Introduction to Two-Dimensional Motion
Dear student, in Grade 11 you studied motion in one dimension — objects moving along a straight line. But in real life, most objects move along curved paths. A ball kicked by a football player, a stone thrown at an angle, a bicycle turning a corner, and planets orbiting the Sun — all these are examples of two-dimensional motion.
In this unit, we will extend what you already know about one-dimensional kinematics to motion in a plane. The key idea is simple: any two-dimensional motion can be broken into two independent one-dimensional motions — usually along the x-axis and y-axis.
Are you ready? Let us begin step by step!
2.1 Projectile Motion
What is a Projectile?
A projectile is any object that is thrown or projected into the air and moves freely under the influence of gravity alone. Once the object is in the air, the only force acting on it is gravity (we ignore air resistance unless told otherwise).
Examples: a ball kicked in football, a stone thrown from a cliff, water from a fountain, a bullet fired from a gun.
Key Assumptions in Projectile Motion
- Air resistance is neglected.
- Acceleration due to gravity g is constant (taken as 9.8 m/s² or 10 m/s² in exams).
- The horizontal component of velocity remains constant (no horizontal force).
- The vertical motion is free fall under gravity.
Breaking Motion into Components
When a projectile is launched with initial velocity v₀ at an angle θ above the horizontal, we resolve this velocity into two components:
Think about it: why do we separate the motion like this? It is because the horizontal and vertical motions are independent of each other. The horizontal velocity never changes, while the vertical velocity changes due to gravity.
Horizontal Motion Equations
Since there is no horizontal acceleration:
Vertical Motion Equations
The vertical motion is free fall (acceleration = −g, taking upward as positive):
Important Derived Formulas
Now let us derive three very important results that appear in almost every exam!
(a) Time of Flight (T)
The projectile returns to the ground when y = 0:
This is the total time the projectile stays in the air.
(b) Maximum Height (H)
At the highest point, the vertical velocity becomes zero: vᵧ = 0
(c) Horizontal Range (R)
The range is the horizontal distance traveled when the projectile lands (at time T):
Can you guess why a football player sometimes kicks the ball at different angles? Yes — the angle changes the height and range of the ball’s path!
Worked Example 1
A ball is thrown with an initial velocity of 20 m/s at an angle of 30° above the horizontal. Find: (a) the time of flight, (b) the maximum height, (c) the horizontal range. (Take g = 10 m/s²)
Given: v₀ = 20 m/s, θ = 30°, g = 10 m/s²
Components:
v₀ₓ = 20 cos 30° = 20 × 0.866 = 17.32 m/s
v₀ᵧ = 20 sin 30° = 20 × 0.5 = 10 m/s
(a) Time of flight:
$$T = \frac{2v_0 \sin\theta}{g} = \frac{2 \times 20 \times \sin 30°}{10} = \frac{2 \times 20 \times 0.5}{10} = \frac{20}{10} = 2 \text{ s}$$
(b) Maximum height:
$$H = \frac{v_0^2 \sin^2\theta}{2g} = \frac{400 \times 0.25}{20} = \frac{100}{20} = 5 \text{ m}$$
(c) Range:
$$R = \frac{v_0^2 \sin 2\theta}{g} = \frac{400 \times \sin 60°}{10} = \frac{400 \times 0.866}{10} = 34.64 \text{ m}$$
Worked Example 2
A stone is thrown horizontally from the top of a cliff 45 m high with a speed of 15 m/s. Find: (a) the time it takes to reach the ground, (b) the horizontal distance from the base of the cliff where it lands, (c) the velocity just before hitting the ground. (g = 10 m/s²)
Since it is thrown horizontally: θ = 0°
v₀ₓ = 15 m/s, v₀ᵧ = 0
(a) Time to reach ground: Using vertical motion: y = v₀ᵧt + ½gt²
45 = 0 + ½(10)t² → 45 = 5t² → t² = 9 → t = 3 s
(b) Horizontal distance (range):
$$R = v_{0x} \cdot t = 15 \times 3 = 45 \text{ m}$$
(c) Velocity just before hitting:
vₓ = 15 m/s (constant)
vᵧ = v₀ᵧ + gt = 0 + 10 × 3 = 30 m/s (downward)
$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 30^2} = \sqrt{225 + 900} = \sqrt{1125} = 33.54 \text{ m/s}$$
Angle below horizontal: tan α = vᵧ/vₓ = 30/15 = 2 → α = 63.4° below horizontal
Worked Example 3
For what angle of projection is the horizontal range maximum? Prove that complementary angles give the same range.
Range: R = v₀² sin 2θ / g
R is maximum when sin 2θ is maximum. The maximum value of sine is 1, which occurs when 2θ = 90°, so θ = 45°.
Complementary angles: Let θ₁ and θ₂ be complementary, so θ₂ = 90° − θ₁.
sin 2θ₁ = sin(2θ₁)
sin 2θ₂ = sin[2(90° − θ₁)] = sin(180° − 2θ₁) = sin 2θ₁
Since sin 2θ₁ = sin 2θ₂, the ranges are equal! ✓
For example: 30° and 60° give the same range.
- At the highest point: vᵧ = 0, but vₓ ≠ 0 (the projectile still moves horizontally!)
- The trajectory is a parabola.
- Time to reach max height = T/2 = v₀ sinθ / g
- Horizontal range formula: R = v₀² sin 2θ / g
- For horizontal projection from height h: T = √(2h/g), R = v₀√(2h/g)
- Complementary angles give equal ranges but different heights.
- 45° gives maximum range for a given initial speed.
Quick Practice: A ball is projected at 40 m/s at an angle of 45°. What is its range? Try it before checking the answer!
