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Chapter 3 Motion In A Plane
Introduction
Describing motion in two or three dimensions requires accounting for the direction of physical quantities like position, displacement, velocity, and acceleration. Unlike one-dimensional motion where direction can be indicated by simple positive and negative signs, motion in a plane or space necessitates the use of vectors. This chapter introduces the concept of vectors and vector algebra to describe motion in two dimensions (a plane).
The chapter will cover vector operations (addition, subtraction, multiplication by scalars), the description of position, displacement, velocity, and acceleration using vectors in a plane, motion with constant acceleration in a plane (including projectile motion), and uniform circular motion.
Scalars And Vectors
Physical quantities can be classified into two types:
- Scalars: Quantities that have magnitude only. They are completely specified by a single number with the appropriate unit. Examples: distance, mass, temperature, time, volume, density. Scalars are combined using ordinary algebraic rules.
- Vectors: Quantities that have both magnitude and direction. They must obey the triangle law or parallelogram law of vector addition. Examples: displacement, velocity, acceleration, force. Vectors are represented using boldface type (e.g., v) or an arrow over the symbol ($\vec{v}$). The magnitude of a vector is its absolute value (e.g., $|\mathbf{v}| = v$).
Position And Displacement Vectors
In a coordinate system with origin O, the position vector (r) of a point P is the vector from the origin O to the point P (OP). It indicates the location of the object at a specific time. As an object moves from position P (at time t, position vector r) to position P' (at time t', position vector r'), its change in position is described by the displacement vector ($\Delta \mathbf{r}$). The displacement vector is the straight line vector from the initial position P to the final position P' (PP').
$\Delta \mathbf{r} = \mathbf{r}' - \mathbf{r}$
The magnitude of the displacement vector is the shortest distance between the initial and final points. The path length (actual distance traveled) between two points can be equal to or greater than the magnitude of the displacement. The displacement vector depends only on the initial and final positions, not the path taken.
Equality Of Vectors
Two vectors, A and B, are considered equal if and only if they have the same magnitude and the same direction. Vectors in our study are often considered "free vectors," meaning their location in space does not define the vector itself; a vector can be moved parallel to itself without changing its identity.
Multiplication Of Vectors By Real Numbers
Multiplying a vector A by a real number $\lambda$ results in a new vector ($\lambda$A) with the following properties:
- The magnitude of the new vector is the magnitude of A multiplied by the absolute value of $\lambda$: $|\lambda\mathbf{A}| = |\lambda||\mathbf{A}|$.
- The direction of the new vector is the same as A if $\lambda$ is positive ($\lambda > 0$).
- The direction of the new vector is opposite to A if $\lambda$ is negative ($\lambda < 0$).
If the scalar $\lambda$ has dimensions, the dimensions of the resulting vector are the product of the dimensions of $\lambda$ and A.
Addition And Subtraction Of Vectors — Graphical Method
Vectors obey specific laws of addition and subtraction, distinct from scalar arithmetic.
Vector Addition:
- Triangle law (Head-to-tail method): To add two vectors A and B, place the tail of vector B at the head of vector A. The sum, or resultant vector R, is the vector drawn from the tail of A to the head of B. ($\mathbf{R} = \mathbf{A} + \mathbf{B}$).
- Parallelogram law: To add two vectors A and B, place their tails at a common origin. Draw a parallelogram with A and B as adjacent sides. The resultant vector R is the vector drawn from the common origin along the diagonal of the parallelogram. ($\mathbf{R} = \mathbf{A} + \mathbf{B}$). The parallelogram method is equivalent to the triangle method.
Properties of Vector Addition:
- Commutative Law: The order of addition does not matter: $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$.
- Associative Law: Grouping of vectors does not matter when adding more than two vectors: $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$.
- Null Vector (Zero Vector, 0): A vector with zero magnitude. $\mathbf{A} + \mathbf{0} = \mathbf{A}$, $\lambda \mathbf{0} = \mathbf{0}$, $0 \mathbf{A} = \mathbf{0}$. The direction of a null vector is undefined. It represents situations like zero displacement when the initial and final positions coincide.
Vector Subtraction:
Subtracting a vector B from a vector A is defined as adding A to the negative of B ($\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$). The vector -B has the same magnitude as B but the opposite direction.
Example 3.1. Rain is falling vertically with a speed of 35 m s⁻¹. Winds starts blowing after sometime with a speed of 12 m s⁻¹ in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ?
Answer:
We need to find the direction of the resultant velocity of the rain relative to the ground, which is the vector sum of the velocity of the rain (relative to still air) and the velocity of the wind (which affects the air the rain falls through). Let the vertical direction (downwards) be the y-axis and the horizontal direction (East-West) be the x-axis. Let East be the positive x-direction, and West be the negative x-direction. Let Downwards be the positive y-direction, and Upwards be the negative y-direction.
Velocity of rain ($\mathbf{v}_r$) = 35 m s⁻¹ vertically downwards. So, $\mathbf{v}_r = 35 \hat{\mathbf{j}}$ (if y is downwards positive).
Let's stick to the standard convention where +y is upwards and +x is right (East). Then velocity of rain $\mathbf{v}_r = -35 \hat{\mathbf{j}}$. Velocity of wind $\mathbf{v}_w$ is 12 m s⁻¹ in east to west direction. So, $\mathbf{v}_w = -12 \hat{\mathbf{i}}$.
The resultant velocity of rain ($\mathbf{R}$) relative to the ground is the vector sum: $\mathbf{R} = \mathbf{v}_r + \mathbf{v}_w = -12 \hat{\mathbf{i}} - 35 \hat{\mathbf{j}}$.
The boy should hold his umbrella in the direction opposite to the resultant velocity of the rain, to block the rain effectively. The direction opposite to $\mathbf{R}$ is $-( -12 \hat{\mathbf{i}} - 35 \hat{\mathbf{j}}) = 12 \hat{\mathbf{i}} + 35 \hat{\mathbf{j}}$. This vector points in the East (+x) and Upwards (+y) direction. Wait, the rain is falling downwards, so the resultant velocity should have a downward component. The direction of rain relative to the ground is the direction of $\mathbf{R}$. The umbrella should point upwards and into the wind.
