Chapter Notes

Introduction to Motion in a Plane

In the previous chapter, we explored motion along a straight line (one-dimensional motion). We saw that we could use simple positive (+) and negative (-) signs to indicate direction. However, the world around us is not limited to straight lines. Objects fly through the air, planets orbit the sun, and cars turn corners. To describe this kind of motion in two dimensions (a plane) or three dimensions (space), we need a more powerful tool: vectors.

This chapter introduces the language of vectors. We will learn what scalars and vectors are, how to add, subtract, and multiply them, and how to use them to describe physical quantities like position, velocity, and acceleration in a plane. We will then apply these concepts to understand two important types of two-dimensional motion: projectile motion and uniform circular motion.

Scalars and Vectors

In physics, all measurable quantities can be classified into two main categories: scalars and vectors. The key difference between them is that vectors have a direction, while scalars do not.

Scalar Quantity A scalar is a quantity that is fully described by its magnitude (a single number) and a proper unit.

• Examples: Distance (5 km), mass (2 kg), temperature (25°C), and time (10 s).
• Combining Scalars: Scalars follow the rules of ordinary algebra. You can add, subtract, multiply, and divide them just like regular numbers. For example, if a rectangle is 1.0 m long and 0.5 m wide, its perimeter is simply the sum of the lengths of its four sides: $1.0 \text{ m} + 0.5 \text{ m} + 1.0 \text{ m} + 0.5 \text{ m} = 3.0 \text{ m}$.

Vector Quantity A vector is a quantity that has both magnitude and direction. To be a vector, a quantity must also obey the specific rules of vector addition, such as the triangle law or the parallelogram law.

• Examples: Displacement (10 m, East), velocity (20 m/s, North), acceleration (9.8 m/s², downwards), and force (50 N, at 30°).
• Representing Vectors: In textbooks, vectors are shown in boldface, like v. When writing by hand, we draw an arrow over the letter, like $\vec{v}$. The magnitude of a vector is its absolute value, written as $|\mathbf{v}|$ or simply $v$.

Position and Displacement Vectors

To describe an object's motion in a plane, we first need to define its position.

• Position Vector (r): This is a vector drawn from a chosen origin (O) to the object's location (P). The length of the position vector represents the distance from the origin, and its direction points from the origin to the object. We denote it as $\mathbf{OP} = \mathbf{r}$.

• Displacement Vector ($\Delta$r): If an object moves from an initial position P to a final position P', the displacement vector is the straight-line vector from P to P'. It represents the change in position.

A crucial point is that the displacement vector only depends on the starting and ending points, not the actual path taken. For example, whether you walk in a straight line or a winding path from your home to school, your displacement is the same: a straight line from home to school. Because of this, the magnitude of the displacement is always less than or equal to the total path length.

Equality of Vectors

Two vectors, A and B, are considered equal if and only if they have the same magnitude and the same direction.

A vector can be moved parallel to itself without changing what it is. This is because, for many applications in physics, a vector's location doesn't matter, only its length and direction. These are called free vectors.

Multiplication of Vectors by Real Numbers

We can multiply a vector by a real number (a scalar), which scales the vector.

• Multiplying by a Positive Number ($\lambda > 0$): When you multiply a vector A by a positive number $\lambda$, the result is a new vector, $\lambda\mathbf{A}$. Its magnitude becomes $\lambda$ times the original magnitude, and its direction remains the same. For example, $2\mathbf{A}$ is a vector in the same direction as A but twice as long.

• Multiplying by a Negative Number ($-\lambda$): When you multiply a vector A by a negative number $-\lambda$, the new vector has a magnitude of $\lambda|\mathbf{A}|$, but its direction is opposite to the original vector. For example, $-1.5\mathbf{A}$ is a vector pointing in the opposite direction of A and is 1.5 times as long.

Addition and Subtraction of Vectors - Graphical Method

Vectors do not add like ordinary numbers. We must use graphical methods that account for their direction.

Head-to-Tail Method (Triangle Law of Addition)

This is the most intuitive way to add vectors. To find the sum of two vectors A and B, follow these steps:

1. Draw vector A.
2. Place the tail of vector B at the head of vector A.
3. The resultant vector R is the vector drawn from the tail of A to the head of B.

This gives us the vector sum: $\mathbf{R} = \mathbf{A} + \mathbf{B}$.