Practice Question 1: A projectile is fired at 50 m/s at an angle of 60°. Find its maximum height and time of flight. (g = 10 m/s²)
Maximum height:
$$H = \frac{(43.3)^2}{2 \times 10} = \frac{1874.89}{20} = 93.74 \text{ m}$$
Time of flight:
$$T = \frac{2 \times 43.3}{10} = 8.66 \text{ s}$$
Practice Question 2: A ball is thrown from the top of a building 20 m high with a horizontal velocity of 8 m/s. How far from the building does it land? (g = 10 m/s²)
Range: R = v₀ₓ × t = 8 × 2 = 16 m
Practice Question 3: Show that for a projectile, the horizontal range is the same when the angle of projection is 25° and 65°.
R₂ = v₀² sin(2 × 65°)/g = v₀² sin 130°/g
Since sin 130° = sin(180° − 130°) = sin 50°
Therefore R₁ = R₂ ✓
Note: 25° + 65° = 90°, so they are complementary angles, and complementary angles always give equal ranges.
2.2 Rotational Motion
From Linear to Angular
Dear student, so far you have been describing motion using quantities like displacement, velocity, and acceleration. These are linear (translational) quantities. But what about a wheel spinning, a fan rotating, or the Earth rotating on its axis? For these, we need angular quantities.
The good news is that every linear quantity has an angular “partner”!
| Linear Quantity | Symbol | Angular Quantity | Symbol |
|---|---|---|---|
| Displacement | x | Angular displacement | θ (theta) |
| Velocity | v | Angular velocity | ω (omega) |
| Acceleration | a | Angular acceleration | α (alpha) |
Angular Displacement (θ)
This is the angle through which an object rotates. It is measured in radians (rad).
where s = arc length, r = radius. One complete revolution = 2π rad = 360°.
Angular Velocity (ω)
Angular velocity is the rate of change of angular displacement:
Unit: rad/s (radians per second).
Angular Acceleration (α)
Unit: rad/s².
Relationship Between Linear and Angular Quantities
Notice there are two types of acceleration in circular motion: tangential (aₜ) changes the speed, and centripetal (aᶜ) changes the direction. The total acceleration is:
Uniform Circular Motion
When an object moves in a circle at constant speed, we call it uniform circular motion. The speed is constant, but the velocity is NOT constant because the direction keeps changing. Therefore, there is acceleration!
Centripetal Acceleration
The acceleration that points toward the center of the circle is called centripetal acceleration:
Centripetal Force
By Newton’s second law, F = ma. The force that keeps an object moving in a circle is:
Period and Frequency
Worked Example 4
A car of mass 1200 kg moves around a circular track of radius 80 m at a constant speed of 20 m/s. Find: (a) the centripetal acceleration, (b) the centripetal force, (c) the period of one revolution.
Given: m = 1200 kg, r = 80 m, v = 20 m/s
(a) Centripetal acceleration:
$$a_c = \frac{v^2}{r} = \frac{400}{80} = 5 \text{ m/s}^2$$
(b) Centripetal force:
$$F_c = ma_c = 1200 \times 5 = 6000 \text{ N}$$
(c) Period:
$$T = \frac{2\pi r}{v} = \frac{2\pi \times 80}{20} = 8\pi \approx 25.13 \text{ s}$$
Worked Example 5
A wheel of radius 0.3 m starts from rest and accelerates uniformly at 2 rad/s². Find: (a) the angular velocity after 5 s, (b) the linear speed of a point on the rim at t = 5 s, (c) the total angle turned in 5 s.
Given: r = 0.3 m, ω₀ = 0, α = 2 rad/s², t = 5 s
(a) Angular velocity:
$$\omega = \omega_0 + \alpha t = 0 + 2 \times 5 = 10 \text{ rad/s}$$
(b) Linear speed:
$$v = r\omega = 0.3 \times 10 = 3 \text{ m/s}$$
(c) Angle turned:
$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(2)(25) = 25 \text{ rad}$$
Worked Example 6 (Conical Pendulum)
A string of length 1 m has a mass of 0.5 kg tied to its end. The mass is set into circular motion in a horizontal circle so that the string makes an angle of 30° with the vertical. Find the speed of the mass and the tension in the string. (g = 10 m/s²)
The radius of the circle: r = L sin θ = 1 × sin 30° = 0.5 m
Forces on the mass: Tension T along the string, weight mg downward.
Vertically: T cos θ = mg → T cos 30° = 0.5 × 10 = 5
$$T = \frac{5}{\cos 30°} = \frac{5}{0.866} = 5.77 \text{ N}$$
Horizontally (centripetal direction): T sin θ = mv²/r
$$5.77 \times \sin 30° = \frac{0.5 \times v^2}{0.5}$$ $$5.77 \times 0.5 = v^2$$ $$v^2 = 2.885 \rightarrow v = 1.70 \text{ m/s}$$
- In uniform circular motion: speed is constant but velocity changes (direction changes).
- Centripetal force always points toward the CENTER — never outward.
- “Centrifugal force” is a fictitious force — it does NOT exist in an inertial frame!
- v = rω, aᶜ = v²/r = rω², Fᶜ = mv²/r = mrω²
- T = 2π/ω = 2πr/v
- The angular equations mirror linear equations: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ
Practice Question 4: A particle moves in a circle of radius 2 m with a speed of 4 m/s. What is its centripetal acceleration and angular velocity?
Practice Question 5: A disc of radius 0.5 m rotates at 300 rev/min. Find the angular velocity in rad/s and the linear speed of a point on the edge.
Practice Question 6: A 1500 kg car rounds a flat curve of radius 60 m at 15 m/s. What minimum coefficient of friction is needed to prevent the car from skidding? (g = 10 m/s²)
μmg = mv²/r
μ = v²/(rg) = 225/(60 × 10) = 225/600 = 0.375
2.3 Rotational Dynamics
What is Torque?