Let's redraw the vectors. Vertical rain speed $v_r = 35 \text{ m s}^{-1}$. Horizontal wind speed $v_w = 12 \text{ m s}^{-1}$ (Westwards). Let vertical downwards be positive Y, and horizontal Westwards be positive X.
$\mathbf{v}_r = 35 \hat{\mathbf{j}}$
$\mathbf{v}_w = 12 \hat{\mathbf{i}}$
Resultant velocity $\mathbf{R} = \mathbf{v}_w + \mathbf{v}_r = 12 \hat{\mathbf{i}} + 35 \hat{\mathbf{j}}$. This vector points in the positive X (West) and positive Y (Downwards) direction.
The magnitude of $\mathbf{R} = \sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37 \text{ m s}^{-1}$.
The direction of $\mathbf{R}$ is given by the angle $\theta$ it makes with the vertical (positive Y axis). Let $\theta$ be the angle from the +Y axis towards the +X axis.
$\tan \theta = \frac{\text{Horizontal component}}{\text{Vertical component}} = \frac{v_w}{v_r} = \frac{12}{35}$.
$\tan \theta \approx 0.3428$.
$\theta = \arctan(0.3428) \approx 18.9^\circ$.
The resultant velocity of the rain is at an angle of about 18.9° from the vertical towards the West. The boy should hold his umbrella in the direction opposite to the rain's velocity vector relative to him, which means holding it upwards at an angle of about 18.9° with the vertical towards the West.
The figure in the example solution shows the resultant R pointing West and Down. The angle $\theta$ is shown relative to the vertical. $\tan \theta = v_w / v_r = 12/35$. The solution says $\theta \approx 19^\circ$. The direction is towards the east in the solution? The wind is blowing East to West. So the rain is pushed West. The umbrella should lean into the wind, so towards the West. The figure shows R pointing West. The angle $\theta$ is shown relative to the vertical, leaning towards the West. Why does the solution say "towards the east"? Let's assume the solution figure and calculation are correct, but the final direction description is a typo.
Calculation of magnitude: $R = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37$. Correct.
Calculation of angle: $\tan \theta = 12/35$. $\theta \approx 19^\circ$. Correct.
Direction: The resultant velocity is in the direction of the wind (Westwards) and rain (Downwards). So the rain comes from the East-Up direction towards the West-Down direction. The umbrella should be tilted upwards and towards the West to intercept the rain. So the direction is an angle of about 19° with the vertical towards the West.
Let's re-read the problem statement carefully. "Winds starts blowing... in east to west direction." This means the wind blows from East TO West. So the wind velocity is towards the West. My vector $\mathbf{v}_w = -12 \hat{\mathbf{i}}$ (if East is +x) is correct if East is +x and West is -x. My figure where X is West is also consistent with my vector $12 \hat{\mathbf{i}}$. Let's assume West is positive X for the figure as drawn in the solution figure (implied). Vertical down is positive Y. Then $\mathbf{v}_r = 35 \hat{\mathbf{j}}$, $\mathbf{v}_w = 12 \hat{\mathbf{i}}$. $\mathbf{R} = 12 \hat{\mathbf{i}} + 35 \hat{\mathbf{j}}$. This is towards positive X (West) and positive Y (Down). This matches the figure and calculation. The direction is 19° from vertical towards the West.
Why the solution says "towards the east"? Let's assume the initial choice of coordinates in the solution figure was: Up is +Y, East is +X. Then rain is $\mathbf{v}_r = -35 \hat{\mathbf{j}}$. Wind is from East to West, so $\mathbf{v}_w = -12 \hat{\mathbf{i}}$. Resultant $\mathbf{R} = -12 \hat{\mathbf{i}} - 35 \hat{\mathbf{j}}$. This vector is in the West (-x) and Down (-y) direction. The angle $\theta$ from the vertical (-y axis) towards the West (-x axis) would have $\tan \theta = |-12|/|-35| = 12/35$. This angle is still towards the West relative to the downward vertical. So it seems the solution's final direction is incorrect, or there is some subtle point missed.
Based on vector addition, the resultant velocity of the rain is directed downwards and westwards. The boy should tilt his umbrella upwards and towards the west, at an angle of 19° with the vertical.
Final conclusion based on consistent vector math and diagrams: The direction is 19° with the vertical towards the West.
Let's re-check the solution wording: "...towards the east." Could "east to west direction" mean the wind *direction* is defined as East-West, and it's blowing from East towards West? Yes, that's the standard meaning. So the velocity vector is towards the West. My interpretation seems correct. The solution's final statement about direction must be a typo.
The boy should hold his umbrella at an angle of about 19° with the vertical towards the west.
Resolution Of Vectors
Any vector A lying in a plane can be uniquely expressed as the sum of two component vectors along any two given non-parallel vectors a and b lying in the same plane. This process is called resolution of a vector. If a and b are unit vectors along two perpendicular axes (e.g., x and y axes), the resolution is particularly convenient.
Unit Vector: A vector of unit magnitude (magnitude = 1) used only to specify a direction. It has no dimension or unit. Unit vectors along the positive x, y, and z axes are denoted by $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$, respectively.
$|\hat{\mathbf{i}}| = |\hat{\mathbf{j}}| = |\hat{\mathbf{k}}| = 1$
Any vector A in the x-y plane can be resolved into components along the x and y axes:
$\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}$
where $A_x$ and $A_y$ are the scalar components (real numbers). If the vector A makes an angle $\theta$ with the positive x-axis, the scalar components are given by:
$A_x = |\mathbf{A}| \cos \theta$
$A_y = |\mathbf{A}| \sin \theta$
Conversely, if the components $A_x$ and $A_y$ are known, the magnitude and direction of A can be found:
$|\mathbf{A}| = \sqrt{A_x^2 + A_y^2}$
$\tan \theta = \frac{A_y}{A_x}$ (with $\theta$ being the angle with the positive x-axis).
In three dimensions, a vector A can be resolved into three components along the x, y, and z axes using unit vectors $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$:
$\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}$
The magnitude is $|\mathbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$. $A_x = |\mathbf{A}| \cos \alpha$, $A_y = |\mathbf{A}| \cos \beta$, $A_z = |\mathbf{A}| \cos \gamma$, where $\alpha, \beta, \gamma$ are the angles A makes with the positive x, y, and z axes, respectively.
A position vector r in 3D can be written as $\mathbf{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}$, where x, y, and z are the coordinates of the point.
