Chapter 3: Motion in a Plane - Detailed Notes

Introduction

Chapter focuses on motion in two (plane) and three (space) dimensions.
Utilizes vectors to describe physical quantities like position, displacement, velocity, and acceleration.
Explores vector operations: addition, subtraction, and multiplication.
Discusses motion with constant acceleration and projectile motion as simple cases.
Examines uniform circular motion and its significance.
Notes that equations can be extended to three dimensions.

Scalars and Vectors

Scalar Quantity: Quantity with magnitude only, specified by a single number and unit.
- Examples: distance, mass, temperature, time.
- Follows ordinary algebra rules for combination (addition, subtraction, multiplication, division).
- Perimeter of a rectangle: $1.0 m + 0.5 m + 1.0 m + 0.5 m = 3.0 m$
- Temperature difference: $35.6 °C - 24.2 °C = 11.4 °C$
- Density of aluminum cube: $\frac{2.7 kg}{10^{-3} m^3} = 2.7 \times 10^3 kg/m^3$
Vector Quantity: Quantity with both magnitude and direction; obeys the triangle or parallelogram law of addition.
- Examples: displacement, velocity, acceleration, force.
- Represented by bold face type (v) or an arrow over a letter ($\vec{v}$).
- Magnitude is the absolute value: $|v| = v$

Position and Displacement Vectors

Position Vector:
- Describes the position of an object in a plane relative to an origin (O).
- Represented by a straight line from the origin to the object's position (P) with an arrow at the head ($\vec{OP} = \vec{r}$).
- Vector length represents magnitude, direction indicates position relative to O.
Displacement Vector:
- Represents the change in position as an object moves from P to P' ($\vec{PP'}$).
- Straight line joining initial and final positions, independent of the path taken.
- Magnitude is less than or equal to the actual path length.
- Example paths: PABCQ, PDQ, PBEFQ all have same $\vec{PQ}$.

Equality of Vectors

Two vectors (A and B) are equal if they have the same magnitude and direction ( $A = B$ ).
Equality can be checked by shifting one vector parallel to itself until its tail coincides with the other's; if the tips also coincide, they are equal.
Vectors do not have fixed locations and can be displaced parallel to themselves without changing the vector (free vectors).
In some applications location is important (localized vectors).

Multiplication of Vectors by Real Numbers

Multiplying a vector A by a positive number $\lambda$ changes the magnitude by a factor of $\lambda$ but preserves the direction.
- $|\lambda A| = \lambda |A|$ if \lambda > 0
- Example: Multiplying A by 2 results in 2A, which is twice the magnitude of A in the same direction.
Multiplying A by a negative number $- \lambda$ results in a vector with magnitude $\lambda |A|$ and direction opposite to A.
- Example: Multiplying A by -1 or -1.5 reverses its direction and scales the magnitude.
The factor $\lambda$ can have physical dimensions.
- Example: Multiplying a constant velocity vector by duration (time) yields a displacement vector.

Addition and Subtraction of Vectors - Graphical Method

Vectors obey the triangle law or parallelogram law of addition.

Head-to-Tail Method (Triangle Method)

To add vectors A and B, place the tail of B at the head of A.
The vector from the tail of A to the head of B represents the sum R = A + B.
Vector addition is commutative: $A + B = B + A$
Vector addition is associative: $(A + B) + C = A + (B + C)$

Null Vector

The sum of two equal and opposite vectors is a null vector (0).
- $A + (-A) = 0$
- $|0| = 0$
A null vector has zero magnitude and undefined direction.
Properties of the null vector:
- $A + 0 = A$
- $\lambda 0 = 0$
- $0 A = 0$
Physical meaning: An object moving from P to P' and back to P has a null displacement vector.

Subtraction of Vectors

Defined as adding the negative of a vector: $A - B = A + (-B)$

Parallelogram Method

Place tails of vectors A and B at a common origin O.
Draw lines parallel to A and B to complete a parallelogram OQSP.
The resultant vector R is directed along the diagonal OS from the origin.
Equivalent to the triangle method.