In linear motion, force causes acceleration. In rotational motion, torque (also called moment of force) causes angular acceleration. Torque measures the “turning effect” of a force.
where r is the distance from the axis of rotation to the point where the force is applied, θ is the angle between r and F, and r⊥ = r sin θ is the perpendicular distance (moment arm).
Unit: N·m (Newton-meter).
Moment of Inertia (I)
In linear motion, mass measures resistance to acceleration (F = ma). In rotational motion, moment of inertia measures resistance to angular acceleration:
For a continuous body: I = ∫r² dm. Unit: kg·m².
Common Moments of Inertia
| Object | Axis | Moment of Inertia |
|---|---|---|
| Thin rod (length L, mass M) | Through center, ⊥ to rod | I = ML²/12 |
| Thin rod (length L, mass M) | Through end, ⊥ to rod | I = ML²/3 |
| Solid disc/cylinder (radius R, mass M) | Through center, ⊥ to face | I = MR²/2 |
| Hoop/ring (radius R, mass M) | Through center, ⊥ to plane | I = MR² |
| Solid sphere (radius R, mass M) | Through center | I = 2MR²/5 |
| Hollow sphere (radius R, mass M) | Through center | I = 2MR²/3 |
| Point mass (mass m, distance r) | Through axis | I = mr² |
Newton’s Second Law for Rotation
This is the rotational equivalent of F = ma. The net torque equals moment of inertia times angular acceleration.
Angular Momentum (L)
Unit: kg·m²/s. Angular momentum is the rotational equivalent of linear momentum (p = mv).
Conservation of Angular Momentum
If no external torque acts on a system, the total angular momentum is conserved:
Rotational Kinetic Energy
For an object that both translates and rotates (like a rolling ball), the total KE is:
Work-Energy Theorem for Rotation
Worked Example 7
A force of 20 N is applied tangentially to a disc of radius 0.4 m and mass 5 kg that is free to rotate about its center. Find: (a) the torque, (b) the angular acceleration, (c) the angular velocity after 3 seconds starting from rest.
I (disc) = MR²/2 = 5 × (0.4)²/2 = 5 × 0.16/2 = 0.4 kg·m²
(a) Torque: Force is tangential, so θ = 90°
$$\tau = rF = 0.4 \times 20 = 8 \text{ N·m}$$
(b) Angular acceleration:
$$\alpha = \frac{\tau}{I} = \frac{8}{0.4} = 20 \text{ rad/s}^2$$
(c) Angular velocity at t = 3 s:
$$\omega = \omega_0 + \alpha t = 0 + 20 \times 3 = 60 \text{ rad/s}$$
Worked Example 8
A disc rotating at 10 rad/s is brought to rest by a frictional torque of 4 N·m in 5 seconds. Find the moment of inertia of the disc.
ω₀ = 10 rad/s, ω = 0, t = 5 s, τ = 4 N·m
Angular deceleration: α = (ω − ω₀)/t = (0 − 10)/5 = −2 rad/s²
$$I = \frac{\tau}{|\alpha|} = \frac{4}{2} = 2 \text{ kg·m}^2$$
Worked Example 9
A figure skater with arms extended has a moment of inertia of 4 kg·m² and rotates at 2 rad/s. She pulls her arms in, reducing her moment of inertia to 1.6 kg·m². What is her new angular velocity?
Conservation of angular momentum (no external torque):
$$I_1\omega_1 = I_2\omega_2$$ $$4 \times 2 = 1.6 \times \omega_2$$ $$\omega_2 = \frac{8}{1.6} = 5 \text{ rad/s}$$
Her angular speed increased from 2 to 5 rad/s — she spins 2.5 times faster!
- Torque τ = rF sinθ; maximum when θ = 90°, zero when θ = 0°.
- Newton’s 2nd law for rotation: τ = Iα (compare with F = ma).
- Parallel axis theorem: I = I_cm + Md² (useful when axis is not through center of mass).
- Conservation of angular momentum: I₁ω₁ = I₂ω₂ (when τ_ext = 0).
- Rolling without slipping: v = rω, a = rα.
- For a solid sphere rolling: KE = ½mv² + ½(2mr²/5)(v²/r²) = 7/10 mv²
- For a hollow sphere rolling: KE = ½mv² + ½(2mr²/3)(v²/r²) = 5/6 mv²
Practice Question 7: A uniform rod of mass 3 kg and length 2 m is pivoted at one end. What torque is needed to give it an angular acceleration of 4 rad/s²?
$$\tau = I\alpha = 4 \times 4 = 16 \text{ N·m}$$
Practice Question 8: A flywheel (I = 0.5 kg·m²) rotating at 100 rad/s is slowed to 20 rad/s by a constant frictional torque in 10 s. Find the frictional torque and the work done by friction.
τ = I|α| = 0.5 × 8 = 4 N·m
Work done by friction = ΔKE
$$W = \frac{1}{2}(0.5)(20^2) – \frac{1}{2}(0.5)(100^2) = 100 – 2500 = -2400 \text{ J}$$ (The negative sign means energy is removed from the system.)
Practice Question 9: A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down an incline. At the bottom, its speed is 4 m/s. Find its total kinetic energy at the bottom.
ω = v/r = 4/0.1 = 40 rad/s
$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}(2)(16) + \frac{1}{2}(0.008)(1600)$$ $$= 16 + 6.4 = 22.4 \text{ J}$$
Alternatively: KE = 7/10 mv² = 7/10 × 2 × 16 = 22.4 J ✓
2.4 Planetary Motion and Kepler’s Laws
Historical Background
For centuries, people believed that planets moved in perfect circles. Johannes Kepler (1571–1630), using the careful observations of Tycho Brahe, discovered that planetary orbits are NOT circles — they are ellipses. He formulated three laws that describe planetary motion.