Vector Addition – Analytical Method
Adding vectors graphically can be tedious and imprecise. A more accurate method is to add vectors by combining their corresponding components. If vectors A and B are given in component form:
$\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}$
$\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}}$
Their sum $\mathbf{R} = \mathbf{A} + \mathbf{B}$ is found by adding the corresponding components:
$\mathbf{R} = (A_x + B_x) \hat{\mathbf{i}} + (A_y + B_y) \hat{\mathbf{j}}$
So, the components of the resultant vector R are $R_x = A_x + B_x$ and $R_y = A_y + B_y$. This can be extended to any number of vectors and to three dimensions. The magnitude and direction of the resultant vector R can then be found from its components $R_x$ and $R_y$ using the formulas from vector resolution.
Example 3.2. Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle q between them.
Answer:
Let vectors A and B have magnitudes $A$ and $B$ respectively, and the angle between them be $\theta$. We can use the parallelogram law of vector addition. Place the tails of A and B at a common origin O. The resultant vector R is the diagonal from O.
Draw a perpendicular SN from S to the line extending OP. In the triangle OSN, by Pythagorean theorem:
$OS^2 = ON^2 + SN^2$
$ON = OP + PN$. In triangle PNS, $PN = PS \cos \theta = B \cos \theta$ (since PS = B). And $SN = PS \sin \theta = B \sin \theta$.
$ON = A + B \cos \theta$.
$R^2 = (A + B \cos \theta)^2 + (B \sin \theta)^2$
$R^2 = A^2 + 2AB \cos \theta + B^2 \cos^2 \theta + B^2 \sin^2 \theta$
$R^2 = A^2 + B^2 (\cos^2 \theta + \sin^2 \theta) + 2AB \cos \theta$
Since $\cos^2 \theta + \sin^2 \theta = 1$:
$R^2 = A^2 + B^2 + 2AB \cos \theta$
The magnitude of the resultant vector R is: $\mathbf{R} = \sqrt{A^2 + B^2 + 2AB \cos \theta}$. This is the Law of Cosines for vector addition.
To find the direction of R, let $\alpha$ be the angle R makes with vector A (along OP). In triangle OSN, $\tan \alpha = \frac{SN}{ON} = \frac{B \sin \theta}{A + B \cos \theta}$.
The angle $\alpha$ gives the direction of R relative to A. Similarly, if $\beta$ is the angle R makes with B, $\tan \beta = \frac{A \sin \theta}{B + A \cos \theta}$.
From the Law of Sines applied to triangle OPS (using sides R, A, B and angles opposite to them): $\frac{R}{\sin \theta} = \frac{B}{\sin \alpha} = \frac{A}{\sin \beta}$.
The magnitude of the resultant is $\mathbf{R} = \sqrt{A^2 + B^2 + 2AB \cos \theta}$, and its direction relative to A is given by $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$.
Example 3.3. A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.
Answer:
Let the velocity of the motorboat relative to still water be $\mathbf{v}_b$, and the velocity of the water current be $\mathbf{v}_c$. The resultant velocity of the boat relative to the ground is $\mathbf{R} = \mathbf{v}_b + \mathbf{v}_c$.
$\mathbf{v}_b$ has magnitude 25 km/h, directed North.
$\mathbf{v}_c$ has magnitude 10 km/h, directed 60° East of South.
Let's set up a coordinate system with North along the positive y-axis and East along the positive x-axis.
$\mathbf{v}_b = 25 \hat{\mathbf{j}}$ km/h.
For $\mathbf{v}_c$, the direction is 60° East of South. This means 60° from the negative y-axis (South) towards the positive x-axis (East). The angle with the negative y-axis is 60°. The angle with the positive x-axis is $90^\circ - 60^\circ = 30^\circ$ from the East axis towards the South axis, or $270^\circ - 60^\circ = 210^\circ$ from the positive x-axis (measured counterclockwise).
Components of $\mathbf{v}_c$: $v_{cx} = v_c \cos(210^\circ) = 10 \cos(210^\circ) = 10 (-\cos 30^\circ) = 10 (-\frac{\sqrt{3}}{2}) = -5\sqrt{3} \approx -8.66$ km/h.
$v_{cy} = v_c \sin(210^\circ) = 10 \sin(210^\circ) = 10 (-\sin 30^\circ) = 10 (-\frac{1}{2}) = -5$ km/h.
Alternatively, consider the angle relative to South (negative y-axis). The East component is $v_c \sin 60^\circ$ (since 60° is with the South axis). The South component is $v_c \cos 60^\circ$. So, $v_{cx} = +10 \sin 60^\circ = 10 (\frac{\sqrt{3}}{2}) = 5\sqrt{3}$ km/h. $v_{cy} = -10 \cos 60^\circ = -10 (\frac{1}{2}) = -5$ km/h. Wait, the direction is 60° East of South. This means START from South, turn 60° towards East. South is along negative y-axis. Turning 60° towards East (+x). The angle relative to the positive x-axis (East) is $90^\circ - 60^\circ = 30^\circ$ South of East. So, the angle from positive x-axis is $360^\circ - 30^\circ = 330^\circ$ or $270^\circ + 30^\circ = 300^\circ$? No. It's 60° from the negative y-axis, in the quadrant where x is positive. Let's use angles relative to axes directly.
Angle of $\mathbf{v}_c$ with the positive x-axis (East): Let this angle be $\phi$. $\mathbf{v}_c$ is in the fourth quadrant. The angle with the negative y-axis (South) is 60°. The angle with the positive x-axis is $90^\circ + 60^\circ = 150^\circ$ from the negative y-axis? No.
Let's draw the vectors. North is +y, East is +x. South is -y, West is -x.
$\mathbf{v}_b$ is along +y (North).
$\mathbf{v}_c$ is 60° East of South. So, from the -y axis, rotate 60° towards the +x axis. The angle from the +x axis is $90^\circ + 60^\circ = 150^\circ$ if measured from +x towards -y? No. The angle from the -y axis (South) is 60° towards the +x axis (East). The angle from the +x axis (East) is $90^\circ - 60^\circ = 30^\circ$ South of East. So the angle from +x axis is $360^\circ - 30^\circ = 330^\circ$. Or $270^\circ + 30^\circ$? No, it's 60 degrees from South. Angle from South towards East is 60. South is -j. East is +i. So the vector is in (+i, -j) quadrant. The angle with the -y axis is 60. Components: $v_{cx} = 10 \sin 60^\circ$ (opposite side to angle with y axis) $= 10 (\sqrt{3}/2) = 5\sqrt{3}$. $v_{cy} = -10 \cos 60^\circ$ (adjacent side to angle with y axis, negative direction) $= -10 (1/2) = -5$. Correct.
