University Physics Volume 1- ch 4

Motion in Two and Three Dimensions

  • Most objects in the universe move in curved paths, not straight lines.
  • Kinematics in one dimension provides a foundation for understanding motion in two and three dimensions.
  • Concepts of position, displacement, velocity, and acceleration can be expanded to two and three dimensions by including vectors.
  • Two special types of motion in two dimensions: projectile motion and circular motion.
  • Relative motion considers the motion of objects with respect to each other or observers.

4.1 Displacement and Velocity Vectors

  • Displacement and velocity in two or three dimensions are vector quantities that extend one-dimensional definitions.
  • Calculations with displacement and velocity must follow the rules of vector algebra.

Position Vector

  • A coordinate system is established using x, y, and z axes to locate a particle at point P(x, y, z) in three dimensions.
  • If the particle is moving, the variables x, y, and z are functions of time (t): x = x(t), y = y(t), z = z(t).
  • The position vector from the origin of the coordinate system to point P is \vec{r}(t).
  • In unit vector notation: \vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}.
  • The orientation of the x, y, and z axes is called a right-handed coordinate system.

Three-Dimensional Displacement

  • Displacement vector \Delta \vec{r} is found by subtracting \vec{r}(t1) from \vec{r}(t2): \Delta \vec{r} = \vec{r}(t2) - \vec{r}(t1).

Example 4.1: Polar Orbiting Satellite: A satellite in a circular polar orbit around Earth at an altitude of 400 km, passing directly overhead at the North and South Poles. The magnitude and direction of the displacement vector are calculated from when it is directly over the North Pole to when it is at -45° latitude.

  • Earth's radius is taken as 6370 km, making the length of each position vector 6770 km.
  • Position vectors in unit vector notation:
    • \vec{r}(t_1) = 6770 \text{ km } \hat{j}
    • \vec{r}(t_2) = 6770 \text{ km } (\cos 45°) \hat{i} + 6770 \text{ km } (\sin(-45°)) \hat{j}
  • Evaluating the sine and cosine:
    • \vec{r}(t_1) = 6770 \hat{j}
    • \vec{r}(t_2) = 4787 \hat{i} - 4787 \hat{j}
  • The displacement of the satellite is:
    • \Delta \vec{r} = \vec{r}(t2) - \vec{r}(t1) = 4787 \hat{i} - 11557 \hat{j}
  • The magnitude of the displacement is:
    • |\Delta \vec{r}| = \sqrt{(4787)^2 + (-11557)^2} = 12509 \text{ km}
  • The angle the displacement makes with the x-axis is:
    • \theta = \tan^{-1} \left( \frac{-11557}{4787} \right) = -67.5°
  • The displacement vector gives the shortest path between two points in one, two, or three dimensions.

Example 4.2: Brownian Motion: Demonstration of chaotic random motion of particles suspended in a fluid, resulting from collisions with the molecules of the fluid. The displacements in numerical order of a particle undergoing Brownian motion (in micrometers) are given, and the total displacement of the particle from the origin is calculated.

  • \Delta \vec{r}_1 = 2.0 \hat{i} + \hat{j} + 3.0 \hat{k}
  • \Delta \vec{r}_2 = -\hat{i} + 3.0 \hat{k}
  • \Delta \vec{r}_3 = 4.0 \hat{i} - 2.0 \hat{j} + \hat{k}
  • \Delta \vec{r}_4 = -3.0 \hat{i} + \hat{j} + 2.0 \hat{k}
  • The total displacement of the particle from the origin:
    • \Delta \vec{r}{\text{Total}} = \sum \Delta \vec{r}i = \Delta \vec{r}1 + \Delta \vec{r}2 + \Delta \vec{r}3 + \Delta \vec{r}4
    • = (2.0 - 1.0 + 4.0 - 3.0) \hat{i} + (1.0 + 0 - 2.0 + 1.0) \hat{j} + (3.0 + 3.0 + 1.0 + 2.0) \hat{k}
    • = 2.0 \hat{i} + 0 \hat{j} + 9.0 \hat{k} \text{ } \mu\text{m}
  • The displacement expressed as magnitude and direction:
    • |\Delta \vec{r}_{\text{Total}}| = \sqrt{2.0^2 + 0^2 + 9.0^2} = 9.2 \text{ } \mu\text{m}
    • \theta = \tan^{-1} \left( \frac{9}{2} \right) = 77°