Kepler’s First Law — The Law of Ellipses
An ellipse has two foci. The Sun is at one focus, not at the center! The distance from the center to a focus is called c, and the semi-major axis is a. The eccentricity e = c/a tells us how “stretched” the ellipse is (e = 0 is a circle, e close to 1 is very elongated).
Kepler’s Second Law — The Law of Equal Areas
This law tells us that a planet moves faster when it is closer to the Sun and slower when it is farther away. This is a consequence of conservation of angular momentum!
Kepler’s Third Law — The Law of Periods
For planets orbiting the Sun, the constant is the same for ALL planets. If T₁ and r₁ are for planet 1, and T₂ and r₂ are for planet 2:
Worked Example 10
The Earth orbits the Sun with a period of 1 year at a distance of 1 AU (astronomical unit). Mars orbits at a distance of 1.52 AU. Find the orbital period of Mars in years.
Using Kepler’s Third Law:
$$\frac{T_{Earth}^2}{r_{Earth}^3} = \frac{T_{Mars}^2}{r_{Mars}^3}$$ $$\frac{1^2}{1^3} = \frac{T_{Mars}^2}{(1.52)^3}$$ $$T_{Mars}^2 = (1.52)^3 = 3.512$$ $$T_{Mars} = \sqrt{3.512} = 1.87 \text{ years}$$
Mars takes about 1.87 Earth years to go around the Sun once.
Worked Example 11
A satellite orbits the Earth at a distance of 7000 km from the center of the Earth. Another satellite orbits at 42000 km from the center. If the first satellite has a period of 97 minutes, find the period of the second satellite.
$$\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$$ $$\frac{97^2}{7000^3} = \frac{T_2^2}{42000^3}$$ $$T_2^2 = 97^2 \times \left(\frac{42000}{7000}\right)^3 = 9409 \times 6^3 = 9409 \times 216 = 2,032,344$$ $$T_2 = \sqrt{2,032,344} = 1425.5 \text{ minutes} \approx 23.8 \text{ hours}$$
(This is close to a geostationary orbit period of 24 hours!)
- 1st Law: Orbits are ellipses, Sun at one focus (not center).
- 2nd Law: Equal areas in equal times → faster near Sun, slower far away.
- 3rd Law: T²/r³ = constant (same for all planets around the same star).
- Kepler’s 2nd Law is a result of conservation of angular momentum.
- Kepler’s 3rd Law can be derived from Newton’s law of gravitation.
- For satellites around Earth: T²/r³ = 4π²/(GM_earth).
Practice Question 10: A planet has an orbital radius 4 times that of Earth. What is its orbital period in Earth years?
Practice Question 11: According to Kepler’s second law, at which point in its orbit does a planet have the greatest speed? Explain why.
Reason: By Kepler’s second law, the line joining the Sun and planet sweeps equal areas in equal times. Near the Sun, the radius is shorter, so the planet must move faster to sweep the same area. This is also explained by conservation of angular momentum: when r decreases, ω must increase to keep L = mr²ω constant.
2.5 Newton’s Law of Universal Gravitation
The Law
Isaac Newton proposed that every particle in the universe attracts every other particle with a force that is:
This is a very small number, which is why gravitational force between everyday objects is too small to feel. It only becomes significant when at least one object is very massive (like a planet).
Key Properties of Gravitational Force
- It is always attractive (never repulsive).
- It acts along the line joining the centers of the two masses.
- It obeys Newton’s third law: F₁₂ = −F₂₁ (equal and opposite).
- It is an inverse-square law: F ∝ 1/r².
- It is independent of the medium between the masses.
Gravitational Field Strength (g)
The gravitational field strength at a point is the force per unit mass experienced by a small test mass placed at that point:
On the surface of the Earth (r = R):
At a height h above the surface (r = R + h):
Variation of g with Height
As height increases, g decreases. At very large heights, g approaches zero.
Orbital Velocity
For a satellite in a circular orbit, the gravitational force provides the centripetal force:
For a satellite orbiting close to the Earth’s surface (r ≈ R):
Geostationary Satellite
A geostationary satellite orbits the Earth with a period of exactly 24 hours, so it appears stationary above a fixed point on the equator. Key properties:
- Period T = 24 hours = 86400 s
- Orbit must be in the equatorial plane
- Orbital radius r ≈ 42,400 km (about 35,800 km above surface)
Escape Velocity
The minimum speed needed for an object to escape from a planet’s gravitational pull (without further propulsion):
Gravitational Potential Energy
The gravitational potential energy of two masses separated by distance r (taking U = 0 at r = ∞):
The negative sign means you need to ADD energy to separate the objects to infinity.
Worked Example 12
Calculate the gravitational force between the Earth (M = 6 × 10²⁴ kg) and the Moon (m = 7.4 × 10²² kg). The distance between their centers is 3.84 × 10⁸ m. (G = 6.67 × 10⁻¹¹ N·m²/kg²)
$$F = G\frac{Mm}{r^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 7.4 \times 10^{22}}{(3.84 \times 10^8)^2}$$ $$F = \frac{6.67 \times 6 \times 7.4 \times 10^{35}}{14.75 \times 10^{16}}$$ $$F = \frac{296.11 \times 10^{35}}{14.75 \times 10^{16}} = 20.08 \times 10^{19} \approx 2 \times 10^{20} \text{ N}$$
Worked Example 13
At what height above the Earth’s surface does the acceleration due to gravity become 25% of its value on the surface? (R_earth = 6400 km)
g_h = 0.25g = g[R/(R+h)]²
0.25 = [R/(R+h)]²
√0.25 = R/(R+h)
0.5 = R/(R+h)
R + h = 2R
h = R = 6400 km
At a height equal to the Earth’s radius, g drops to 25% of its surface value!