$\mathbf{v}_c = (5\sqrt{3}) \hat{\mathbf{i}} - 5 \hat{\mathbf{j}}$ km/h.
Resultant velocity $\mathbf{R} = \mathbf{v}_b + \mathbf{v}_c = 25 \hat{\mathbf{j}} + (5\sqrt{3}) \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} = (5\sqrt{3}) \hat{\mathbf{i}} + (25 - 5) \hat{\mathbf{j}} = (5\sqrt{3}) \hat{\mathbf{i}} + 20 \hat{\mathbf{j}}$ km/h.
Magnitude of $\mathbf{R} = \sqrt{(5\sqrt{3})^2 + 20^2} = \sqrt{(25 \times 3) + 400} = \sqrt{75 + 400} = \sqrt{475}$.
$\sqrt{475} = \sqrt{25 \times 19} = 5\sqrt{19} \approx 5 \times 4.359 \approx 21.798$ km/h.
Magnitude is approximately 21.8 km/h.
Direction of $\mathbf{R}$ is given by the angle $\alpha$ with the positive x-axis (East): $\tan \alpha = \frac{R_y}{R_x} = \frac{20}{5\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx \frac{4 \times 1.732}{3} \approx \frac{6.928}{3} \approx 2.309$.
$\alpha = \arctan(2.309) \approx 66.5^\circ$.
The resultant velocity is approximately 21.8 km/h at an angle of 66.5° North of East. The angle with the North (positive y-axis) is $90^\circ - 66.5^\circ = 23.5^\circ$ towards the East.
So the boat's resultant velocity is approximately 21.8 km/h in the direction 23.5° East of North.
Let's compare with the example's solution method (Law of Cosines). The angle between $\mathbf{v}_b$ (North) and $\mathbf{v}_c$ (60° East of South) is $180^\circ - 60^\circ = 120^\circ$. Using Law of Cosines: $R = \sqrt{v_b^2 + v_c^2 + 2 v_b v_c \cos 120^\circ} = \sqrt{25^2 + 10^2 + 2(25)(10)(-1/2)} = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 21.8$. Magnitude matches.
Using Law of Sines to find the angle $\phi$ between R and $\mathbf{v}_b$ (North): $\frac{v_c}{\sin \phi} = \frac{R}{\sin 120^\circ}$.
$\sin \phi = \frac{v_c \sin 120^\circ}{R} = \frac{10 \times (\sqrt{3}/2)}{21.8} = \frac{5\sqrt{3}}{21.8} \approx \frac{8.66}{21.8} \approx 0.397$.
$\phi = \arcsin(0.397) \approx 23.4^\circ$.
So the resultant velocity is at an angle of 23.4° from the North direction. This angle is towards the East (since the resultant vector has a positive x-component). So the direction is 23.4° East of North.
This matches my component method direction (23.5° East of North, slight difference due to rounding). The provided solution angle calculation (f @ 23.4) is the angle with the North direction. The final direction stated in the solution text "23.4o" should be "23.4° East of North".
Motion In A Plane
Describing motion in a plane (two dimensions) involves using vectors for position, displacement, velocity, and acceleration. This allows us to account for changes in both magnitude and direction of these quantities.
Position Vector And Displacement
The position of a particle P in an x-y plane relative to the origin is given by the position vector $\mathbf{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}}$, where x and y are the coordinates. The displacement ($\Delta \mathbf{r}$) from an initial position $\mathbf{r}$ (at time t) to a final position $\mathbf{r}'$ (at time t') is the vector difference $\Delta \mathbf{r} = \mathbf{r}' - \mathbf{r}$. In component form, $\Delta \mathbf{r} = (x'-x)\hat{\mathbf{i}} + (y'-y)\hat{\mathbf{j}} = \Delta x \hat{\mathbf{i}} + \Delta y \hat{\mathbf{j}}$.
Velocity
The average velocity ($\bar{\mathbf{v}}$) over a time interval $\Delta t$ is the ratio of displacement to the time interval: $\bar{\mathbf{v}} = \frac{\Delta \mathbf{r}}{\Delta t}$. The instantaneous velocity (v) at time t is the limit of the average velocity as $\Delta t \to 0$: $\mathbf{v} = \lim_{\Delta t \to 0} \frac{\Delta \mathbf{r}}{\Delta t} = \frac{d\mathbf{r}}{dt}$. The direction of instantaneous velocity is always tangential to the path of motion at that point and in the direction of motion.
In component form, $\mathbf{v} = \frac{d}{dt}(x \hat{\mathbf{i}} + y \hat{\mathbf{j}}) = \frac{dx}{dt} \hat{\mathbf{i}} + \frac{dy}{dt} \hat{\mathbf{j}} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}$. The magnitude is $v = \sqrt{v_x^2 + v_y^2}$, and the direction angle $\theta$ with the x-axis is $\tan \theta = v_y/v_x$.
Acceleration
The average acceleration ($\bar{\mathbf{a}}$) over a time interval $\Delta t$ is the change in velocity divided by the time interval: $\bar{\mathbf{a}} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\mathbf{v}' - \mathbf{v}}{\Delta t}$. The instantaneous acceleration (a) at time t is the limit of the average acceleration as $\Delta t \to 0$: $\mathbf{a} = \lim_{\Delta t \to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$. The direction of instantaneous acceleration is the direction of $\Delta \mathbf{v}$ as $\Delta t \to 0$.
In component form, $\mathbf{a} = \frac{d}{dt}(v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}) = \frac{dv_x}{dt} \hat{\mathbf{i}} + \frac{dv_y}{dt} \hat{\mathbf{j}} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}}$.
For motion in two or three dimensions, the velocity and acceleration vectors are generally not along the same straight line. They can have any angle between 0° and 180° between them.