Example 3.1

Rain falling vertically at $35 m/s$ , wind blowing east to west at $12 m/s$ .
Resultant velocity R: $R = \sqrt{vr^2 + vw^2} = \sqrt{35^2 + 12^2} = 37 m/s$
Direction $\theta$ from vertical: $\theta = \tan^{-1}(\frac{vw}{vr}) = \tan^{-1}(\frac{12}{35}) = 19°$
Boy should hold umbrella at 19 degrees to the vertical towards the east.

Resolution of Vectors

Any vector A in a plane can be expressed as a sum of two vectors along two non-zero, non-collinear vectors a and b: $A = \lambda a + \mu b$
$\lambda $and$ \mu $are real numbers.</li></ul><h4 id="6ccbf88a-af99-4c0a-967e-79d9b171beca" data-toc-id="6ccbf88a-af99-4c0a-967e-79d9b171beca" collapsed="false" seolevelmigrated="true">Unit Vectors</h4><ul><li>A vector with unit magnitude pointing in a specific direction.</li><li>Dimensionless and unitless, used only to specify direction.</li><li>Unit vectors along x, y, and z axes are denoted as$ \hat{i} $,$ \hat{j} $, and$ \hat{k} $respectively.<ul><li>$ |\hat{i}| = |\hat{j}| = |\hat{k}| = 1 $</li></ul></li><li>A vector A can be written as$ A = |A| \hat{n} $, where$ \hat{n} $is the unit vector along A.</li><li>A vector A in the x-y plane can be resolved into components using unit vectors$ \hat{i} $and$ \hat{j} $:<ul><li>$ A = Ax \hat{i} + Ay \hat{j} $</li><li>$ Ax $and$ Ay $are the x- and y-components of A, respectively.</li><li>$ A_x = A \cos \theta $</li><li>$ A_y = A \sin \theta $</li></ul></li><li>A component of a vector can be positive, negative, or zero, depending on$ \theta $.</li></ul><h5 id="86b920ce-0b4b-4343-ace0-410ec3ee5d2c" data-toc-id="86b920ce-0b4b-4343-ace0-410ec3ee5d2c" collapsed="false" seolevelmigrated="true">Specifying a Vector</h5><ul><li>By its magnitude (A) and direction ($ \theta $).</li><li>By its components ($ Ax $and$ Ay $).</li><li>If$ Ax $and$ Ay $are given:<ul><li>$ A = \sqrt{Ax^2 + Ay^2} $</li><li>$ \theta = \tan^{-1} \frac{Ay}{Ax} $</li></ul></li><li>In three dimensions, a vector A can be resolved into three components along x, y, and z axes:<ul><li>$ A_x = A \cos \alpha $</li><li>$ A_y = A \cos \beta $</li><li>$ A_z = A \cos \gamma $</li><li>$ A = Ax \hat{i} + Ay \hat{j} + A_z \hat{k} $</li><li>$ A = \sqrt{Ax^2 + Ay^2 + A_z^2} $</li></ul></li><li>Position vector r can be expressed as$ r = x \hat{i} + y \hat{j} + z \hat{k} $, where x, y, and z are components along x, y, and z axes.</li></ul><h3 id="1e1f0c47-cfae-4230-af71-fa72aca8fece" data-toc-id="1e1f0c47-cfae-4230-af71-fa72aca8fece" collapsed="false" seolevelmigrated="true">Vector Addition – Analytical Method</h3><ul><li>Easier and more accurate than the graphical method.</li><li>Consider two vectors A and B in the x-y plane:<ul><li>$ A = Ax \hat{i} + Ay \hat{j} $</li><li>$ B = Bx \hat{i} + By \hat{j} $</li></ul></li><li>Their sum is R = A + B:<ul><li>$ R = (Ax + Bx) \hat{i} + (Ay + By) \hat{j} $</li><li>$ Rx = Ax + B_x $</li><li>$ Ry = Ay + B_y $</li></ul></li><li>In three dimensions:<ul><li>$ A = Ax \hat{i} + Ay \hat{j} + A_z \hat{k} $</li><li>$ B = Bx \hat{i} + By \hat{j} + B_z \hat{k} $</li><li>$ R = Rx \hat{i} + Ry \hat{j} + R_z \hat{k} $</li><li>$ Rx = Ax + B_x $</li><li>$ Ry = Ay + B_y $</li><li>$ Rz = Az + B_z $</li></ul></li><li>For multiple vectors, e.g., T = a + b – c:<ul><li>$ Tx = ax + bx - cx $</li><li>$ Ty = ay + by - cy $</li><li>$ Tz = az + bz - cz $</li></ul></li></ul><h5 id="1129e223-ade9-457b-b648-fb572d14756e" data-toc-id="1129e223-ade9-457b-b648-fb572d14756e" collapsed="false" seolevelmigrated="true">Example 3.2</h5><ul><li>Finding magnitude and direction of resultant of two vectors A and B with angle$ \theta $between them.</li><li>Magnitude:$ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} $</li><li>Direction:<ul><li>$ \frac{R}{\sin \theta}= \frac{B}{\sin \alpha} $</li><li>$ \sin \alpha = \frac{B \sin \theta}{R} $</li><li>$ \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} $</li></ul></li><li>Law of Cosines:$ R^2 = A^2 + B^2 + 2AB \cos \theta $</li><li>Law of Sines:$ \frac{R}{\sin \theta} = \frac{A}{\sin \beta} = \frac{B}{\sin \alpha} $</li></ul><h5 id="57b8315c-dfbe-4721-9c97-73d29a149ddf" data-toc-id="57b8315c-dfbe-4721-9c97-73d29a149ddf" collapsed="false" seolevelmigrated="true">Example 3.3</h5><ul><li>Motorboat racing north at$ 25 km/h $, water current at$ 10 km/h $at$ 60° $east of south.</li><li>Resultant velocity:$ R = \sqrt{25^2 + 10^2 + 2 \times 25 \times 10 \cos 120°} \approx 22 km/h $</li><li>Direction:$ \sin \phi = \frac{10 \sin 120°}{21.8} \approx 0.397 $, so$ \phi \approx 23.4° $</li></ul><h3 id="1e4c0592-5700-424f-b1d8-4a68b786be9a" data-toc-id="1e4c0592-5700-424f-b1d8-4a68b786be9a" collapsed="false" seolevelmigrated="true">Motion in a Plane</h3><h4 id="b0573120-5d86-4370-9137-09398ab7332f" data-toc-id="b0573120-5d86-4370-9137-09398ab7332f" collapsed="false" seolevelmigrated="true">Position Vector and Displacement</h4><ul><li>Position vector:$ r = x \hat{i} + y \hat{j} $, where x and y are coordinates of the object.</li><li>Displacement:$ \Delta r = r' - r = (x' - x) \hat{i} + (y' - y) \hat{j} = \Delta x \hat{i} + \Delta y \hat{j} $where$ \Delta x = x' - x $and$ \Delta y = y' - y $</li></ul><h4 id="f45d26a0-1bc7-406d-81ce-be7f11b745bf" data-toc-id="f45d26a0-1bc7-406d-81ce-be7f11b745bf" collapsed="false" seolevelmigrated="true">Velocity</h4><ul><li>Average velocity:$ \vec{v} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\Delta x}{\Delta t} \hat{i} + \frac{\Delta y}{\Delta t} \hat{j} $</li><li>Instantaneous velocity:<ul><li>$ \vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d \vec{r}}{dt} $</li><li>Direction is tangent to the path of the object.