Worked Example 14
Find the orbital velocity and period of a satellite orbiting 300 km above the Earth’s surface. (R = 6400 km, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹)
r = R + h = 6400 + 300 = 6700 km = 6.7 × 10⁶ m
Orbital velocity:
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.7 \times 10^6}}$$ $$v = \sqrt{\frac{40.02 \times 10^{13}}{6.7 \times 10^6}} = \sqrt{5.976 \times 10^7} = 7731 \text{ m/s} \approx 7.73 \text{ km/s}$$
Period:
$$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.7 \times 10^6}{7731} = \frac{4.21 \times 10^7}{7731} = 5445 \text{ s} \approx 90.8 \text{ minutes}$$
Worked Example 15
Calculate the escape velocity from the Earth’s surface. (g = 10 m/s², R = 6400 km)
$$v_{escape} = \sqrt{2gR} = \sqrt{2 \times 10 \times 6.4 \times 10^6}$$ $$= \sqrt{1.28 \times 10^8} = 1.13 \times 10^4 \text{ m/s} = 11.3 \text{ km/s}$$
(More precisely, using g = 9.8 m/s², we get 11.2 km/s.)
- F = GMm/r² — always attractive, inverse-square law.
- G = 6.67 × 10⁻¹¹ N·m²/kg² (very small — know this value!).
- g = GM/R² on surface; g_h = GM/(R+h)² at height h.
- Orbital velocity: v = √(GM/r) = √(gR) near surface ≈ 8 km/s.
- Escape velocity: v_e = √(2GM/r) = √(2gR) ≈ 11.2 km/s.
- v_escape = √2 × v_orbit.
- Geostationary satellite: T = 24 h, orbits above equator, r ≈ 42,400 km.
- Gravitational PE: U = −GMm/r (negative!).
- Kepler’s 3rd law from gravitation: T² = 4π²r³/(GM).
Practice Question 12: Two spheres of masses 10 kg and 20 kg are placed 2 m apart. Find the gravitational force between them. (G = 6.67 × 10⁻¹¹)
This is an extremely small force — which is why we don’t feel gravitational attraction between everyday objects!
Practice Question 13: The acceleration due to gravity on the surface of a planet is 4 m/s². If the planet’s radius is half that of Earth and Earth’s g = 10 m/s², find the mass of the planet in terms of Earth’s mass.
g_E = GM_E/R² = 10
$$\frac{g_p}{g_E} = \frac{4GM_p/R^2}{GM_E/R^2} = \frac{4M_p}{M_E}$$ $$\frac{4}{10} = \frac{4M_p}{M_E}$$ $$M_p = \frac{M_E}{10} = 0.1M_E$$
Practice Question 14: A satellite is in a circular orbit at a height of 600 km above Earth. If the radius of Earth is 6400 km and g on the surface is 10 m/s², find the orbital velocity of the satellite.
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{gR^2}{r}} = \sqrt{\frac{10 \times (6.4 \times 10^6)^2}{7 \times 10^6}}$$ $$= \sqrt{\frac{10 \times 4.096 \times 10^{13}}{7 \times 10^6}} = \sqrt{\frac{4.096 \times 10^{14}}{7 \times 10^6}}$$ $$= \sqrt{5.851 \times 10^7} = 7650 \text{ m/s} \approx 7.65 \text{ km/s}$$
Unit 2 Summary
In this unit, you learned five major topics:
- Projectile Motion: Breaking 2D motion into independent horizontal (constant velocity) and vertical (free fall) components. Key formulas: T = 2v₀sinθ/g, H = v₀²sin²θ/2g, R = v₀²sin2θ/g.
- Rotational Motion: Describing circular motion using angular quantities. Key relationships: v = rω, aᶜ = v²/r = rω², Fᶜ = mv²/r.
- Rotational Dynamics: Torque causes angular acceleration (τ = Iα). Conservation of angular momentum: I₁ω₁ = I₂ω₂.
- Kepler’s Laws: Elliptical orbits, equal areas in equal times, T² ∝ r³.
- Universal Gravitation: F = GMm/r², orbital velocity v = √(GM/r), escape velocity v_e = √(2GM/r).
Quick Revision Notes — Unit 2: Two-Dimensional Motion
2.1 Projectile Motion — Key Formulas
| Quantity | Formula |
|---|---|
| Horizontal velocity | vₓ = v₀ cosθ (constant) |
| Vertical velocity at time t | vᵧ = v₀ sinθ − gt |
| Horizontal displacement | x = v₀ cosθ · t |
| Vertical displacement | y = v₀ sinθ · t − ½gt² |
| Time of flight | T = 2v₀ sinθ / g |
| Maximum height | H = v₀² sin²θ / 2g |
| Horizontal range | R = v₀² sin 2θ / g |
| Time to max height | t_up = v₀ sinθ / g = T/2 |
| Horizontal projection (from height h) | T = √(2h/g), R = v₀√(2h/g) |
- At max height: vᵧ = 0, only vₓ exists.
- Max range at θ = 45°.
- Complementary angles → same range, different heights.
- For horizontal projection: initial vertical velocity = 0.
- Always draw a diagram and resolve velocities into components first!
2.2 Rotational Motion — Key Formulas
| Quantity | Formula |
|---|---|
| Angular velocity | ω = Δθ/Δt |
| Angular acceleration | α = Δω/Δt |
| Linear-angular relation | v = rω |
| Tangential acceleration | aₜ = rα |
| Centripetal acceleration | aᶜ = v²/r = rω² |
| Centripetal force | Fᶜ = mv²/r = mrω² |
| Period | T = 2π/ω = 2πr/v |
| Frequency | f = 1/T = ω/2π |
| Angular kinematics | ω = ω₀ + αt; θ = ω₀t + ½αt²; ω² = ω₀² + 2αθ |
- Centripetal force is NOT a separate force — it is the net inward force.