Example 3.4. The position of a particle is given by $\mathbf{r} = 3.0t \hat{\mathbf{i}} + 2.0t^2 \hat{\mathbf{j}} + 5.0 \hat{\mathbf{k}}$ where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find $\mathbf{v}(t)$ and $\mathbf{a}(t)$ of the particle. (b) Find the magnitude and direction of $\mathbf{v}(t)$ at t = 1.0 s.
Answer:
(a) To find the velocity $\mathbf{v}(t)$, we differentiate the position vector $\mathbf{r}(t)$ with respect to time $t$.
$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(3.0t \hat{\mathbf{i}} + 2.0t^2 \hat{\mathbf{j}} + 5.0 \hat{\mathbf{k}})$
$\mathbf{v}(t) = \frac{d}{dt}(3.0t) \hat{\mathbf{i}} + \frac{d}{dt}(2.0t^2) \hat{\mathbf{j}} + \frac{d}{dt}(5.0) \hat{\mathbf{k}}$
$\mathbf{v}(t) = 3.0 \hat{\mathbf{i}} + (2.0 \times 2t) \hat{\mathbf{j}} + 0 \hat{\mathbf{k}}$
$\mathbf{v}(t) = 3.0 \hat{\mathbf{i}} + 4.0t \hat{\mathbf{j}}$ m/s.
To find the acceleration $\mathbf{a}(t)$, we differentiate the velocity vector $\mathbf{v}(t)$ with respect to time $t$.
$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3.0 \hat{\mathbf{i}} + 4.0t \hat{\mathbf{j}})$
$\mathbf{a}(t) = \frac{d}{dt}(3.0) \hat{\mathbf{i}} + \frac{d}{dt}(4.0t) \hat{\mathbf{j}}$
$\mathbf{a}(t) = 0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}$
$\mathbf{a}(t) = 4.0 \hat{\mathbf{j}}$ m/s².
So, the velocity is $\mathbf{v}(t) = (3.0 \hat{\mathbf{i}} + 4.0t \hat{\mathbf{j}}) \text{ m/s}$, and the acceleration is $\mathbf{a}(t) = 4.0 \hat{\mathbf{j}} \text{ m/s}^2$. The acceleration is constant and directed along the positive y-axis.
(b) Find the magnitude and direction of $\mathbf{v}(t)$ at $t = 1.0 \text{ s}$.
Substitute $t = 1.0 \text{ s}$ into the expression for $\mathbf{v}(t)$:
$\mathbf{v}(1.0) = 3.0 \hat{\mathbf{i}} + 4.0(1.0) \hat{\mathbf{j}} = 3.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}$ m/s.
The magnitude of this velocity vector is $v = |\mathbf{v}(1.0)| = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0 \text{ m/s}$.
The magnitude of the velocity at t = 1.0 s is 5.0 m/s.
The direction of the velocity vector at $t=1.0 \text{ s}$ is given by the angle $\theta$ it makes with the positive x-axis:
$\tan \theta = \frac{v_y}{v_x} = \frac{4.0}{3.0} = \frac{4}{3}$.
$\theta = \arctan(4/3) \approx 53.1^\circ$.
The direction of the velocity vector at t = 1.0 s is approximately 53.1° with the positive x-axis.
Motion In A Plane With Constant Acceleration
If an object moves in a plane with constant acceleration $\mathbf{a}$, its motion can be described using vector forms of the kinematic equations derived for one-dimensional motion. Let the initial velocity be $\mathbf{v}_0$ at time $t=0$ and the initial position be $\mathbf{r}_0$ at time $t=0$.
The velocity $\mathbf{v}$ at time $t$ is given by:
$\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$
The position vector $\mathbf{r}$ at time $t$ is given by:
$\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0t + \frac{1}{2}\mathbf{a}t^2$
These vector equations can be separated into component equations along perpendicular axes (e.g., x and y axes). This shows that motion in a plane with constant acceleration is equivalent to two independent, simultaneous one-dimensional motions with constant acceleration along perpendicular directions.
In component form, if $\mathbf{r}_0 = x_0\hat{\mathbf{i}} + y_0\hat{\mathbf{j}}$, $\mathbf{v}_0 = v_{0x}\hat{\mathbf{i}} + v_{0y}\hat{\mathbf{j}}$, and $\mathbf{a} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}}$ (where $a_x$ and $a_y$ are constant):
$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$
$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$
$v_x = v_{0x} + a_xt$
$v_y = v_{0y} + a_yt$
Example 3.5. A particle starts from origin at t = 0 with a velocity 5.0 $\hat{\mathbf{i}}$ m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0$\hat{\mathbf{i}}$+2.0$\hat{\mathbf{j}}$ ) m/s². (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ?
Answer:
Initial position $\mathbf{r}_0 = 0$ (starts from origin). Initial velocity $\mathbf{v}_0 = 5.0 \hat{\mathbf{i}}$ m/s. Constant acceleration $\mathbf{a} = (3.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}}) \text{ m/s}^2$.
From $\mathbf{r}_0 = 0$, we have $x_0 = 0$ and $y_0 = 0$.
From $\mathbf{v}_0 = 5.0 \hat{\mathbf{i}}$, we have $v_{0x} = 5.0 \text{ m/s}$ and $v_{0y} = 0 \text{ m/s}$.
From $\mathbf{a} = 3.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}}$, we have $a_x = 3.0 \text{ m/s}^2$ and $a_y = 2.0 \text{ m/s}^2$. Both $a_x$ and $a_y$ are constant.
The position at time $t$ is given by $x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$ and $y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$.
$x = 0 + 5.0t + \frac{1}{2}(3.0)t^2 \implies x = 5.0t + 1.5t^2$.
$y = 0 + 0t + \frac{1}{2}(2.0)t^2 \implies y = t^2$.
(a) What is the y-coordinate when the x-coordinate is 84 m? First, find the time $t$ when $x = 84 \text{ m}$.
$84 = 5.0t + 1.5t^2$.
$1.5t^2 + 5.0t - 84 = 0$.
Multiply by 2 to remove decimal: $3t^2 + 10t - 168 = 0$.
Solve the quadratic equation for $t$: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{10^2 - 4(3)(-168)}}{2(3)} = \frac{-10 \pm \sqrt{100 + 2016}}{6} = \frac{-10 \pm \sqrt{2116}}{6}$.
$\sqrt{2116} = 46$.
$t = \frac{-10 \pm 46}{6}$.