</li><li>$ \vec{v} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} = vx \hat{i} + vy \hat{j} $</li><li>Magnitude:$ v = \sqrt{vx^2 + vy^2} $</li><li>Direction:$ \theta = \tan^{-1} (\frac{vy}{vx}) $</li></ul></li></ul><h4 id="b6b16430-d4af-435e-a569-8ef35193a850" data-toc-id="b6b16430-d4af-435e-a569-8ef35193a850" collapsed="false" seolevelmigrated="true">Acceleration</h4><ul><li>Average acceleration:$ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\Delta vx}{\Delta t} \hat{i} + \frac{\Delta vy}{\Delta t} \hat{j} = ax \hat{i} + ay \hat{j} $</li><li>Instantaneous acceleration:<ul><li>$ \vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d \vec{v}}{dt} $</li><li>$ \vec{a} = \frac{dvx}{dt} \hat{i} + \frac{dvy}{dt} \hat{j} = ax \hat{i} + ay \hat{j} $</li><li>$ ax = \frac{dvx}{dt}, ay = \frac{dvy}{dt} $</li></ul></li></ul><h5 id="2534a534-fa04-4f2a-bb7d-7477daabace2" data-toc-id="2534a534-fa04-4f2a-bb7d-7477daabace2" collapsed="false" seolevelmigrated="true">Example 3.4</h5><ul><li>Position of a particle:$ \vec{r} = 3.0t \hat{i} - 2.0t^2 \hat{j} + 5.0 \hat{k} $</li><li>Velocity:$ \vec{v(t)} = \frac{d \vec{r}}{dt} = 3.0 \hat{i} - 4.0t \hat{j} $</li><li>Acceleration:$ \vec{a(t)} = \frac{d \vec{v}}{dt} = -4.0 \hat{j} $</li><li>At t = 1.0 s,$ \vec{v} = 3.0 \hat{i} - 4.0 \hat{j} $<ul><li>Magnitude:$ v = \sqrt{3^2 + (-4)^2} = 5.0 m/s $</li><li>Direction:$ \theta = \tan^{-1} (\frac{-4}{3}) \approx -53° $</li></ul></li></ul><h3 id="a61ba370-3770-404f-9dc3-c37e472c040c" data-toc-id="a61ba370-3770-404f-9dc3-c37e472c040c" collapsed="false" seolevelmigrated="true">Motion in a Plane with Constant Acceleration</h3><ul><li>Constant acceleration:$ \vec{a} $is constant.</li><li>Velocity at time t:$ \vec{v} = \vec{v_0} + \vec{a}t $<ul><li>$ vx = v{0x} + a_x t $</li><li>$ vy = v{0y} + a_y t $</li></ul></li><li>Position at time t:$ \vec{r} = \vec{r0} + \vec{v0}t + \frac{1}{2} \vec{a}t^2 $<ul><li>$ x = x0 + v{0x}t + \frac{1}{2} a_x t^2 $</li><li>$ y = y0 + v{0y}t + \frac{1}{2} a_y t^2 $</li></ul></li><li>Motion in x- and y-directions can be treated independently.</li></ul><h5 id="672e78b8-4cd9-4946-b7e1-a52981630097" data-toc-id="672e78b8-4cd9-4946-b7e1-a52981630097" collapsed="false" seolevelmigrated="true">Example 3.5</h5><ul><li>Particle starts at origin with velocity$ 5.0 \hat{i} m/s $, acceleration$ (3.0 \hat{i} + 2.0 \hat{j}) m/s^2 $</li><li>Position:$ \vec{r} = 5.0t \hat{i} + \frac{1}{2}(3.0 \hat{i} + 2.0 \hat{j})t^2 = (5.0t + 1.5t^2) \hat{i} + t^2 \hat{j} $</li><li>x-coordinate is 84 m:$ 5.0t + 1.5t^2 = 84 \implies t = 6 s $</li><li>y-coordinate at t = 6 s:$ y = (6)^2 = 36.0 m $</li><li>Velocity at t = 6 s:$ \vec{v} = (5.0 + 3.0 \times 6) \hat{i} + (2.0 \times 6) \hat{j} = 23.0 \hat{i} + 12.0 \hat{j} $</li><li>Speed at t = 6 s:$ v = \sqrt{23^2 + 12^2} \approx 26 m/s $</li></ul><h3 id="df1c202f-947d-4f12-b3b2-8c72369b048d" data-toc-id="df1c202f-947d-4f12-b3b2-8c72369b048d" collapsed="false" seolevelmigrated="true">Projectile Motion</h3><ul><li>Object in flight after being thrown or projected is a projectile.</li><li>Motion can be thought of as two separate components: horizontal (no acceleration) and vertical (constant acceleration due to gravity).</li><li>Air resistance is neglected.</li></ul><ul><li>Projectile launched with velocity$ v0 $at angle$ \theta0 $with x-axis.<ul><li>Acceleration:$ ax = 0, ay = -g $</li><li>Initial velocity components:$ v{0x} = v0 \cos \theta0, v{0y} = v0 \sin \theta0 $</li></ul></li><li>Position at time t:<ul><li>$ x = (v0 \cos \theta0)t $</li><li>$ y = (v0 \sin \theta0)t - \frac{1}{2}gt^2 $</li></ul></li><li>Velocity components at time t:<ul><li>$ vx = v0 \cos \theta_0 $</li><li>$ vy = v0 \sin \theta_0 - gt $</li></ul></li></ul><ul><li>Position at time t:<ul><li>$ x = (v0 \cos \theta0)t $</li><li>$ y = (v0 \sin \theta0)t - \frac{1}{2}gt^2 $</li></ul></li><li>Velocity components at time t:<ul><li>$ vx = v0 \cos \theta_0 $</li><li>$ vy = v0 \sin \theta_0 - gt $</li></ul></li></ul><h4 id="346e558e-aa82-4794-9f88-2aaba1761038" data-toc-id="346e558e-aa82-4794-9f88-2aaba1761038" collapsed="false" seolevelmigrated="true">Equation of Path</h4><ul><li>By eliminating time:<ul><li>$ y = x \tan \theta0 - \frac{g}{2(v0 \cos \theta_0)^2} x^2 $</li></ul></li><li>Equation of a parabola.</li></ul><h4 id="79b75ede-a19e-432f-96a1-178343712e3d" data-toc-id="79b75ede-a19e-432f-96a1-178343712e3d" collapsed="false" seolevelmigrated="true">Time of Maximum Height</h4><ul><li>At maximum height,$ v_y = 0 $</li><li>$ tm = \frac{v0 \sin \theta_0}{g} $</li></ul><h4 id="f37cc248-6e68-427b-a9b6-811ad44c01bc" data-toc-id="f37cc248-6e68-427b-a9b6-811ad44c01bc" collapsed="false" seolevelmigrated="true">Total Time of Flight</h4><ul><li>Time during which the projectile is in flight:<ul><li>$ Tf = \frac{2(v0 \sin \theta0)}{g} = 2tm $</li></ul></li></ul><h4 id="fe3f61fe-183a-4c72-932b-231beaf681c5" data-toc-id="fe3f61fe-183a-4c72-932b-231beaf681c5" collapsed="false" seolevelmigrated="true">Maximum Height</h4><ul><li>$ hm = \frac{(v0 \sin \theta_0)^2}{2g} $</li></ul><h4 id="e0262f5d-e4e7-402f-8609-105127e4d7b2" data-toc-id="e0262f5d-e4e7-402f-8609-105127e4d7b2" collapsed="false" seolevelmigrated="true">Horizontal Range</h4><ul><li>Horizontal distance traveled by the projectile:<ul><li>$ R = \frac{v0^2 \sin 2\theta0}{g} $</li></ul></li><li>Maximum range when$ \theta_0 = 45° $:<ul><li>$ Rm = \frac{v0^2}{g} $</li></ul></li></ul><h5 id="900bed15-d87b-4f0e-b1e7-99e22d325152" data-toc-id="900bed15-d87b-4f0e-b1e7-99e22d325152" collapsed="false" seolevelmigrated="true">Example 3.6</h5><ul><li>Galileo’s statement: Ranges are equal for elevations exceeding or falling short of 45° by equal amounts.