Vector addition has two important properties:

• Commutative Law: The order of addition doesn't matter. $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$.
• Associative Law: When adding three or more vectors, it doesn't matter how you group them. $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$.

Null Vector (Zero Vector)

What happens when you add a vector A to its negative, $-\mathbf{A}$? The result is a null vector, denoted by 0.

• A null vector has zero magnitude.
• Because its magnitude is zero, its direction cannot be specified.

Properties of the null vector:

• $\mathbf{A} + \mathbf{0} = \mathbf{A}$
• $\lambda\mathbf{0} = \mathbf{0}$ (where $\lambda$ is a scalar)
• $0\mathbf{A} = \mathbf{0}$

Subtraction of Vectors

Subtracting a vector is defined as adding its negative. The difference between two vectors A and B is: $\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$ To do this graphically, you first reverse the direction of vector B to get $-\mathbf{B}$, and then add $-\mathbf{B}$ to A using the head-to-tail method.

Parallelogram Law of Addition

This is an alternative but equivalent method for adding two vectors.

1. Place the tails of both vectors A and B at a common origin, O.
2. Complete the parallelogram using A and B as adjacent sides.
3. The resultant vector R is the diagonal of the parallelogram that starts from the common origin O.

Given

• Velocity of rain, $v_r = 35 \text{ m s}^{-1}$ (vertically downward)
• Velocity of wind, $v_w = 12 \text{ m s}^{-1}$ (from east to west, i.e., horizontally)

To Find

The direction in which the boy should hold his umbrella. This will be opposite to the direction of the resultant velocity of the rain.

Formula

The magnitude of the resultant velocity R is found using the Pythagorean theorem, as the vectors are perpendicular. $R = \sqrt{v_r^2 + v_w^2}$ The direction angle $\theta$ with the vertical is given by: $\tan\theta = \frac{v_w}{v_r}$

Solution

The boy needs to hold the umbrella to block the rain, which appears to be coming at him with a resultant velocity R, which is the vector sum of the rain's velocity and the wind's velocity.

First, calculate the magnitude of the resultant velocity: $R = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37 \text{ m s}^{-1}$

Next, find the direction. Let $\theta$ be the angle the resultant velocity R makes with the vertical. $\tan\theta = \frac{v_w}{v_r} = \frac{12}{35} \approx 0.343$ $\theta = \tan^{-1}(0.343) \approx 19^\circ$

The resultant velocity of the rain is $19^\circ$ from the vertical towards the west. To protect himself, the boy must hold his umbrella at an angle of $19^\circ$ with the vertical towards the east.

Final Answer The boy should hold his umbrella at an angle of about $19^\circ$ with the vertical, tilted towards the east.

Resolution of Vectors

Just as we can add vectors to get a single resultant, we can also break down a single vector into multiple parts, called component vectors. This process is known as resolution of vectors.

It is most useful to resolve a vector into components that are perpendicular to each other, typically along the x and y-axes of a rectangular coordinate system. For this, we use unit vectors.

A unit vector is a vector with a magnitude of exactly 1. It has no dimensions or units; its only purpose is to specify a direction.

• $\hat{\mathbf{i}}$ is the unit vector in the positive x-direction.
• $\hat{\mathbf{j}}$ is the unit vector in the positive y-direction.
• $\hat{\mathbf{k}}$ is the unit vector in the positive z-direction.

Any vector A can be written as the sum of its rectangular components: $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}$ Here, $A_x$ and $A_y$ are the scalar components of A along the x and y-axes. $A_x\hat{\mathbf{i}}$ is the vector component along the x-axis.

If a vector A has a magnitude $A$ and makes an angle $\theta$ with the positive x-axis, its components are:

• x-component: $A_x = A \cos\theta$
• y-component: $A_y = A \sin\theta$

Conversely, if we know the components $A_x$ and $A_y$, we can find the magnitude and direction of the vector A:

• Magnitude: $A = \sqrt{A_x^2 + A_y^2}$
• Direction: $\tan\theta = \frac{A_y}{A_x}$

This can be extended to three dimensions: $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}$ $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$

Vector Addition - Analytical Method

While the graphical method helps visualize vector addition, it's often tedious and not very precise. The analytical method, using components, is much easier and more accurate.