Possible values for t are $t_1 = \frac{-10 + 46}{6} = \frac{36}{6} = 6 \text{ s}$, and $t_2 = \frac{-10 - 46}{6} = \frac{-56}{6} = -9.33 \text{ s}$.
Since time must be positive, $t = 6 \text{ s}$.
Now find the y-coordinate at $t = 6 \text{ s}$.
$y = t^2 = (6 \text{ s})^2 = 36 \text{ m}$.
The y-coordinate of the particle when its x-coordinate is 84 m is 36 m.
(b) What is the speed of the particle at this time? Speed is the magnitude of the velocity vector $\mathbf{v}$. First, find the velocity vector at $t = 6 \text{ s}$.
The velocity components are $v_x = v_{0x} + a_xt = 5.0 + 3.0t$ and $v_y = v_{0y} + a_yt = 0 + 2.0t = 2.0t$.
At $t = 6 \text{ s}$:
$v_x = 5.0 + 3.0(6) = 5.0 + 18.0 = 23.0 \text{ m/s}$.
$v_y = 2.0(6) = 12.0 \text{ m/s}$.
The velocity vector is $\mathbf{v}(6) = 23.0 \hat{\mathbf{i}} + 12.0 \hat{\mathbf{j}}$ m/s.
The speed is the magnitude of the velocity vector: $v = |\mathbf{v}(6)| = \sqrt{v_x^2 + v_y^2} = \sqrt{(23.0)^2 + (12.0)^2} = \sqrt{529.0 + 144.0} = \sqrt{673.0}$.
$v \approx 25.94 \text{ m/s}$.
The speed of the particle at this time is approximately 25.9 m/s.
Projectile Motion
Projectile motion is a classic example of motion in a plane with constant acceleration. It describes the motion of an object (a projectile) launched into the air, subjected only to the acceleration due to gravity (neglecting air resistance). The acceleration due to gravity $\mathbf{g}$ is directed vertically downwards, so if we choose the x-axis horizontally and the y-axis vertically upwards, the acceleration vector is $\mathbf{a} = -g \hat{\mathbf{j}}$. Thus, $a_x = 0$ and $a_y = -g$ are constant.
If a projectile is launched from the origin (x₀=0, y₀=0) with initial velocity $\mathbf{v}_0$ at an angle $\theta_0$ with the horizontal x-axis, the initial velocity components are $v_{0x} = v_0 \cos \theta_0$ and $v_{0y} = v_0 \sin \theta_0$.
The equations of motion derived from constant acceleration in components are:
$x = v_{0x}t = (v_0 \cos \theta_0)t$ (Horizontal motion with constant velocity)
$y = v_{0y}t + \frac{1}{2}a_yt^2 = (v_0 \sin \theta_0)t - \frac{1}{2}gt^2$ (Vertical motion with constant downward acceleration)
The velocity components at time t are:
$v_x = v_{0x} = v_0 \cos \theta_0$
$v_y = v_{0y} + a_yt = v_0 \sin \theta_0 - gt$
Equation Of Path Of A Projectile
The shape of the path (trajectory) is found by eliminating time 't' from the equations for x and y. From $x = (v_0 \cos \theta_0)t$, we get $t = \frac{x}{v_0 \cos \theta_0}$. Substituting this into the equation for y:
$y = (v_0 \sin \theta_0) \left(\frac{x}{v_0 \cos \theta_0}\right) - \frac{1}{2}g \left(\frac{x}{v_0 \cos \theta_0}\right)^2$
$y = (\tan \theta_0) x - \left(\frac{g}{2v_0^2 \cos^2 \theta_0}\right) x^2$
This equation is of the form $y = ax - bx^2$, which is the equation of a parabola. Thus, the path of a projectile is parabolic.
Time Of Maximum Height
At the maximum height ($h_m$), the vertical component of velocity ($v_y$) is zero. Let $t_m$ be the time taken to reach the maximum height. Using $v_y = v_0 \sin \theta_0 - gt$, set $v_y = 0$:
$0 = v_0 \sin \theta_0 - gt_m$
$t_m = \frac{v_0 \sin \theta_0}{g}$
Maximum Height Of A Projectile
The maximum height ($h_m$) is the vertical position 'y' at time $t_m$. Substitute $t_m$ into the equation for y:
$h_m = (v_0 \sin \theta_0) t_m - \frac{1}{2}gt_m^2 = (v_0 \sin \theta_0) \left(\frac{v_0 \sin \theta_0}{g}\right) - \frac{1}{2}g \left(\frac{v_0 \sin \theta_0}{g}\right)^2$
$h_m = \frac{v_0^2 \sin^2 \theta_0}{g} - \frac{1}{2}g \frac{v_0^2 \sin^2 \theta_0}{g^2} = \frac{v_0^2 \sin^2 \theta_0}{g} - \frac{v_0^2 \sin^2 \theta_0}{2g}$
$h_m = \frac{v_0^2 \sin^2 \theta_0}{2g}$
Horizontal Range Of A Projectile
The horizontal range (R) is the horizontal distance traveled when the projectile returns to its initial vertical level (y=0). The time taken for this is the time of flight ($T_f$). Due to the parabolic symmetry, $T_f = 2t_m = \frac{2v_0 \sin \theta_0}{g}$. The horizontal distance is $x = (v_0 \cos \theta_0)t$. Substitute $T_f$ for $t$:
$R = (v_0 \cos \theta_0) T_f = (v_0 \cos \theta_0) \left(\frac{2v_0 \sin \theta_0}{g}\right) = \frac{v_0^2 (2 \sin \theta_0 \cos \theta_0)}{g}$
Using the trigonometric identity $2 \sin \theta \cos \theta = \sin 2\theta$:
$R = \frac{v_0^2 \sin 2\theta_0}{g}$
The range R is maximum for a given initial speed $v_0$ when $\sin 2\theta_0 = 1$, which means $2\theta_0 = 90^\circ$, or $\theta_0 = 45^\circ$. The maximum range is $R_{max} = \frac{v_0^2}{g}$.
Example 3.6. Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Answer:
Let the initial velocity be $v_0$. The range R is given by the formula $R = \frac{v_0^2 \sin 2\theta_0}{g}$.
Consider two projection angles: $\theta_1 = 45^\circ + \alpha$ and $\theta_2 = 45^\circ - \alpha$. Both angles differ from 45° by the same amount $\alpha$.