- “Centrifugal force” does not exist in inertial frames — do not use it!
- For a car on a curve: friction provides centripetal force → μmg = mv²/r.
- For conical pendulum: T sinθ = mv²/r (horizontal), T cosθ = mg (vertical).
2.3 Rotational Dynamics — Key Formulas
| Quantity | Formula |
|---|---|
| Torque | τ = rF sinθ |
| Moment of inertia (point mass) | I = mr² |
| Solid disc/cylinder | I = MR²/2 |
| Thin hoop/ring | I = MR² |
| Solid sphere | I = 2MR²/5 |
| Hollow sphere | I = 2MR²/3 |
| Thin rod (center) | I = ML²/12 |
| Thin rod (end) | I = ML²/3 |
| Newton’s 2nd law (rotation) | τ_net = Iα |
| Angular momentum | L = Iω |
| Conservation of L | I₁ω₁ = I₂ω₂ |
| Rotational KE | KE = ½Iω² |
| Parallel axis theorem | I = I_cm + Md² |
| Rolling KE (solid sphere) | KE = 7/10 mv² |
- τ = Iα is the rotational version of F = ma — always identify all torques first.
- For rolling without slipping: v = rω and a = rα always hold.
- Conservation of angular momentum: only when net external torque = 0.
- When comparing rolling objects down an incline: the one with the smallest I/MR² ratio reaches the bottom first (solid sphere > solid disc > hollow sphere > hoop).
2.4 Kepler’s Laws — Quick Summary
| Law | Statement | Formula |
|---|---|---|
| 1st (Ellipses) | Planets move in elliptical orbits; Sun at one focus | e = c/a |
| 2nd (Equal Areas) | Equal areas swept in equal times | dA/dt = L/(2m) = constant |
| 3rd (Periods) | T² proportional to r³ | T²/r³ = constant |
2.5 Universal Gravitation — Key Formulas
| Quantity | Formula |
|---|---|
| Gravitational force | F = GMm/r² |
| g on surface | g = GM/R² |
| g at height h | g_h = GM/(R+h)² = g[R/(R+h)]² |
| Orbital velocity | v = √(GM/r) = √(gR) near surface |
| Escape velocity | v_e = √(2GM/r) = √(2gR) |
| Period of satellite | T = 2π√(r³/GM) |
| Gravitational PE | U = −GMm/r |
| Kepler’s 3rd (from gravitation) | T² = 4π²r³/(GM) |
- G = 6.67 × 10⁻¹¹ N·m²/kg²
- g = 9.8 m/s² (or 10 m/s² in exams)
- R_Earth = 6400 km = 6.4 × 10⁶ m
- M_Earth = 6 × 10²⁴ kg
- v_orbit (near surface) ≈ 8 km/s
- v_escape (from surface) ≈ 11.2 km/s
- Geostationary height ≈ 35,800 km above surface
Common Mistakes to Avoid
- Projectile: Forgetting that horizontal velocity is constant. Do NOT use g in horizontal equations!
- Projectile: Confusing sinθ and cosθ — always draw the diagram first.
- Projectile: Taking g as positive in upward direction. Always define your positive direction clearly.
- Circular motion: Writing centripetal force as an additional force. It is the NET force, not a separate force.
- Circular motion: Using “centrifugal force” as a real force in inertial frames.
- Rotational dynamics: Using wrong moment of inertia formula — check the axis of rotation!
- Rotational dynamics: Forgetting units — torque is in N·m, I is in kg·m².
- Gravitation: Using surface g formula at height without adjustment.
- Gravitation: Forgetting that r in F = GMm/r² is the distance between CENTERS, not surfaces.
- Kepler’s 3rd law: Forgetting that T²/r³ is constant only for objects orbiting the SAME central body.
Important Definitions
- Projectile: Any object launched into the air that moves freely under gravity alone.
- Trajectory: The path followed by a projectile (a parabola when air resistance is neglected).
- Angular displacement: The angle through which a body rotates, measured in radians.
- Angular velocity: The rate of change of angular displacement.
- Centripetal acceleration: The acceleration directed toward the center of a circular path, aᶜ = v²/r.
- Centripetal force: The net force directed toward the center that keeps an object in circular motion.
- Torque: The turning effect of a force, τ = rF sinθ.
- Moment of inertia: A measure of an object’s resistance to angular acceleration, I = Σmr².
- Angular momentum: The rotational analog of linear momentum, L = Iω.
- Perihelion: The closest point of a planet’s orbit to the Sun.
- Aphelion: The farthest point of a planet’s orbit from the Sun.
- Geostationary orbit: An orbit where the satellite has the same period as Earth’s rotation (24 h), appearing stationary.
- Escape velocity: The minimum speed needed to escape a planet’s gravitational field.
Challenge Exam Questions — Unit 2
Section A: Multiple Choice Questions
Question 1: A projectile is fired at an angle of 45° with the horizontal. The horizontal range is 200 m. If the initial speed is doubled, the new range will be:
R = v₀² sin 2θ / g. At 45°, sin 90° = 1, so R = v₀²/g.
If v₀ is doubled: R_new = (2v₀)²/g = 4v₀²/g = 4R = 4 × 200 = 800 m.
Range is proportional to the SQUARE of initial velocity. This is a common trap!
Question 2: The centripetal force on a car rounding a curve is provided by:
On a flat (unbanked) curve, the only horizontal force is friction. This friction provides the centripetal force needed to keep the car moving in a circle. Without friction, the car would slide outward in a straight line (Newton’s first law).
Question 3: Two satellites A and B orbit the Earth. Satellite A is at distance r from the center and satellite B is at distance 2r. The ratio of their orbital periods T_A : T_B is:
By Kepler’s 3rd law: T² ∝ r³
T_A² / T_B² = r³ / (2r)³ = 1/8
T_A / T_B = 1/√8 = 1/(2√2)
So T_A : T_B = 1 : 2√2
Question 4: A hollow sphere and a solid sphere of the same mass and radius are released from the same height on an inclined plane. Which reaches the bottom first?
For rolling objects, more rotational KE means less translational KE, so slower speed.
Solid sphere: I = 2MR²/5, KE_total = 7/10 mv²
Hollow sphere: I = 2MR²/3, KE_total = 5/6 mv²
The solid sphere has a smaller I/MR² ratio (0.4 vs 0.667), so more energy goes into translational motion. It reaches the bottom first.
Question 5: At what angle of projection does a projectile have equal horizontal range and maximum height? (Given: tan θ = 4)
Set R = H:
v₀² sin 2θ / g = v₀² sin²θ / 2g
2 sin 2θ = sin²θ
2(2 sinθ cosθ) = sin²θ
4 cosθ = sinθ
tanθ = 4
θ = arctan(4) ≈ 76°
Question 6: The escape velocity from the Earth is 11.2 km/s. If the radius of a planet is double that of Earth and its density is the same, the escape velocity from this planet is:
v_e = √(2GM/R). If density ρ is the same and R’ = 2R:
M’ = ρ(4/3)πR’³ = ρ(4/3)π(2R)³ = 8M
v_e’ = √(2G × 8M / 2R) = √(8GM/R) = √4 × √(2GM/R) = 2v_e = 2 × 11.2 = 22.4 km/s
Key insight: Same density + double radius → mass increases by factor of 8, but radius only doubles.
Section B: Fill in the Blanks
Question 7: The angle of projection for which the horizontal range is maximum is ________.
Since R = v₀² sin 2θ / g, R is maximum when sin 2θ = 1, i.e., 2θ = 90°, so θ = 45°.
Question 8: The centripetal acceleration of a body moving in a circle of radius r with angular velocity ω is ________.
aᶜ = rω² = v²/r, both are correct expressions for centripetal acceleration.
Question 9: The SI unit of moment of inertia is ________.
I = Σmr², so the unit is kg × m² = kg·m².
Question 10: According to Kepler’s third law, T²/r³ = ________ for all planets orbiting the Sun.
This constant is the same for all planets because they all orbit the same central body (the Sun).
Question 11: The escape velocity from a planet depends on its ________ and ________.
v_e = √(2GM/R) — it depends only on the mass and radius of the planet, NOT on the mass of the escaping object.
Section C: Short Answer Questions
Question 12: Why does a projectile follow a parabolic path? Explain in terms of the horizontal and vertical components of motion.
(1) The horizontal motion is uniform (constant velocity, no acceleration) — so x increases linearly with time: x = v₀cosθ · t.
(2) The vertical motion is uniformly accelerated (constant acceleration g downward) — so y varies quadratically with time: y = v₀sinθ · t − ½gt².
Combining these: eliminating t from the two equations gives y = x tanθ − (g/(2v₀²cos²θ))x², which is the equation of a parabola (y = Ax − Bx² form).
Question 13: Distinguish between centripetal force and centrifugal force. Is centrifugal force a real force?
Centrifugal force: An apparent outward force felt by an object in circular motion when observed from a rotating (non-inertial) reference frame. It is NOT a real force — it is a fictitious force that appears only in non-inertial frames.
In inertial reference frames (like the ground), centrifugal force does NOT exist. An object moving in a circle is accelerating toward the center, and the inward force (centripetal) is what causes this acceleration.
Question 14: State and explain Kepler’s second law of planetary motion. What physical principle underlies it?
Explanation: When a planet is closer to the Sun (perihelion), it moves faster. When it is farther (aphelion), it moves slower. But in equal time intervals, the area swept is always the same.
Underlying principle: This is a consequence of the conservation of angular momentum. Since the gravitational force on the planet is always directed toward the Sun (central force), it produces zero torque about the Sun. With zero external torque, L = mvr sinθ = constant. As r decreases, v must increase, and vice versa.
Question 15: Explain why the moment of inertia of a hollow sphere (2MR²/3) is greater than that of a solid sphere (2MR²/5) of the same mass and radius.
In a solid sphere, the mass is distributed throughout the volume, with much of it close to the center (small r values). This gives a smaller average r².
In a hollow sphere, ALL the mass is at the outer surface (at r = R), so every mass element contributes MR² to the sum. This gives a larger average r².
Since I depends on r², having mass farther from the axis gives a larger I. This is why it is harder to start or stop a hollow sphere from spinning compared to a solid one of the same mass and size.
Section D: Calculation Questions
Question 16: A projectile is launched from the ground with an initial velocity of 40 m/s at an angle of 53° above the horizontal. (Take g = 10 m/s², sin 53° = 0.8, cos 53° = 0.6). Find:
(a) The maximum height reached.
(b) The time of flight.
(c) The horizontal range.
