\ \text{rad}}{1\ \text{rev}} = 52.36\ \text{rad/s} \]

Apply uniform angular acceleration equation:

\[ \omega = \omega_0 + \alpha t \]
\[ 52.36 = 0 + \alpha(20) \]
\[ \alpha = 2.62\ \text{rad/s}^2 \]

Answer: \(2.62\ \text{rad/s}^2\)

3. Curvilinear Motion

Curvilinear motion occurs when a particle moves along a curved path.

3.1 Projectile Motion

A projectile follows a parabolic trajectory under gravity with constant horizontal velocity and vertical uniform acceleration.

Projectile Motion Equations

\[ x = (v_0 \cos\theta) t \]
\[ y = (v_0 \sin\theta) t - \dfrac{1}{2}gt^2 \]
\[ y = x \tan\theta - \dfrac{g x^2}{2 v_0^2 \cos^2\theta} \]
\[ R = \dfrac{v_0^2 \sin 2\theta}{g} \]
\[ H = \dfrac{v_0^2 \sin^2\theta}{2g} \]

Where: \(v_0\) = launch velocity, \(\theta\) = launch angle, \(x\) = horizontal distance, \(y\) = vertical height, \(R\) = range, \(H\) = maximum height

CE Board May 1995

Example. A projectile is launched at \(50\ \text{m/s}\) at an angle of \(30^\circ\) above the horizontal. Calculate its maximum height.

Solution. Apply the maximum height formula:

\[ H = \dfrac{v_0^2 \sin^2\theta}{2g} \]
\[ H = \dfrac{(50)^2 (\sin 30^\circ)^2}{2(9.81)} = 31.86\ \text{m} \]

Answer: \(31.86\ \text{m}\)

4. Relation of Angular and Peripheral Motion

The linear components along the circular arc are related to angular components by the radius of rotation.

  • Arc distance: \(s = r\theta\)
  • Linear velocity: \(v = r\omega\)
  • Tangential acceleration: \(a_t = r\alpha\)
  • Normal acceleration: \(a_n = \dfrac{v^2}{r} = r\omega^2\)

Linear and Angular Relationships

\[ s = r\theta \qquad\qquad v = r\omega \qquad\qquad a_t = r\alpha \]
\[ a_n = \dfrac{v^2}{r} = r\omega^2 \]
\[ a_{\text{total}} = \sqrt{a_t^2 + a_n^2} \]

Where: \(r\) = radius, \(s\) = arc length, \(v\) = tangential speed, \(\omega\) = angular speed, \(a_t\) = tangential acceleration, \(a_n\) = normal acceleration

ME Board April 1996

Example. A cyclist travels around a circular track of radius \(r = 800\ \text{ft}\) at a speed of \(27\ \text{ft/s}\). If his tangential speed increases at \(3\ \text{ft/s}^2\), determine his total acceleration.

Solution. Calculate normal acceleration:

\[ a_n = \dfrac{v^2}{r} = \dfrac{(27)^2}{800} = 0.91125\ \text{ft/s}^2 \]

Combine with tangential acceleration \(a_t = 3\ \text{ft/s}^2\):

\[ a_{\text{total}} = \sqrt{a_t^2 + a_n^2} = \sqrt{(3)^2 + (0.91125)^2} = 3.1\ \text{ft/s}^2 \]

Answer: \(3.1\ \text{ft/s}^2\)

5. D’Alembert’s Principle

D’Alembert’s principle converts a dynamic system into an equivalent static equilibrium system by introducing a reverse effective force.

Reverse Effective Force

\[ \text{REF} = ma = \dfrac{W}{g}a \]

Where: \(\text{REF}\) = reverse effective force directed opposite to acceleration, \(m\) = mass, \(W\) = weight, \(a\) = acceleration

ME Board April 1998

Example. An elevator weighing \(2,000\ \text{lb}\) reaches an upward velocity of \(16\ \text{ft/s}\) in \(4\ \text{seconds}\) under uniform acceleration. Find the tension in the supporting cable.