For $\theta_1 = 45^\circ + \alpha$, the argument for the sine function in the range formula is $2\theta_1 = 2(45^\circ + \alpha) = 90^\circ + 2\alpha$.
$R_1 = \frac{v_0^2 \sin (90^\circ + 2\alpha)}{g}$.
For $\theta_2 = 45^\circ - \alpha$, the argument for the sine function is $2\theta_2 = 2(45^\circ - \alpha) = 90^\circ - 2\alpha$.
$R_2 = \frac{v_0^2 \sin (90^\circ - 2\alpha)}{g}$.
Using the trigonometric identities $\sin(90^\circ + x) = \cos x$ and $\sin(90^\circ - x) = \cos x$, we have:
$\sin (90^\circ + 2\alpha) = \cos (2\alpha)$.
$\sin (90^\circ - 2\alpha) = \cos (2\alpha)$.
So, $\sin (90^\circ + 2\alpha) = \sin (90^\circ - 2\alpha)$.
Therefore, $R_1 = \frac{v_0^2 \cos (2\alpha)}{g}$ and $R_2 = \frac{v_0^2 \cos (2\alpha)}{g}$.
$R_1 = R_2$.
This proves Galileo's statement that for projection angles that exceed or fall short of 45° by equal amounts, the horizontal ranges are equal, assuming the initial speed $v_0$ is the same.
Example 3.7. A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s⁻¹. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m s⁻²).
Answer:
Let's set the origin ($x=0, y=0$) at the point where the stone is thrown (edge of the cliff). Let the positive x-axis be horizontal in the direction of the throw, and the positive y-axis be vertically downwards. The initial height above the ground is 490 m. Since we chose the origin at the cliff edge and +y downwards, the ground is at $y = +490 \text{ m}$.
Initial position: $x_0 = 0$, $y_0 = 0$.
Initial velocity: The stone is thrown horizontally with speed 15 m s⁻¹. So, $v_{0x} = 15 \text{ m s}^{-1}$ and $v_{0y} = 0$.
Acceleration: Acceleration due to gravity is downwards. Since +y is downwards, $a_x = 0$ and $a_y = +g = +9.8 \text{ m s}^{-2}$. Both are constant.
We need to find the time $t$ taken to reach the ground, which is at $y = 490 \text{ m}$.
Use the equation for vertical motion: $y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$.
$490 \text{ m} = 0 + (0)t + \frac{1}{2}(9.8 \text{ m s}^{-2})t^2$.
$490 = 4.9 t^2$.
$t^2 = \frac{490}{4.9} = 100$.
$t = \sqrt{100} = 10 \text{ s}$ (since time must be positive).
The time taken by the stone to reach the ground is 10 s.
Now, find the speed with which it hits the ground. Speed is the magnitude of the velocity vector at $t = 10 \text{ s}$. The velocity components are $v_x = v_{0x} + a_xt$ and $v_y = v_{0y} + a_yt$.
$v_x = 15 \text{ m s}^{-1} + (0)(10 \text{ s}) = 15 \text{ m s}^{-1}$. (Horizontal velocity is constant).
$v_y = 0 \text{ m s}^{-1} + (9.8 \text{ m s}^{-2})(10 \text{ s}) = 98 \text{ m s}^{-1}$.
The velocity vector at impact is $\mathbf{v} = 15 \hat{\mathbf{i}} + 98 \hat{\mathbf{j}}$ m/s. (Remember +y is downwards).
The speed is $v = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(15)^2 + (98)^2} = \sqrt{225 + 9604} = \sqrt{9829}$.
$v \approx 99.14 \text{ m/s}$.
The speed with which it hits the ground is approximately 99.1 m/s.
Example 3.8. A cricket ball is thrown at a speed of 28 m s⁻¹ in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.
Answer:
Let the origin (x=0, y=0) be the point where the ball is thrown. Let the positive x-axis be horizontal and the positive y-axis be vertically upwards. Initial speed $v_0 = 28 \text{ m s}^{-1}$. Projection angle $\theta_0 = 30^\circ$ above the horizontal.
Acceleration is due to gravity downwards: $a_x = 0$, $a_y = -g = -9.8 \text{ m s}^{-2}$.
(a) Calculate the maximum height ($h_m$). Using the formula $h_m = \frac{v_0^2 \sin^2 \theta_0}{2g}$.
$h_m = \frac{(28 \text{ m s}^{-1})^2 (\sin 30^\circ)^2}{2(9.8 \text{ m s}^{-2})} = \frac{(784 \text{ m}^2 \text{ s}^{-2}) (0.5)^2}{19.6 \text{ m s}^{-2}} = \frac{784 \times 0.25}{19.6} \text{ m}$.
$h_m = \frac{196}{19.6} \text{ m} = 10 \text{ m}$.
The maximum height is 10 m.
(b) Calculate the time taken by the ball to return to the same level (time of flight, $T_f$). Using the formula $T_f = \frac{2v_0 \sin \theta_0}{g}$.
$T_f = \frac{2(28 \text{ m s}^{-1})(\sin 30^\circ)}{9.8 \text{ m s}^{-2}} = \frac{2 \times 28 \times 0.5}{9.8} \text{ s} = \frac{28}{9.8} \text{ s}$.
$T_f \approx 2.857 \text{ s}$.
The time of flight is approximately 2.9 s.
(c) Calculate the distance from the thrower to the point where the ball returns to the same level (horizontal range, R). Using the formula $R = \frac{v_0^2 \sin 2\theta_0}{g}$.
$R = \frac{(28 \text{ m s}^{-1})^2 (\sin (2 \times 30^\circ))}{9.8 \text{ m s}^{-2}} = \frac{784 (\sin 60^\circ)}{9.8} \text{ m} = \frac{784 \times (\sqrt{3}/2)}{9.8} \text{ m}$.
$R = \frac{784 \times 0.866}{9.8} \text{ m} \approx \frac{678.944}{9.8} \text{ m} \approx 69.28 \text{ m}$.
The horizontal range is approximately 69 m.
Uniform Circular Motion
Uniform circular motion is the motion of an object following a circular path at a constant speed. Even though the speed is constant, the velocity vector is continuously changing direction (it's always tangent to the circle), so the object is accelerating.