To add two vectors $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}$ and $\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}}$, simply add their corresponding components: $\mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x)\hat{\mathbf{i}} + (A_y + B_y)\hat{\mathbf{j}}$ So, the components of the resultant vector R are:

• $R_x = A_x + B_x$
• $R_y = A_y + B_y$

This method easily extends to three dimensions and to the addition or subtraction of any number of vectors.

Given

• Magnitude of vector A is $A$.
• Magnitude of vector B is $B$.
• Angle between A and B is $\theta$.

To Find

• The magnitude $R$ of the resultant vector $\mathbf{R} = \mathbf{A} + \mathbf{B}$.
• The direction $\alpha$ of the resultant vector R with respect to vector A.

Formula

• Law of Cosines for magnitude: $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$
• Law of Sines for direction: $\frac{R}{\sin\theta} = \frac{B}{\sin\alpha}$
• Alternative formula for direction: $\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$

Solution

We use the parallelogram law. Let vector A be along the x-axis (OP). Vector B (OQ) makes an angle $\theta$ with A. The resultant R is the diagonal OS. We drop a perpendicular SN from S to the extended line OP.

In the right-angled triangle OSN: $OS^2 = ON^2 + SN^2$

We know that $ON = OP + PN$. From triangle PSN, $PN = PS\cos\theta = B\cos\theta$ and $SN = PS\sin\theta = B\sin\theta$. Since $OP = A$, we have $ON = A + B\cos\theta$.

Substituting these into the first equation: $R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2$ $R^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta$ $R^2 = A^2 + B^2(\cos^2\theta + \sin^2\theta) + 2AB\cos\theta$ Since $\cos^2\theta + \sin^2\theta = 1$: $R^2 = A^2 + B^2 + 2AB\cos\theta$ $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ This is the Law of Cosines.

To find the direction, let $\alpha$ be the angle that R makes with A. In triangle OSN: $\tan\alpha = \frac{SN}{ON} = \frac{B\sin\theta}{A + B\cos\theta}$

Final Answer The magnitude of the resultant is $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ and its direction relative to vector A is given by $\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$.

Given

• Velocity of the boat, $v_b = 25 \text{ km/h}$ (North)
• Velocity of the current, $v_c = 10 \text{ km/h}$ ($60^\circ$ East of South)

To Find

The magnitude and direction of the resultant velocity R.

Formula

The angle $\theta$ between the vector $\mathbf{v_b}$ (North) and $\mathbf{v_c}$ ($60^\circ$ E of S) is $180^\circ - 60^\circ = 120^\circ$. We use the Law of Cosines to find the magnitude: $R = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos\theta}$ And the Law of Sines to find the direction: $\frac{v_c}{\sin\varphi} = \frac{R}{\sin\theta}$

Solution

First, calculate the magnitude of the resultant velocity R: $R = \sqrt{25^2 + 10^2 + 2(25)(10)\cos(120^\circ)}$ Since $\cos(120^\circ) = -1/2$: $R = \sqrt{625 + 100 + 500(-1/2)}$ $R = \sqrt{725 - 250} = \sqrt{475} \approx 21.8 \text{ km/h}$ The text approximates this to $22 \text{ km/h}$. Let's use $21.8 \text{ km/h}$ for better precision.

Next, find the direction. Let $\varphi$ be the angle the resultant R makes with the boat's velocity direction (North). $\sin\varphi = \frac{v_c \sin\theta}{R} = \frac{10 \times \sin(120^\circ)}{21.8}$ Since $\sin(120^\circ) = \frac{\sqrt{3}}{2}$: $\sin\varphi = \frac{10 \times \sqrt{3}/2}{21.8} = \frac{5\sqrt{3}}{21.8} \approx \frac{8.66}{21.8} \approx 0.397$ $\varphi = \sin^{-1}(0.397) \approx 23.4^\circ$

This angle is towards the East of North.

Final Answer The resultant velocity of the boat is approximately $22 \text{ km/h}$ in a direction $23.4^\circ$ east of north.

Motion in a Plane

Now we can use vectors to describe motion in two dimensions.