Velocity Vector

  • The instantaneous velocity vector is found by calculating the derivative of the position function with respect to time: \vec{v}(t) = \lim_{\Delta t \to 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t} = \frac{d\vec{r}}{dt}.
  • The velocity vector is tangent to the path of the particle at time t.
  • Equation 4.4 can also be written in terms of components: \vec{v}(t) = vx(t) \hat{i} + vy(t) \hat{j} + v_z(t) \hat{k}.
  • vx(t) = \frac{dx(t)}{dt}, vy(t) = \frac{dy(t)}{dt}, v_z(t) = \frac{dz(t)}{dt}.
  • The average velocity is \vec{v}{\text{avg}} = \frac{\vec{r}(t2) - \vec{r}(t1)}{t2 - t_1}.

Example 4.3: Calculating the Velocity Vector: The position function of a particle is given as \vec{r}(t) = 2.0t^2 \hat{i} + (2.0 + 3.0t) \hat{j} + 5.0t \hat{k} \text{ m}.

  • The instantaneous velocity and speed at t = 2.0 s are calculated.
    • \vec{v}(t) = \frac{d\vec{r}(t)}{dt} = 4.0t \hat{i} + 3.0 \hat{j} + 5.0 \hat{k} \text{ m/s}
    • \vec{v}(2.0 \text{s}) = 8.0 \hat{i} + 3.0 \hat{j} + 5.0 \hat{k} \text{ m/s}
    • \text{Speed } |\vec{v}(2.0 \text{ s})| = \sqrt{8^2 + 3^2 + 5^2} = 9.9 \text{ m/s}
  • The average velocity between 1.0 s and 3.0 s is calculated.
    • \vec{v}{\text{avg}} = \frac{\vec{r}(t2) - \vec{r}(t1)}{t2 - t_1} = \frac{\vec{r}(3.0 \text{ s}) - \vec{r}(1.0 \text{ s})}{3.0 \text{ s} - 1.0 \text{ s}}
    • = \frac{(18 \hat{i} + 11 \hat{j} + 15 \hat{k}) \text{ m} - (2 \hat{i} + 5 \hat{j} + 5 \hat{k}) \text{ m}}{2.0 \text{ s}}
    • = \frac{(16 \hat{i} + 6 \hat{j} + 10 \hat{k}) \text{ m}}{2.0 \text{ s}} = 8.0 \hat{i} + 3.0 \hat{j} + 5.0 \hat{k} \text{ m/s}

Independence of Perpendicular Motions

  • The components of position and velocity equations are separate and unique functions of time that do not depend on one another.
  • Motion along one direction has no part of its motion along perpendicular directions.
  • The motion of an object in two or three dimensions can be divided into separate, independent motions along the perpendicular axes of the coordinate system.
  • In the kinematic description of motion, the horizontal and vertical components are treated separately.
  • Motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.

4.2 Acceleration Vector

  • The acceleration vector is the instantaneous acceleration and can be obtained from the derivative with respect to time of the velocity function.
  • The only difference in two or three dimensions is that these are now vector quantities.
  • Taking the derivative with respect to time \vec{v}(t), we find \vec{a}(t) = \lim_{t \to 0} \frac{\vec{v}(t + \Delta t) - \vec{v}(t)}{\Delta t} = \frac{d\vec{v}(t)}{dt}.
  • The acceleration in terms of components is \vec{a}(t) = \frac{dvx(t)}{dt} \hat{i} + \frac{dvy(t)}{dt} \hat{j} + \frac{dv_z(t)}{dt} \hat{k}.

Example 4.4: Finding an Acceleration Vector: If a particle has a velocity \vec{v}(t) = 5.0t \hat{i} + t^2 \hat{j} - 2.0t^3 \hat{k} \text{ m/s}, the acceleration function is \vec{a}(t) = 5.0 \hat{i} + 2.0t \hat{j} - 6.0t^2 \hat{k} \text{ m/s}^2 and the acceleration vector at t = 2.0 s is \vec{a}(2.0 \text{ s}) = 5.0 \hat{i} + 4.0 \hat{j} - 24.0 \hat{k} \text{ m/s}^2.

  • The magnitude of the acceleration is |\vec{a}(2.0 \text{ s})| = \sqrt{5.0^2 + 4.0^2 + (-24.0)^2} = 24.8 \text{ m/s}^2.

Example 4.5: Finding a Particle Acceleration: For a particle with position function \vec{r}(t) = (10t - t^2) \hat{i} + 5t \hat{j} + 5t \hat{k} \text{ m}, the velocity is \vec{v}(t) = (10 - 2t) \hat{i} + 5 \hat{j} + 5 \hat{k} \text{ m/s} and the acceleration is \vec{a}(t) = -2 \hat{i} \text{ m/s}^2.

Constant Acceleration

  • Multidimensional motion with constant acceleration can be treated the same way as one-dimensional motion.
  • Three-dimensional motion is equivalent to three one-dimensional motions, each along an axis perpendicular to the others.
  • 2D problem of a particle moving in the xy plane with constant acceleration, ignoring the z-component.
  • The acceleration vector is \vec{a} = a{0x} \hat{i} + a{0y} \hat{j}.
  • Each component of the motion has a separate set of equations similar to those of one-dimensional motion: in the x- and y-directions.
  • x(t) = x0 + (vx)_{\text{avg}} t
  • vx(t) = v{0x} + a_x t
  • x(t) = x0 + v{0x} t + \frac{1}{2} a_x t^2
  • vx^2(t) = v{0x}^2 + 2ax(x - x0)
  • y(t) = y0 + (vy)_{\text{avg}} t
  • vy(t) = v{0y} + a_y t
  • y(t) = y0 + v{0y} t + \frac{1}{2} a_y t^2
  • vy^2(t) = v{0y}^2 + 2ay(y - y0)

Example 4.6: A Skier: A skier moves with an acceleration of 2.1 m/s² down a slope of 15° at t = 0. Her initial position and velocity are given as \vec{r}(0) = (75.0 \hat{i} - 50.0 \hat{j}) \text{ m} and \vec{v}(0) = (4.1 \hat{i} - 1.1 \hat{j}) \text{ m/s}. The x- and y-components of the skier’s position and velocity are determined as functions of time.

  • The components of the acceleration are:
    • a_x = (2.1 \text{ m/s}^2) \cos(15°) = 2.0 \text{ m/s}^2
    • a_y = (-2.1 \text{ m/s}^2) \sin 15° = -0.54 \text{ m/s}^2
  • Inserting the initial position and velocity:
    • x(t) = 75.0 \text{ m} + (4.1 \text{ m/s})t + \frac{1}{2} (2.0 \text{ m/s}^2) t^2
    • v_x(t) = 4.1 \text{ m/s} + (2.0 \text{ m/s}^2) t
    • y(t) = -50.0 \text{ m} + (-1.1 \text{ m/s})t + \frac{1}{2} (-0.54 \text{ m/s}^2) t^2
    • v_y(t) = -1.1 \text{ m/s} + (-0.54 \text{ m/s}^2) t
  • Evaluating her position and velocity at t = 10.0 s:
    • x(10.0 \text{ s}) = 75.0 \text{ m} + (4.1 \text{ m/s})(10.0 \text{ s}) + \frac{1}{2} (2.0 \text{ m/s}^2) (10.0 \text{ s})^2 = 216.0 \text{ m}
    • v_x(10.0 \text{ s}) = 4.1 \text{ m/s} + (2.0 \text{ m/s}^2) (10.0 \text{ s}) = 24.1 \text{ m/s}
    • y(10.0 \text{ s}) = -50.0 \text{ m} + (-1.1 \text{ m/s})(10.0 \text{ s}) + \frac{1}{2} (-0.54 \text{ m/s}^2) (10.0 \text{ s})^2 = -88.0 \text{ m}
    • v_y(10.0 \text{ s}) = -1.1 \text{ m/s} + (-0.54 \text{ m/s}^2) (10.0 \text{ s}) = -6.5 \text{ m/s}
  • The position and velocity at t = 10.0 s: \vec{r}(10.0 \text{ s}) = (216.0 \hat{i} - 88.0 \hat{j}) \text{ m} and \vec{v}(10.0 \text{ s}) = (24.1 \hat{i} - 6.5 \hat{j}) \text{ m/s}.

4.3 Projectile Motion

  • Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration due to gravity.
  • Applications in physics and engineering are numerous (meteors, fireworks, sports balls).
  • Such objects are called projectiles and their path is called a trajectory.
  • Motion Along a Straight Line is a simple one-dimensional type of projectile motion (no horizontal movement).
  • Two-dimensional projectile motion, the effects of air resistance are neglected.
  • Motions along perpendicular axes are independent and can be analyzed separately.

Components of Projectile Motion

  • Motion is broken into two motions: one along the horizontal axis and the other along the vertical axis.
  • The horizontal axis is the x-axis and the vertical axis is the y-axis.
  • \vec{s} is the total displacement, and \vec{x} and \vec{y} are component vectors along the horizontal and vertical axes, respectively.
  • The magnitudes of these vectors are s, x, and y.
  • The components of acceleration are: ay = -g = -9.8 \text{ m/s}^2 (-32 ft/s²) and ax = 0.
  • If ax = 0, then the initial velocity in the x-direction is equal to the final velocity in the x-direction, or vx = v_{0x}.

Equations for Projectile Motion

  • Horizontal Motion:
    • v{0x} = vx, x = x0 + vx t
  • Vertical Motion:
    • y = y0 + \frac{1}{2} (v{0y} + v_y)t
    • vy = v{0y} - gt
    • y = y0 + v{0y} t - \frac{1}{2} gt^2
    • vy^2 = v{0y}^2 - 2g(y - y_0)

Problem-Solving Strategy: Projectile Motion

  1. Resolve the motion into horizontal and vertical components along the x- and y-axes.
  2. Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical.
  3. Solve for the unknowns in the two separate motions: one horizontal and one vertical (time t is the only common variable between the motions).
  4. Recombine quantities in the horizontal and vertical directions to find the total displacement \vec{s} and velocity \vec{v}. Solve for the magnitude and direction of the displacement and velocity using s = \sqrt{x^2 + y^2}, \theta = \tan^{-1}(y/x), v = \sqrt{vx^2 + vy^2}, where \theta is the direction of the displacement \vec{s}.

Example 4.7: A Fireworks Projectile Explodes High and Away: A shell is shot with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal.

  • The height at which the shell explodes, is calculated as y = 233 \text{ m}.
    • y = \frac{v_{0y}^2}{2g}
  • The component of the initial velocity in the y direction is determined by: v{0y} = v0 \sin \theta_0 = (70.0 \text{ m/s}) \sin 75° = 67.6 \text{ m/s}
  • The time between the launch of the shell and the explosion, is calculated as t = 6.90 \text{ s}.
    • t = \frac{v_{0y}}{g} = \frac{67.6 \text{ m/s}}{9.80 \text{ m/s}^2} = 6.90 \text{ s}
  • The horizontal displacement of the shell, x = 125 \text{ m}.
    • x = vx t, where vx = v_0 \cos \theta = (70.0 \text{ m/s})\cos 75° = 18.1 \text{ m/s}.
  • The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point:
    • \vec{s} = 125 \hat{i} + 233 \hat{j}
    • |\vec{s} | = \sqrt{125^2 + 233^2} = 264 \text{ m}
    • \theta = \tan^{-1} (\frac{233}{125}) = 61.8°
  • The maximum height of a projectile above its launch position: h = \frac{v_{0y}^2}{2g}.

Example 4.8: Calculating Projectile Motion: Tennis Player: A tennis player hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal. The ball is caught by a spectator 10 m above the point where the ball was hit.

  • The time it takes the tennis ball to reach the spectator, t = 3.79 \text{ s}.
    • Using y = y0 + v{0y} t - \frac{1}{2} gt^2, where y_0 = 0 and y = 10 \text{ m},
    • v{0y} = v0 \sin \theta_0 = (30.0 \text{ m/s}) \sin 45° = 21.2 \text{ m/s},
    • 10.0 \text{ m} = (21.2 \text{ m/s}) t - (4.90 \text{ m/s}^2) t^2.
  • The final horizontal and vertical velocities vx and vy with the use of the result is:
    • Resulting in the quadratic equation, (4.90 \text{ m/s}^2) t^2 - (21.2 \text{ m/s}) t + 10.0 \text{ m} = 0.
  • The magnitude of the final velocity \vec{v} is
    • v = \sqrt{vx^2 + vy^2} = \sqrt{(21.2 \text{ m/s})^2 + (-15.9 \text{ m/s})^2} = 26.5 \text{ m/s}.
    • The final vertical velocity is given by vy = v{0y} - gt.
      • Therefore, v_y = 21.2 \text{ m/s} - 9.8 \text{ m/s}^2(3.79 \text{ s}) = -15.9 \text{ m/s}.
      • Since vx is constant, vx = v0 \cos \theta0 = (30 \text{ m/s}) \cos 45° = 21.2 \text{ m/s}.
  • The direction \theta_v is found using the inverse tangent:
    • \thetav = \tan^{-1}(\frac{vy}{v_x}) = \tan^{-1} (\frac{-15.9}{21.2}) = -53.1°.

Time of Flight, Trajectory, and Range

  • Time of flight of a projectile that is both launched and impacts on a flat horizontal surface:
    • T{\text{tof}} = \frac{2(v0 \sin \theta_0)}{g}.
  • The trajectory with t = \frac{x}{v0 \cos \theta0} is:
    • y = (v0 \sin \theta0)(\frac{x}{v0 \cos \theta0}) - \frac{1}{2} g(\frac{x}{v0 \cos \theta0})^2.
    • y = (\tan \theta0)x - [\frac{g}{2(v0 \cos \theta_0)^2}]x^2.
  • Finding the range:
    • y = x[\tan \theta0 - \frac{g}{2(v0 \cos \theta_0)^2}x].
  • R = \frac{v0^2 \sin 2\theta0}{g}.

Example 4.9: Comparing Golf Shots: A golfer is in two different situations. On the second hole he is 120 m from the green and wants to hit the ball 90 m and let it run onto the green angling the shot low to the ground at 30° to the horizontal, letting the ball roll after impact. On the fourth hole he is 90 m from the green and wants to let the ball drop with a minimum amount of rolling after impact angling the shot at 70° to the horizontal, minimizing rolling after impact.

  • For the shot at 30°:
    • R = \frac{v0^2 \sin 2\theta0}{g} \Rightarrow v0 = \sqrt{\frac{Rg}{\sin 2\theta0}} = \sqrt{\frac{90.0 \text{ m} (9.8 \text{ m/s}^2)}{\sin(2(70°))}} = 37.0 \text{ m/s}
  • For the shot at 70°:
    • R = \frac{v0^2 \sin 2\theta0}{g} \Rightarrow v0 = \sqrt{\frac{Rg}{\sin 2\theta0}} = \sqrt{\frac{90.0 \text{ m} (9.8 \text{ m/s}^2)}{\sin(2(30°))}} = 31.9 \text{ m/s}
  • Trajectory equations:
    • y = x[\tan \theta0 - \frac{g}{2(v0 \cos \theta_0)^2}x]
    • Second hole: y = x[\tan 70° - \frac{9.8 \text{ m/s}^2}{2[(37.0 \text{ m/s})(\cos 70°)]^2}x]
    • Fourth hole: y = x[\tan 30° - \frac{9.8 \text{ m/s}^2}{2[(31.9 \text{ m/s})(\cos 30°)]^2}x]

4.4 Uniform Circular Motion

  • Uniform circular motion occurs when an object travels in a circle with a constant speed.

Centripetal Acceleration

  • Even with constant speed, a particle can have acceleration if it moves along a curved trajectory like a circle.
  • The velocity vector is changing, or \frac{d\vec{v}}{dt} \neq 0.
  • In time \Delta t, a particle moves counterclockwise from \vec{r}(t) to \vec{r}(t + \Delta t).
  • Velocity vector has constant magnitude and is tangent to the path as it changes from \vec{v}(t) to \vec{v}(t + \Delta t).
  • Magnitude of the acceleration can be found from a = \lim{\Delta t \to 0} (\frac{\Delta v}{\Delta t}) = \frac{v}{r} (\lim{\Delta t \to 0} \frac{\Delta r}{\Delta t}) = \frac{v^2}{r}.
  • The direction of the acceleration is toward the center of the circle.
  • A particle moving in a circle at a constant speed has an acceleration with magnitude a_C = \frac{v^2}{r}.

Example 4.10: Creating an Acceleration of 1 g: A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. The radius of the circle that will produce a centripetal acceleration of 1 g, r = 1835 \text{ m} = 1.835 \text{ km}.

Equations of Motion for Uniform Circular Motion

  • A particle executing circular motion can be described by its position vector \vec{r}(t) = A \cos \omega t \hat{i} + A \sin \omega t \hat{j}.
  • \omega is angular frequency (radians/second), the number of radians of angular measure through which the particle passes per second.
  • For periodic motion T (time to complete one revolution = 2π rad):
    • \omega = \frac{2\pi}{T}
  • Velocity and acceleration can be obtained from the position function by differentiation:
    • \vec{v}(t) = \frac{d\vec{r}(t)}{dt} = -A \omega \sin \omega t \hat{i} + A \omega \cos \omega t \hat{j}
    • \vec{a}(t) = \frac{d\vec{v}(t)}{dt} = -A \omega^2 \cos \omega t \hat{i} - A \omega^2 \sin \omega t \hat{j}.

Example 4.11: Circular Motion of a Proton: A proton has speed 5 \times 10^6 \text{ m/s} and is moving in a circle in the xy plane of radius r = 0.175 \text{ m}. Its position in the xy plane at time t = 2.0 \times 10^{-7} \text{ s} = 200 \text{ ns} is evaluated as follows:
* T = \frac{2\pi r}{v} = \frac{2\pi(0.175 \text{ m})}{5.0 \times 10^6 \text{ m/s}} = 2.20 \times 10^{-7} \text{ s}
* \omega = \frac{2\pi}{T} = \frac{2\pi}{2.20 \times 10^{-7} \text{ s}} = 2.856 \times 10^7 \text{ rad/s}
* \vec{r}(2.0 \times 10^{-7} \text{ s}) = A \cos \omega (2.0 \times 10^{-7} \text{ s}) \hat{i} + A \sin \omega (2.0 \times 10^{-7} \text{ s}) \hat{j} \text{ m}
* = 0.175 \cos[(2.856 \times 10^7 \text{ rad/s})(2.0 \times 10^{-7} \text{ s})] \hat{i} + 0.175 \sin[(2.856 \times 10^7 \text{ rad/s})(2.0 \times 10^{-7} \text{ s})] \hat{j} \text{ m}
* = 0.175 \cos(5.712 \text{ rad}) \hat{i} + 0.175 \sin(5.712 \text{ rad}) \hat{j} = 0.147 \hat{i} - 0.095 \hat{j} \text{ m}

Nonuniform Circular Motion

  • Circular motion does not have to be at a constant speed.
  • A particle can travel in a circle and speed up or slow down.
  • Introducing an additional acceleration in the direction tangential to the circle.
  • Centripetal acceleration is the time rate of change of the direction of the velocity vector.
  • Tangential acceleration is the time rate of change of the magnitude of the velocity: a_T = \frac{d|\vec{v}|}{dt}.
  • Total