Solution. Determine upward acceleration:

\[ a = \dfrac{v - v_0}{t} = \dfrac{16 - 0}{4} = 4\ \text{ft/s}^2 \]

Apply dynamic equilibrium using D’Alembert’s principle:

\[ T - W - \text{REF} = 0 \]
\[ T = W + \dfrac{W}{g}a = 2000 + \dfrac{2000}{32.2}(4) = 2248.4\ \text{lb} \approx 2250\ \text{lb} \]

Answer: \(2250\ \text{lb}\)

6. Newton’s Laws of Motion

Newton’s laws of motion describe the relationship between forces acting on a body and its motion.

  • First Law (Law of Inertia): A body remains at rest or in uniform motion unless acted upon by an unbalanced force.
  • Second Law (Law of Acceleration): An unbalanced force produces acceleration proportional to the force and inversely proportional to mass.
  • Third Law (Law of Reaction): For every action force, there is an equal and opposite reaction force.

Newton’s Second Law

\[ F = ma = \dfrac{W}{g}a \]

Where: \(F\) = resultant net force, \(m\) = mass, \(a\) = acceleration, \(W\) = weight, \(g\) = acceleration due to gravity

EE Board April 1996

Example. A car with a mass of \(1700\ \text{kg}\) travels at \(70\ \text{km/h}\). What retarding force is needed to decelerate it at \(40\ \text{cm/s}^2\)?

Solution. Convert acceleration to meters per second squared:

\[ a = 40\ \text{cm/s}^2 = 0.40\ \text{m/s}^2 \]

Apply Newton’s second law:

\[ F = ma = 1700(0.40) = 680\ \text{N} \]

Answer: \(680\ \text{N}\)

7. Centrifugal and Centripetal Forces

Centripetal force is the inward force directing a body along a curved path, while centrifugal force is the apparent outward inertia force.

Centrifugal and Centripetal Force

\[ F_c = \dfrac{m v^2}{r} = \dfrac{W v^2}{g r} = m \omega^2 r \]

Where: \(F_c\) = centrifugal or centripetal force, \(m\) = mass, \(v\) = linear speed, \(r\) = radius of curvature, \(\omega\) = angular velocity

ME Board April 1991

Example. A student rotates an \(80\ \text{g}\) stone tied to a string in a circle of \(1000\ \text{mm}\) diameter. If the stone leaves the path at \(25\ \text{m/s}\), calculate the pull exerted on the string.

Solution. Identify mass and radius in standard units:

\[ m = 0.08\ \text{kg} \qquad\qquad r = 0.5\ \text{m} \]

Compute centrifugal force:

\[ F_c = \dfrac{m v^2}{r} = \dfrac{0.08(25)^2}{0.5} = 100\ \text{N} \]

Answer: \(100\ \text{N}\)

8. Pendulums

A pendulum consists of a mass suspended from a pivot that oscillates or rotates under the influence of gravity.

8.1 Simple Pendulum

A simple pendulum is a point mass suspended by a light string swinging back and forth in a vertical plane.

Simple Pendulum Formulas

\[ T = 2\pi \sqrt{\dfrac{L}{g}} \]
\[ v = \sqrt{2gh} \]

Where: \(T\) = period for one complete oscillation, \(L\) = length of pendulum, \(g\) = acceleration due to gravity, \(v\) = speed at lowest position, \(h\) = vertical drop

8.2 Conical Pendulum

A conical pendulum consists of a mass attached to a string sweeping out a cone while moving in a horizontal circle.

Conical Pendulum Formulas

\[ \tan\theta = \dfrac{F_c}{W} = \dfrac{v^2}{g r} = \dfrac{\omega^2 r}{g} \]
\[ T_{\text{period}} = 2\pi \sqrt{\dfrac{h}{g}} = 2\pi \sqrt{\dfrac{L \cos\theta}{g}} \]

Where: \(\theta\) = angle with vertical, \(h = L \cos\theta\) = height of cone, \(r = L \sin\theta\) = radius of horizontal circle, \(T_{\text{period}}\) = time for one revolution

Example. A \(4\ \text{N}\) ball is attached to a \(1.2\ \text{m}\) long cord revolving about a vertical axis such that the cord makes an angle of \(20^\circ\) with the vertical. Calculate the speed of rotation in rpm.