</li><li>For angles (45° + α) and (45° – α), the ranges are equal because$ \sin(90° + 2\alpha) = \sin(90° - 2\alpha) = \cos 2\alpha $</li></ul><h5 id="f646b839-46ed-4511-9822-6c7caeb9ec48" data-toc-id="f646b839-46ed-4511-9822-6c7caeb9ec48" collapsed="false" seolevelmigrated="true">Example 3.7</h5><ul><li>Hiker throws a stone horizontally from a cliff 490 m above the ground with$ v_0 = 15 m/s $</li><li>$ x(t) = v{0x}t, y(t) = y0 + v{0y}t + (1/2) ay t^2 $</li><li>Time to reach the ground:<ul><li>$ -490 = -\frac{1}{2}(9.8)t^2 \implies t = 10 s $</li></ul></li><li>Velocity components when hitting the ground:<ul><li>$ v_x = 15 m/s $</li><li>$ v_y = -9.8 \times 10 = -98 m/s $</li></ul></li><li>Speed when hitting the ground:<ul><li>$ v = \sqrt{15^2 + (-98)^2} \approx 99 m/s $</li></ul></li></ul><h5 id="d3395d80-d364-4982-b7ea-97fddfd1243c" data-toc-id="d3395d80-d364-4982-b7ea-97fddfd1243c" collapsed="false" seolevelmigrated="true">Example 3.8</h5><ul><li>Cricket ball thrown at$ 28 m/s $at 30° above the horizontal.</li><li>Maximum height:<ul><li>$ h_m = \frac{(28 \sin 30°)^2}{2 \times 9.8} \approx 10.0 m $</li></ul></li><li>Time to return to the same level:<ul><li>$ T_f = \frac{2 \times 28 \times \sin 30°}{9.8} \approx 2.9 s $</li></ul></li><li>Horizontal distance:<ul><li>$ R = \frac{28^2 \sin 60°}{9.8} \approx 69 m $</li></ul></li></ul><h3 id="928844ea-d8d4-40ca-a9a3-ad7170ffb4b6" data-toc-id="928844ea-d8d4-40ca-a9a3-ad7170ffb4b6" collapsed="false" seolevelmigrated="true">Uniform Circular Motion</h3><ul><li>Object follows a circular path at a constant speed.</li><li>Speed is uniform but direction changes.</li><li>Acceleration is always directed towards the center.</li><li>Acceleration is called centripetal acceleration<ul><li>Magnitude:$ a = v^2/R $</li></ul></li><li>Angular speed:$ \omega = \frac{\Delta \theta}{\Delta t} $</li><li>Relationship between linear and angular speed:$ v = R \omega $</li><li>Centripetal acceleration in terms of angular speed:$ a_c = \omega^2 R
Time period (T) is the time to make one revolution, frequency ($\nu) is revolutions per second ( $\nu = 1/T$ ).
$v = \frac{2 \pi R}{T} = 2 \pi R \nu$
$\omega = 2 \pi \nu$
$a_c = 4 \pi^2 \nu^2 R$

Example 3.9

Insect in circular groove of radius 12 cm, completes 7 revolutions in 100 s.
Angular speed: $\omega = \frac{2 \pi \times 7}{100} \approx 0.44 rad/s$
Linear speed: $v = 0.44 \times 12 \approx 5.3 cm/s$
Acceleration is not a constant vector, but its magnitude is constant:
- $a = (0.44)^2 \times 12 \approx 2.3 cm/s^2$