Position Vector and Displacement

The position of a particle P in an x-y plane is given by the position vector: $\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}$ If the particle moves from position $\mathbf{r}$ to $\mathbf{r'}$, the displacement $\Delta\mathbf{r}$ is: $\Delta\mathbf{r} = \mathbf{r'} - \mathbf{r} = (x' - x)\hat{\mathbf{i}} + (y' - y)\hat{\mathbf{j}} = \Delta x\hat{\mathbf{i}} + \Delta y\hat{\mathbf{j}}$

Velocity

• Average Velocity ($\overline{\mathbf{v}}$): The ratio of displacement to the time interval. $\overline{\mathbf{v}} = \frac{\Delta\mathbf{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta y}{\Delta t}\hat{\mathbf{j}}$ The direction of average velocity is the same as the direction of the displacement vector $\Delta\mathbf{r}$.

• Instantaneous Velocity (v): The velocity at a specific moment in time. It is the limiting value of the average velocity as the time interval $\Delta t$ approaches zero. $\mathbf{v} = \lim_{\Delta t \to 0} \frac{\Delta\mathbf{r}}{\Delta t} = \frac{d\mathbf{r}}{dt}$ In component form: $\mathbf{v} = v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}} \quad \text{where} \quad v_x = \frac{dx}{dt} \text{ and } v_y = \frac{dy}{dt}$

Acceleration

• Average Acceleration ($\overline{\mathbf{a}}$): The ratio of the change in velocity to the time interval. $\overline{\mathbf{a}} = \frac{\Delta\mathbf{v}}{\Delta t} = \frac{\Delta v_x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta v_y}{\Delta t}\hat{\mathbf{j}}$

• Instantaneous Acceleration (a): The acceleration at a specific moment. It is the limiting value of the average acceleration as $\Delta t$ approaches zero. $\mathbf{a} = \lim_{\Delta t \to 0} \frac{\Delta\mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}$ In component form: $\mathbf{a} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}} \quad \text{where} \quad a_x = \frac{dv_x}{dt} \text{ and } a_y = \frac{dv_y}{dt}$

Unlike one-dimensional motion, in two or three dimensions, the velocity and acceleration vectors can have any angle between $0^\circ$ and $180^\circ$ relative to each other.

Given

• Position vector: $\mathbf{r}(t) = 3.0t\hat{\mathbf{i}} + 2.0t^2\hat{\mathbf{j}} + 5.0\hat{\mathbf{k}}$ m

To Find

(a) Velocity vector $\mathbf{v}(t)$ and acceleration vector $\mathbf{a}(t)$. (b) Magnitude and direction of $\mathbf{v}(t)$ at $t = 1.0 \text{ s}$.

Formula

$\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}$ $\mathbf{a}(t) = \frac{d\mathbf{v}}{dt}$

Solution

(a) Find velocity and acceleration vectors

To find the velocity, we take the derivative of the position vector with respect to time: $\mathbf{v}(t) = \frac{d}{dt}(3.0t\hat{\mathbf{i}} + 2.0t^2\hat{\mathbf{j}} + 5.0\hat{\mathbf{k}})$ $\mathbf{v}(t) = 3.0\hat{\mathbf{i}} + 4.0t\hat{\mathbf{j}}$

To find the acceleration, we take the derivative of the velocity vector: $\mathbf{a}(t) = \frac{d}{dt}(3.0\hat{\mathbf{i}} + 4.0t\hat{\mathbf{j}})$ $\mathbf{a}(t) = 4.0\hat{\mathbf{j}}$ The acceleration is constant, with a magnitude of $4.0 \text{ m s}^{-2}$ along the positive y-direction.

Answer for part (a) = $\mathbf{v}(t) = 3.0\hat{\mathbf{i}} + 4.0t\hat{\mathbf{j}} \text{ m/s}$ and $\mathbf{a}(t) = 4.0\hat{\mathbf{j}} \text{ m/s}^2$.

(b) Find magnitude and direction of velocity at t = 1.0 s

First, substitute $t = 1.0 \text{ s}$ into the velocity equation: $\mathbf{v}(1.0) = 3.0\hat{\mathbf{i}} + 4.0(1.0)\hat{\mathbf{j}} = 3.0\hat{\mathbf{i}} + 4.0\hat{\mathbf{j}}$

The magnitude is: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0 \text{ m s}^{-1}$

The direction (angle $\theta$ with the x-axis) is: $\tan\theta = \frac{v_y}{v_x} = \frac{4.0}{3.0}$ $\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ$

Answer for part (b) = The magnitude is $5.0 \text{ m s}^{-1}$ and the direction is approximately $53^\circ$ with the x-axis.

Motion in a Plane with Constant Acceleration

Let's consider an object moving in a plane with a constant acceleration a. If the object has an initial velocity $\mathbf{v}_0$ at time $t=0$ and velocity v at time $t$, we can write: $\mathbf{a} = \frac{\mathbf{v} - \mathbf{v}_0}{t}$ Rearranging this gives the first kinematic equation in vector form: $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$

To find the position, we can use the average velocity. The displacement is the average velocity multiplied by time: $\mathbf{r} - \mathbf{r}_0 = \left(\frac{\mathbf{v} + \mathbf{v}_0}{2}\right)t$ Substituting the first equation into this one gives the second kinematic equation: $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$

In component form, these equations are:

• $v_x = v_{ox} + a_x t$
• $v_y = v_{oy} + a_y t$
• $x = x_0 + v_{ox}t + \frac{1}{2}a_x t^2$
• $y = y_0 + v_{oy}t + \frac{1}{2}a_y t^2$

Given

• Initial position, $\mathbf{r}_0 = \mathbf{0}$
• Initial velocity, $\mathbf{v}_0 = 5.0\hat{\mathbf{i}} \text{ m/s}$
• Constant acceleration, $\mathbf{a} = (3.0\hat{\mathbf{i}} + 2.0\hat{\mathbf{j}}) \text{ m/s}^2$

To Find

(a) The value of $y$ when $x = 84 \text{ m}$. (b) The speed $|\mathbf{v}|$ at that time.

Formula

Position vector as a function of time: $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$ Velocity vector as a function of time: $\mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a}t$

Solution

(a) Find the y-coordinate

First, write the position vector equation with the given values: $\mathbf{r}(t) = \mathbf{0} + (5.0\hat{\mathbf{i}})t + \frac{1}{2}(3.0\hat{\mathbf{i}} + 2.0\hat{\mathbf{j}})t^2$ $\mathbf{r}(t) = 5.0t\hat{\mathbf{i}} + (1.5t^2\hat{\mathbf{i}} + 1.0t^2\hat{\mathbf{j}})$ $\mathbf{r}(t) = (5.0t + 1.5t^2)\hat{\mathbf{i}} + (1.0t^2)\hat{\mathbf{j}}$

From this, we can get the separate equations for the x and y coordinates: $x(t) = 5.0t + 1.5t^2$ $y(t) = 1.0t^2$

We are given that $x(t) = 84 \text{ m}$. We need to find the time $t$ when this happens. $84 = 5.0t + 1.5t^2$ Rearranging into a standard quadratic equation ($1.5t^2 + 5.0t - 84 = 0$): Solving this quadratic equation gives $t = 6 \text{ s}$ (the negative solution is not physically meaningful).

Now, find the y-coordinate at $t = 6 \text{ s}$: $y(6) = 1.0(6)^2 = 36.0 \text{ m}$

Answer for part (a) = The y-coordinate is $36.0 \text{ m}$.

(b) Find the speed at this time

First, find the velocity vector at $t = 6 \text{ s}$. We can find the general velocity vector by differentiating $\mathbf{r}(t)$ or using $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$. $\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = (5.0 + 3.0t)\hat{\mathbf{i}} + 2.0t\hat{\mathbf{j}}$ At $t = 6 \text{ s}$: $\mathbf{v}(6) = (5.0 + 3.0(6))\hat{\mathbf{i}} + 2.0(6)\hat{\mathbf{j}} = 23.0\hat{\mathbf{i}} + 12.0\hat{\mathbf{j}}$

The speed is the magnitude of this velocity vector: $\text{speed} = |\mathbf{v}| = \sqrt{23.0^2 + 12.0^2} = \sqrt{529 + 144} = \sqrt{673} \approx 25.9 \text{ m s}^{-1}$ The textbook approximates this to $26 \text{ m s}^{-1}$.

Answer for part (b) = The speed of the particle is approximately $26 \text{ m s}^{-1}$.

Projectile Motion

A projectile is any object that is thrown or projected into the air and then moves under the influence of gravity alone (we will ignore air resistance). Examples include a thrown baseball, a kicked football, or a cannonball.

As Galileo first realized, the motion of a projectile can be analyzed as two independent components:

1. Horizontal motion: There is no acceleration in this direction ($a_x = 0$). The horizontal velocity is constant.
2. Vertical motion: The object is in free fall. It has a constant downward acceleration due to gravity, $g$ ($a_y = -g$).

Let's analyze the motion of a projectile launched from the origin ($x_0=0, y_0=0$) with an initial velocity $\mathbf{v}_0$ at an angle $\theta_0$ above the horizontal.

The initial velocity components are:

• $v_{ox} = v_o \cos\theta_o$
• $v_{oy} = v_o \sin\theta_o$

The components of position and velocity at any time $t$ are:

• $x(t) = (v_o \cos\theta_o)t$
• $y(t) = (v_o \sin\theta_o)t - \frac{1}{2}gt^2$
• $v_x(t) = v_o \cos\theta_o$ (constant)
• $v_y(t) = v_o \sin\theta_o - gt$

Equation of Path of a Projectile

By eliminating time $t$ from the position equations, we can find the shape of the projectile's path. From the x-equation, $t = \frac{x}{v_o \cos\theta_o}$. Substituting this into the y-equation gives: $y = (v_o \sin\theta_o)\left(\frac{x}{v_o \cos\theta_o}\right) - \frac{1}{2}g\left(\frac{x}{v_o \cos\theta_o}\right)^2$ $y = (\tan\theta_o)x - \left(\frac{g}{2(v_o \cos\theta_o)^2}\right)x^2$ This equation is of the form $y = ax + bx^2$, which is the equation of a parabola. The path of a projectile is a parabola.

Time of Maximum Height ($t_m$)

At the maximum height of its trajectory, the vertical component of the projectile's velocity is momentarily zero ($v_y = 0$). $v_y = v_o \sin\theta_o - gt_m = 0$ $t_m = \frac{v_o \sin\theta_o}{g}$

Maximum Height ($h_m$)

To find the maximum height, we substitute the time $t_m$ into the y-position equation: $h_m = (v_o \sin\theta_o)\left(\frac{v_o \sin\theta_o}{g}\right) - \frac{1}{2}g\left(\frac{v_o \sin\theta_o}{g}\right)^2$ $h_m = \frac{(v_o \sin\theta_o)^2}{2g}$

Time of Flight ($T_f$)

The time of flight is the total time the projectile is in the air. This occurs when it returns to its initial height, so $y=0$. $y(T_f) = (v_o \sin\theta_o)T_f - \frac{1}{2}gT_f^2 = 0$ $T_f \left(v_o \sin\theta_o - \frac{1}{2}gT_f\right) = 0$ This gives two solutions: $T_f=0$ (the start) and: $T_f = \frac{2v_o \sin\theta_o}{g}$ Notice that the total time of flight is exactly twice the time to reach the maximum height ($T_f = 2t_m$).

Horizontal Range (R)

The horizontal range is the total horizontal distance traveled during the time of flight. $R = v_x T_f = (v_o \cos\theta_o)\left(\frac{2v_o \sin\theta_o}{g}\right)$ Using the trigonometric identity $2\sin\theta_o\cos\theta_o = \sin(2\theta_o)$, we get: $R = \frac{v_o^2 \sin(2\theta_o)}{g}$

Given

• Initial height, $y_0 = 490 \text{ m}$ (We will set the origin at the cliff edge, so the ground is at $y = -490 \text{ m}$)
• Initial horizontal velocity, $v_{ox} = 15 \text{ m s}^{-1}$
• Initial vertical velocity, $v_{oy} = 0 \text{ m s}^{-1}$ (thrown horizontally)
• Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$ (so $a_y = -9.8 \text{ m s}^{-2}$)

To Find

• Time to reach the ground, $t$.
• Speed on hitting the ground, $v$.

Formula

• Vertical position: $y(t) = y_0 + v_{oy}t + \frac{1}{2}a_y t^2$
• Velocity components: $v_x = v_{ox}$, $v_y = v_{oy} + a_y t$
• Final speed: $v = \sqrt{v_x^2 + v_y^2}$

Solution

Let the origin (0,0) be the point where the stone is thrown. The ground is at $y = -490 \text{ m}$.

First, find the time of flight using the vertical motion equation: $-490 = 0 + (0)t + \frac{1}{2}(-9.8)t^2$ $-490 = -4.9t^2$ $t^2 = \frac{490}{4.9} = 100$ $t = 10 \text{ s}$

Now, find the components of the velocity when the stone hits the ground at $t = 10 \text{ s}$. The horizontal component remains constant: $v_x = 15 \text{ m s}^{-1}$ The vertical component is: $v_y = 0 + (-9.8)(10) = -98 \text{ m s}^{-1}$

Finally, calculate the speed (magnitude of the final velocity): $v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + (-98)^2}$ $v = \sqrt{225 + 9604} = \sqrt{9829} \approx 99 \text{ m s}^{-1}$

Final Answer The stone takes 10 s to reach the ground and hits it with a speed of approximately $99 \text{ m s}^{-1}$.