Solution. Determine the vertical height of the conical path:

\[ h = L \cos\theta = 1.2 \cos 20^\circ = 1.1276\ \text{m} \]

Apply conical pendulum rotational relation:

\[ \omega = \sqrt{\dfrac{g}{h}} = \sqrt{\dfrac{9.81}{1.1276}} = 2.949\ \text{rad/s} \]

Convert angular velocity to rpm:

\[ N = 2.949\ \text{rad/s} \times \dfrac{60\ \text{s}}{2\pi\ \text{rad}} = 28.16\ \text{rpm} \]

Answer: \(28.16\ \text{rpm}\)

9. Banking of Curves

Banking of curves involves tilting a roadway at an angle \(\theta\) to prevent lateral skidding using the normal force component.

Highway Banking Formulas

\[ \tan\theta = \dfrac{v^2}{g R} \]
\[ \tan(\theta + \phi) = \dfrac{v^2}{g R} \]

Where: \(\theta\) = ideal banking angle, \(\phi\) = friction angle, \(v\) = speed, \(R\) = radius of curve

CE Board November 1998

Example. A highway curve is banked at an angle of \(7^\circ\). Determine the radius of the curve so that no lateral pressure acts on the tires at \(40\ \text{mph}\).

Solution. Convert velocity to meters per second:

\[ v = 40\ \text{mph} \times \dfrac{5280\ \text{ft}}{1\ \text{mi}} \times \dfrac{1\ \text{m}}{3.281\ \text{ft}} \times \dfrac{1\ \text{hr}}{3600\ \text{s}} = 17.88\ \text{m/s} \]

Apply ideal banking equation:

\[ \tan 7^\circ = \dfrac{v^2}{g R} = \dfrac{(17.88)^2}{9.81 R} \]
\[ R = 265.41\ \text{m} \]

Answer: \(265.41\ \text{m}\)

10. Horizontal Rotating Platform

A horizontal rotating platform holds an object by static friction until centrifugal force exceeds the maximum frictional resistance.

Impending Motion on Rotating Platform

\[ \mu = \dfrac{v^2}{g R} = \dfrac{\omega^2 R}{g} \]

Where: \(\mu\) = coefficient of static friction, \(\omega\) = angular speed in rad/s, \(R\) = radial distance from rotation axis, \(v\) = tangential speed

Example. An object sits \(3\ \text{ft}\) from the center of a horizontal rotating platform. If the coefficient of static friction is \(0.30\), find the rotational speed in rpm when the object begins to slide.

Solution. Set frictional force equal to centrifugal force:

\[ \mu = \dfrac{\omega^2 R}{g} \]
\[ 0.30 = \dfrac{\omega^2 (3)}{32.2} \]
\[ \omega = 1.794\ \text{rad/s} \]

Convert to revolutions per minute:

\[ N = 1.794 \times \dfrac{60}{2\pi} = 17\ \text{rpm} \]

Answer: \(17\ \text{rpm}\)

11. Work and Energy

Work is energy transferred by a force acting over a displacement, while energy is the capacity to do work.

  • Work-energy theorem: \(W_{\text{net}} = \Delta \text{KE} = \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}\)
  • Power: \(P = \dfrac{W}{t} = Fv\)

11.1 Kinetic Energy

Kinetic energy is the energy possessed by a body due to its motion.

Translational Kinetic Energy

\[ \text{KE} = \dfrac{1}{2}m v^2 = \dfrac{1}{2}\dfrac{W}{g}v^2 \]

Where: \(m\) = mass, \(W\) = weight, \(v\) = linear velocity

Example. A \(5\ \text{kg}\) cart starting from rest accelerates at \(3\ \text{m/s}^2\) for \(4\ \text{seconds}\). Calculate its kinetic energy after this time.

Solution. Find the final velocity:

\[ v = v_0 + at = 0 + 3(4) = 12\ \text{m/s} \]

Calculate kinetic energy:

\[ \text{KE} = \dfrac{1}{2}m v^2 = \dfrac{1}{2}(5)(12)^2 = 360\ \text{J} \]

Answer: \(360\ \text{J}\)

11.2 Potential Energy

Potential energy is the energy stored in a body by virtue of its position or configuration.

Gravitational Potential Energy

\[ \text{PE} = mgh = Wh \]

Where: \(m\) = mass, \(g\) = gravitational acceleration, \(h\) = height above reference plane, \(W\) = weight

ME Board April 1995, ME Board April 1998

Example. A \(10\ \text{kg}\) block is lifted vertically through \(3\ \text{meters}\). Determine the change in potential energy.