The acceleration in uniform circular motion is always directed towards the center of the circle and is called centripetal acceleration (ac). Its magnitude is given by:
$a_c = \frac{v^2}{R}$
where $v$ is the constant speed and $R$ is the radius of the circle. Since $v$ and $R$ are constant, the magnitude of centripetal acceleration is constant, but its direction is always changing (towards the center), so it is not a constant vector.
The angular speed ($\omega$) is the rate of change of the angular position. It is related to the linear speed ($v$) by $v = R\omega$. The centripetal acceleration can also be expressed in terms of angular speed: $a_c = R\omega^2$.
The time period ($T$) for one revolution is related to frequency ($n=1/T$) and angular speed by $\omega = 2\pi n = 2\pi/T$. Linear speed $v = 2\pi Rn = 2\pi R/T$. Centripetal acceleration $a_c = 4\pi^2n^2R = 4\pi^2R/T^2$.
Example 3.9. An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ?
Answer:
This is an example of uniform circular motion since the insect moves steadily along a circular path at a constant speed.
Radius of the groove $R = 12 \text{ cm} = 0.12 \text{ m}$.
Number of revolutions = 7. Time taken = 100 s.
Frequency of revolution $n = \frac{\text{Number of revolutions}}{\text{Time taken}} = \frac{7}{100 \text{ s}} = 0.07 \text{ s}^{-1}$.
Time period $T = \frac{1}{n} = \frac{1}{0.07 \text{ s}^{-1}} \approx 14.286 \text{ s}$.
(a) Angular speed ($\omega$) is given by $\omega = 2\pi n$ or $\omega = 2\pi/T$.
$\omega = 2\pi \times 0.07 \text{ s}^{-1} \approx 0.4398 \text{ rad/s}$.
Linear speed ($v$) is given by $v = R\omega$.
$v = (0.12 \text{ m}) \times (0.4398 \text{ rad/s}) \approx 0.05278 \text{ m/s}$.
Converting to cm/s: $v = 0.05278 \text{ m/s} \times 100 \text{ cm/m} \approx 5.278 \text{ cm/s}$.
The angular speed is approximately 0.44 rad/s, and the linear speed is approximately 5.3 cm/s.
(b) Is the acceleration vector a constant vector? The acceleration in uniform circular motion is the centripetal acceleration, which is directed towards the center of the circle. As the object moves around the circle, the direction towards the center continuously changes. Therefore, the acceleration vector is not a constant vector, even though its magnitude is constant.
Magnitude of acceleration ($a_c$) is given by $a_c = R\omega^2$ or $a_c = v^2/R$.
Using $\omega \approx 0.4398 \text{ rad/s}$ and $R = 12 \text{ cm}$:
$a_c = (12 \text{ cm}) \times (0.4398 \text{ s}^{-1})^2 = 12 \times 0.193424 \text{ cm s}^{-2} \approx 2.321 \text{ cm s}^{-2}$.
The magnitude of the acceleration is approximately 2.3 cm s⁻².
The statement about the magnitude in the example solution (2.3 cm/s²) matches our calculation.
Exercises
Question 3.1. State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:
Question 3.2. Pick out the two scalar quantities in the following list :
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Question 3.3. Pick out the only vector quantity in the following list :
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Question 3.4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :
(a) adding any two scalars,
(b) adding a scalar to a vector of the same dimensions ,
(c) multiplying any vector by any scalar,
(d) multiplying any two scalars,
(e) adding any two vectors,
(f) adding a component of a vector to the same vector.
Answer:
Question 3.5. Read each statement below carefully and state with reasons, if it is true or false :
(a) The magnitude of a vector is always a scalar,
(b) each component of a vector is always a scalar,
(c) the total path length is always equal to the magnitude of the displacement vector of a particle.
(d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time,
(e) Three vectors not lying in a plane can never add up to give a null vector.
Answer:
Question 3.6. Establish the following vector inequalities geometrically or otherwise :
(a) $|a+b| \le |a| + |b|$
(b) $|a+b| \ge ||a| - |b||$
(c) $|a-b| \le |a| + |b|$
(d) $|a-b| \ge ||a| - |b||$
When does the equality sign above apply?
Answer:
Question 3.7. Given a + b + c + d = 0, which of the following statements are correct :
(a) a, b, c, and d must each be a null vector,
(b) The magnitude of (a + c) equals the magnitude of (b + d),
(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d,
(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ?
Answer:
Question 3.8. Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?
Answer:
Question 3.9. A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?
Answer:
Question 3.10. On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer:
Question 3.11. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ?
Answer:
Question 3.12. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s$^{-1}$ can go without hitting the ceiling of the hall ?
Answer:
Question 3.13. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?
Answer:
Question 3.14. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?
Answer:
Question 3.15. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Answer:
Question 3.16. Read each statement below carefully and state, with reasons, if it is true or false :
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector
Answer:
Question 3.17. The position of a particle is given by
$$r = 3.0t \hat{i} - 2.0t^2 \hat{j} + 4.0 \hat{k} \ m$$
where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?
Answer:
Question 3.18. A particle starts from the origin at t = 0 s with a velocity of 10.0 $\hat{j}$ m/s and moves in the x-y plane with a constant acceleration of (8.0 $\hat{i}$ + 2.0 $\hat{j}$) m s$^{-2}$. (a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ?
Answer:
Question 3.19. $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{i} + \hat{j}$, and $\hat{i} - \hat{j}$ ? What are the components of a vector A= 2$\hat{i}$ + 3$\hat{j}$ along the directions of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$? [You may use graphical method]
Answer:
Question 3.20. For any arbitrary motion in space, which of the following relations are true :
(a) $v_{average} = (1/2) (v (t_1) + v (t_2))$
(b) $v_{average} = [r(t_2) - r(t_1) ] /(t_2 – t_1)$
(c) $v (t) = v (0) + a t$
(d) $r (t) = r (0) + v (0) t + (1/2) a t^2$
(e) $a_{average} =[ v (t_2) - v (t_1 )] /( t_2 – t_1)$
(The ‘average’ stands for average of the quantity over the time interval $t_1$ to $t_2$)
Answer:
Question 3.21. Read each statement below carefully and state, with reasons and examples, if it is true or false :
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
Answer:
Question 3.22. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, what is the speed of the aircraft ?
Answer: