"Motion in a Plane"

About

This chapter extends kinematics from one dimension to two dimensions. You will learn about projectile motion — objects moving under gravity following a parabolic path — and uniform circular motion where an object moves at constant speed along a circular path with acceleration directed toward the centre.

Key Concepts

4.1 Motion in Two Dimensions

When a body moves in a plane, its position, velocity, and acceleration are vector quantities with two components (x and y).

Position vector: $\vec{r}=x\hat{i}+y\hat{j}$

Velocity: $\vec{v}=v_x\hat{i}+v_y\hat{j}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$

Acceleration: $\vec{a}=a_x\hat{i}+a_y\hat{j}=\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}$

The motion along x and y axes can be treated independently.

4.2 Projectile Motion

A projectile is an object launched into the air and moving under the influence of gravity alone (neglecting air resistance).

Characteristics:

• Horizontal component of velocity ($v_x$) remains constant (no horizontal acceleration)
• Vertical component of velocity ($v_y$) changes due to gravity
• The path is a parabola

Key parameters for a projectile launched with speed $v_0$ at angle ${\theta }_0$:

Maximum range occurs at ${\theta }_0=45°$: $R_{\text{max}}=\frac{v_0^2}{g}$

Complementary angles ($\theta$ and $90°-\theta$) give the same range.

4.3 Uniform Circular Motion

When an object moves in a circular path with constant speed, it is said to be in uniform circular motion.

Key quantities:

Centripetal acceleration: $a_c=\frac{v^2}{r}={\omega }^{2}r$

• Directed toward the centre of the circle
• Magnitude is constant but direction continuously changes
• Velocity is NOT constant (direction changes continuously)
• Speed IS constant in uniform circular motion

Centripetal force: $F_c=\frac{m{v}^2}{r}=m{\omega }^{2}r$

4.4 Non-Uniform Circular Motion

In vertical circular motion, gravity causes the speed to vary:

• Speed increases as the object moves downward
• Speed decreases as the object moves upward
• Angular velocity ($\omega$) changes because speed changes

INTEXT QUESTIONS 4.1

Q1. Identify examples of projectile motion from among the following situations:

(a) An archer shoots an arrow at a target.

Ans: Yes, it is an example of projectile motion.

(b) Rocks are ejected from an exploding volcano.

Ans: Yes, it is an example of projectile motion.

(c) A truck moves on a mountainous road.

Ans: No, it is not an example of projectile motion.

(d) A bomb is released from a bomber plane.

Ans: Yes, it is an example of projectile motion. (At the time of release, the bomb shares the horizontal motion of the plane.)

(e) A boat sails in a river.

Ans: No, it is not an example of projectile motion.

Q2. Three balls thrown at different angles reach the same maximum height:

(a) Are the vertical components of the initial velocity the same for all the balls? If not, which one has the least vertical component?

Ans: Yes, the vertical components are the same for all three balls. Since $H=\frac{v_{0y}^2}{2g}$, the same maximum height means the same vertical component of initial velocity.

(b) Will they all have the same time of flight?

Ans: Yes, they will all have the same time of flight. $T=\frac{2v_0\sin {\theta }_0}{g}=\frac{2v_{0y}}{g}$, and since $v_{0y}$ is the same for all, the time of flight is the same.

(c) Which one has the greatest horizontal velocity component?

Ans: The ball thrown at the smallest angle has the largest $v_{0x}=v_0\cos {\theta }_0$. Since $v_{0y}=v_0\sin {\theta }_0$ is the same for all, a smaller ${\theta }_0$ means a larger $v_0$ and therefore a larger $v_{0x}$.

Q3. An athlete set the record for the long jump with a jump of 8.90 m. Assume his initial speed on take off to be 9.5 m s⁻¹. How close did he come to the maximum possible range in the absence of air resistance? Take g = 9.78 m s⁻².

Ans: Maximum possible range (at 45°): $R_{\text{max}}=\frac{v_0^2}{g}=\frac{(9.5{)}^2}{9.78}=\frac{90.25}{9.78}=9.23\text{ m}$

Difference = 9.23 − 8.90 = 0.33 m

Hence the record is only 0.33 m short of the ideal value.

INTEXT QUESTIONS 4.2

Q1. In uniform circular motion:

(a) Is the speed constant?

Ans: Yes. In uniform circular motion, the object moves along a circular path with constant speed (magnitude of velocity).

(b) Is the velocity constant?

Ans: No. Velocity is a vector, and though speed is constant, the direction of motion continuously changes, so the velocity is not constant.

(c) Is the magnitude of the acceleration constant?

Ans: Yes. The acceleration always points toward the centre of the circle and its magnitude is given by $a=\frac{v^2}{r}$ which stays constant if speed and radius remain constant.

(d) Is acceleration constant? Explain.

Ans: No. Although the magnitude is constant, the direction of acceleration keeps changing (always pointing toward the centre), so the acceleration vector is not constant.

Q2. In a vertical circular motion does the angular velocity of the body change? Explain.

Ans: Yes, in vertical circular motion angular velocity changes. This is because gravitational force affects the speed of the object — speed increases as the object moves downward and decreases as it goes upward. Since angular velocity depends on speed ($\omega =\frac{v}{r}$), it also changes.

Q3. An athlete runs around a circular track with a speed of 9.0 m s⁻¹ and a centripetal acceleration of 3 m s⁻². What is the radius of the track?

Ans: Given: $v=9.0$ m/s, $a_c=3$ m/s²

$a_c=\frac{v^2}{r}⟹r=\frac{v^2}{a_c}=\frac{(9.0{)}^2}{3}=\frac{81}{3}=27\text{ m}$

Answer: Radius = 27 m

Q4. The Fermi lab accelerator is one of the largest particle accelerators. In this accelerator, protons are forced to travel in an evacuated tube in a circular orbit of diameter 2.0 km at a speed which is nearly equal to 99.99995% of the speed of light. What is the centripetal acceleration of these protons? Take c = 3 × 10⁸ m s⁻¹.

Ans:

• Diameter = 2.0 km = 2000 m → Radius $r=1000$ m
• $v=0.9999995\times c=0.9999995\times 3\times 10^8\approx 3\times 10^8$ m/s

$a_c=\frac{v^2}{r}=\frac{(3\times 10^8{)}^2}{1000}=\frac{9\times 10^{16}}{10^3}=9\times 10^{13}{\text{ m/s}}^2$

Terminal Exercise

1. What is a projectile? Derive expressions for: (a) time of flight, (b) maximum height, (c) horizontal range of a projectile fired at an angle $\theta$ with the horizontal.

2. Prove that the path of a projectile is a parabola.

3. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the same ball?

4. Show that the range of a projectile is the same for complementary angles of projection.

5. A football is kicked at a velocity of 20 m/s at an angle of 30° with the horizontal. Calculate: (a) time of flight, (b) maximum height, (c) horizontal range. (g = 10 m/s²)

6. Define angular velocity and angular acceleration. Derive the relation $v=r\omega$.

7. Derive an expression for centripetal acceleration: $a_c=\frac{v^2}{r}$.

8. A body of mass 0.5 kg is moving on a circular path of radius 2 m with a speed of 4 m/s. Find the centripetal force acting on it.

9. Distinguish between uniform circular motion and non-uniform circular motion. Give one example of each.

10. The moon revolves around the earth in a nearly circular orbit of radius $3.84\times 10^8$ m and takes 27.3 days to complete one revolution. Calculate the centripetal acceleration of the moon.

Worked Examples

Example 1: Projectile Range

Problem: A ball is projected with a velocity of 40 m/s at an angle of 30° with the horizontal. Find the horizontal range. (g = 10 m/s²)

Solution: $R=\frac{v_0^2\sin 2{\theta }_0}{g}=\frac{(40{)}^2\times \sin 60°}{10}=\frac{1600\times 0.866}{10}=138.6\text{ m}$

Example 2: Maximum Height

Problem: For the same ball in Example 1, find the maximum height reached.

Solution: $H=\frac{v_0^2{\sin }^2{\theta }_0}{2g}=\frac{(40{)}^2\times {\sin }^{2}30°}{2\times 10}=\frac{1600\times 0.25}{20}=20\text{ m}$

Example 3: Centripetal Force

Problem: A car of mass 1000 kg takes a circular turn of radius 50 m at a speed of 10 m/s. Find the centripetal force required.

Solution: $F_c=\frac{m{v}^2}{r}=\frac{1000\times (10{)}^2}{50}=\frac{1000\times 100}{50}=2000\text{ N}$

Example 4: Angular Velocity

Problem: A particle moves in a circle of radius 0.5 m with a constant speed of 3 m/s. Find its angular velocity and time period.

Solution: $\omega =\frac{v}{r}=\frac{3}{0.5}=6\text{ rad/s}$

$T=\frac{2\pi }{\omega }=\frac{2\pi }{6}=\frac{\pi }{3}\approx 1.05\text{ s}$

Common Mistakes

1. Forgetting that horizontal velocity is constant in projectile motion: $v_x$ never changes (ignoring air resistance).
2. Using the range formula when launch and landing heights differ: $R=\frac{v_0^2\sin 2\theta }{g}$ works only when launch and landing are at the same height.
3. Confusing speed with velocity in circular motion: Speed is constant but velocity is NOT (direction changes).
4. Thinking centripetal acceleration is a separate force: It's the acceleration caused by the net force toward the centre — not a new type of force.
5. Using degrees instead of radians in angular quantities: $\omega$ must be in rad/s for $v=r\omega$ to work.

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