Solution. Compute gravitational potential energy:

\[ \Delta \text{PE} = mgh = 10(9.81)(3) = 294.3\ \text{J} \]

Answer: \(294\ \text{J}\)

11.3 Kinetic Energy of Rotating Body

Rotational kinetic energy is the kinetic energy stored in a body rotating about an axis.

Rotational Kinetic Energy

\[ \text{KE}_{\text{rot}} = \dfrac{1}{2}I \omega^2 \]
\[ I = mk^2 = \dfrac{W}{g}k^2 \]

Where: \(I\) = mass moment of inertia, \(k\) = radius of gyration, \(\omega\) = angular velocity in rad/s

REE Board March 2015

Example. Compute the rotational kinetic energy of a \(25\ \text{kg}\) wheel rotating at \(6.0\ \text{rev/s}\) if its radius of gyration is \(22\ \text{cm}\).

Solution. Convert angular velocity and radius of gyration to standard units:

\[ \omega = 6.0(2\pi) = 12\pi\ \text{rad/s} \qquad\qquad k = 0.22\ \text{m} \]

Calculate mass moment of inertia and rotational kinetic energy:

\[ I = mk^2 = 25(0.22)^2 = 1.21\ \text{kg}\cdot\text{m}^2 \]
\[ \text{KE}_{\text{rot}} = \dfrac{1}{2}I \omega^2 = \dfrac{1}{2}(1.21)(12\pi)^2 = 859.8\ \text{J} = 0.86\ \text{kJ} \]

Answer: \(0.86\ \text{kJ}\)

12. Impulse and Momentum

Linear momentum is the product of mass and velocity, while impulse is the product of force and time.

Impulse-Momentum Principle

\[ p = mv \]
\[ \text{Impulse} = F \Delta t = \Delta p = m v_{\text{final}} - m v_{\text{initial}} \]

Where: \(p\) = linear momentum, \(F\) = force, \(\Delta t\) = time interval, \(v_{\text{initial}}\) and \(v_{\text{final}}\) = velocities

ME Board April 1997

Example. Calculate the linear momentum possessed by a \(40\ \text{lbm}\) projectile moving at \(420\ \text{mph}\).

Solution. Convert speed to feet per second:

\[ v = 420 \times \dfrac{5280}{3600} = 616\ \text{ft/s} \]

Calculate momentum in \(\text{lbf-sec}\):

\[ p = \dfrac{m v}{g} = \dfrac{40(616)}{32.2} = 765\ \text{lbf-sec} \]

Answer: \(765\ \text{lbf-sec}\)

13. Law of Conservation of Momentum

The law of conservation of momentum states that in the absence of net external forces, the total momentum of a system remains constant.

Conservation of Linear Momentum

\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]

Where: \(m_1, m_2\) = masses of colliding bodies, \(v_1, v_2\) = initial velocities, \(v_1', v_2'\) = final velocities

13.1 Types of Collision

Collisions are categorized based on whether total mechanical kinetic energy is conserved.

  • Elastic collision: both momentum and kinetic energy are conserved (\(e = 1\)).
  • Inelastic collision: momentum is conserved but kinetic energy is lost (\(0 < e < 1\)).
  • Perfectly inelastic collision: bodies stick together after impact (\(e = 0\)).

13.2 Coefficient of Restitution

The coefficient of restitution \(e\) is the negative ratio of relative velocity after collision to relative velocity before collision.

Coefficient of Restitution

\[ e = -\dfrac{v_2' - v_1'}{v_2 - v_1} = \dfrac{v_1' - v_2'}{v_2 - v_1} \]
\[ e = \sqrt{\dfrac{h'}{h}} \]

Where: \(v_1, v_2\) = velocities before collision, \(v_1', v_2'\) = velocities after collision, \(h\) = drop height, \(h'\) = rebound height

Example. A \(16\ \text{kg}\) mass moving at \(0.3\ \text{m/s}\) collides head-on with a \(4\ \text{kg}\) mass traveling in the opposite direction at \(0.5\ \text{m/s}\). If they stick together after impact, determine their common velocity.

Solution. Apply the law of conservation of momentum for a completely inelastic collision:

\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2)v \]
\[ 16(0.3) + 4(-0.5) = (16 + 4)v \]
\[ 4.8 - 2.0 = 20v \]
\[ v = 0.14\ \text{m/s} \]

Answer: \(0.14\ \text{m/s}\)