Uniform Circular Motion

When an object moves in a circle at a constant speed, its motion is called uniform circular motion.

Even though the speed is constant, the object is still accelerating. Why? Because acceleration is the rate of change of velocity, and velocity is a vector. In circular motion, the direction of the velocity vector is constantly changing (it's always tangent to the circle). This change in direction means there must be an acceleration.

This acceleration is called centripetal acceleration ($a_c$), which means "center-seeking."

• Direction: The centripetal acceleration is always directed towards the center of the circle.
• Magnitude: For an object moving with speed $v$ in a circle of radius $R$, the magnitude of the centripetal acceleration is: $a_c = \frac{v^2}{R}$

Because the direction of $a_c$ is always changing (always pointing to the center), centripetal acceleration is not a constant vector, even though its magnitude might be constant.

Angular Speed and Other Quantities

We can also describe circular motion using angular quantities.

• Angular Speed ($\omega$): The rate at which the angle changes. If an object sweeps through an angle $\Delta\theta$ in time $\Delta t$, its angular speed is $\omega = \frac{\Delta\theta}{\Delta t}$. The unit is radians per second (rad/s).
• Relation between Linear and Angular Speed: The linear speed $v$ is related to the angular speed $\omega$ by: $v = R\omega$
• Centripetal Acceleration in terms of $\omega$: $a_c = \frac{v^2}{R} = \frac{(R\omega)^2}{R} = \omega^2 R$

Other useful terms:

• Time Period (T): The time taken to complete one full revolution.
• Frequency ($\nu$): The number of revolutions completed in one second. $\nu = 1/T$.

We can relate these to speed and acceleration:

• $v = \frac{2\pi R}{T} = 2\pi R\nu$
• $\omega = \frac{2\pi}{T} = 2\pi\nu$
• $a_c = 4\pi^2\nu^2 R$

Given

• Radius, $R = 12 \text{ cm}$
• Number of revolutions = 7
• Total time = 100 s

To Find

(a) Angular speed $\omega$ and linear speed $v$. (b) Whether the acceleration vector is constant, and its magnitude $a$.

Formula

• Time period, $T = \frac{\text{Total time}}{\text{Number of revolutions}}$
• Angular speed, $\omega = \frac{2\pi}{T}$
• Linear speed, $v = \omega R$
• Acceleration magnitude, $a = \omega^2 R$

Solution

(a) Calculate angular and linear speed

First, find the time period for one revolution: $T = \frac{100 \text{ s}}{7 \text{ rev}} \approx 14.3 \text{ s/rev}$

Now, calculate the angular speed. The source calculates this using frequency: Frequency $\nu = \frac{7 \text{ rev}}{100 \text{ s}} = 0.07 \text{ s}^{-1}$. $\omega = 2\pi\nu = 2\pi \left(\frac{7}{100}\right) \approx 0.44 \text{ rad/s}$

Next, calculate the linear speed: $v = \omega R = (0.44 \text{ s}^{-1})(12 \text{ cm}) = 5.28 \text{ cm s}^{-1} \approx 5.3 \text{ cm s}^{-1}$

Answer for part (a) = The angular speed is $0.44 \text{ rad/s}$, and the linear speed is $5.3 \text{ cm s}^{-1}$.

(b) Analyze the acceleration vector

The acceleration vector is the centripetal acceleration, which is always directed towards the center of the circle. Since the insect is moving, its position changes, and therefore the direction of the "center" relative to the insect changes. The direction of the acceleration vector is continuously changing. Therefore, the acceleration vector is not a constant vector.

However, its magnitude is constant: $a = \omega^2 R = (0.44 \text{ s}^{-1})^2 (12 \text{ cm})$ $a = (0.1936)(12) \text{ cm s}^{-2} \approx 2.3 \text{ cm s}^{-2}$

Answer for part (b) = The acceleration vector is not constant because its direction changes. Its magnitude is constant and is equal to $2.3 \text{ cm s}^{-2}$.