(d) The velocity (magnitude and direction) when the projectile is at half its maximum height on the way up.
v₀ᵧ = 40 × 0.8 = 32 m/s
(a) Maximum height:
$$H = \frac{v_{0y}^2}{2g} = \frac{1024}{20} = 51.2 \text{ m}$$
(b) Time of flight:
$$T = \frac{2v_{0y}}{g} = \frac{64}{10} = 6.4 \text{ s}$$
(c) Range:
$$R = v_{0x} \times T = 24 \times 6.4 = 153.6 \text{ m}$$
(d) At half maximum height (y = 25.6 m) on the way up:
vᵧ² = v₀ᵧ² − 2gy = 1024 − 2(10)(25.6) = 1024 − 512 = 512
vᵧ = √512 = 22.63 m/s (upward)
vₓ = 24 m/s (unchanged)
$$v = \sqrt{24^2 + 22.63^2} = \sqrt{576 + 512} = \sqrt{1088} = 32.98 \text{ m/s}$$
Angle above horizontal: tan α = 22.63/24 = 0.943 → α = 43.3°
Question 17: A string of length 80 cm has a 200 g ball tied to its end. The ball is whirled in a horizontal circle so that the string makes an angle of 37° with the vertical. Find: (a) the tension in the string, (b) the speed of the ball, (c) the period of revolution. (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8)
Radius of circle: r = L sinθ = 0.8 × 0.6 = 0.48 m
(a) Tension:
Vertically: T cosθ = mg
T × 0.8 = 0.2 × 10 = 2
T = 2/0.8 = 2.5 N
(b) Speed:
Horizontally: T sinθ = mv²/r
2.5 × 0.6 = 0.2v²/0.48
1.5 = 0.417v²
v² = 3.6
v = 1.90 m/s
(c) Period:
$$T_{period} = \frac{2\pi r}{v} = \frac{2\pi \times 0.48}{1.90} = \frac{3.016}{1.90} = 1.59 \text{ s}$$
Question 18: A solid disc of mass 10 kg and radius 0.5 m is mounted on a frictionless axle. A force of 25 N is applied tangentially to the rim for 4 seconds. The disc starts from rest. Find: (a) the angular acceleration, (b) the angular velocity at t = 4 s, (c) the number of revolutions made in 4 s, (d) the work done by the force.
τ = rF = 0.5 × 25 = 12.5 N·m
(a) Angular acceleration:
$$\alpha = \frac{\tau}{I} = \frac{12.5}{1.25} = 10 \text{ rad/s}^2$$
(b) Angular velocity at t = 4 s:
$$\omega = \omega_0 + \alpha t = 0 + 10 \times 4 = 40 \text{ rad/s}$$
(c) Number of revolutions:
$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(10)(16) = 80 \text{ rad}$$ $$\text{Revolutions} = \frac{80}{2\pi} = 12.73 \text{ rev}$$
(d) Work done:
W = τθ = 12.5 × 80 = 1000 J
(Verification: W = ΔKE = ½Iω² = ½(1.25)(1600) = 1000 J ✓)
Question 19: A satellite orbits the Earth at a height of 500 km above the surface. Calculate: (a) the orbital velocity, (b) the period of revolution, (c) the centripetal acceleration. (R = 6400 km, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹)
(a) Orbital velocity:
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.9 \times 10^6}}$$ $$= \sqrt{\frac{40.02 \times 10^{13}}{6.9 \times 10^6}} = \sqrt{5.803 \times 10^7} = 7618 \text{ m/s} \approx 7.62 \text{ km/s}$$
(b) Period:
$$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.9 \times 10^6}{7618} = \frac{4.335 \times 10^7}{7618} = 5689 \text{ s} \approx 94.8 \text{ min}$$
(c) Centripetal acceleration:
$$a_c = \frac{v^2}{r} = \frac{(7618)^2}{6.9 \times 10^6} = \frac{5.803 \times 10^7}{6.9 \times 10^6} = 8.41 \text{ m/s}^2$$
(Note: This is slightly less than g = 9.8 m/s², as expected for a satellite above the surface.)
Question 20: A uniform rod of mass 4 kg and length 2 m is pivoted at its center. A particle of mass 1 kg is attached to one end of the rod. The system is set into rotation. Find: (a) the total moment of inertia of the system, (b) if a torque of 6 N·m is applied, what is the angular acceleration?
I_rod (about center) = ML²/12 = 4 × 4/12 = 1.333 kg·m²
I_particle = mr² = 1 × 1² = 1 kg·m² (r = L/2 = 1 m)
I_total = 1.333 + 1 = 2.333 kg·m²
(b) Angular acceleration:
$$\alpha = \frac{\tau}{I} = \frac{6}{2.333} = 2.57 \text{ rad/s}^2$$
Question 21: The acceleration due to gravity at a height h above the Earth’s surface is the same as at a depth d below the surface. Show that h = d (if h ≪ R). (Hint: g below surface varies as g_d = g(1 − d/R))
For h ≪ R: R + h ≈ R, so g_h ≈ g(1 − 2h/R) (using binomial approximation)
At depth d: g_d = g(1 − d/R)
Setting g_h = g_d:
g(1 − 2h/R) = g(1 − d/R)
1 − 2h/R = 1 − d/R
2h/R = d/R
2h = d, or h = d/2
So actually h = d/2, not h = d! The acceleration decreases faster with height than with depth. At height h = d, g_h would be less than g_d. To get the same g, the height needed is only half the depth.
Question 22: A ball rolls without slipping down an incline of height 5 m and length 10 m. The ball is a solid sphere of mass 2 kg and radius 0.1 m. Find: (a) the speed of the ball at the bottom, (b) the translational and rotational kinetic energies at the bottom. (g = 10 m/s²)
For solid sphere: I = 2MR²/5, ω = v/R
mgh = ½mv² + ½(2MR²/5)(v²/R²) = ½mv² + mv²/5 = 7mv²/10
(a) Speed at bottom:
$$v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10 \times 10 \times 5}{7}} = \sqrt{\frac{500}{7}} = \sqrt{71.43} = 8.45 \text{ m/s}$$
(b) Kinetic energies:
KE_translational = ½mv² = ½ × 2 × 71.43 = 71.43 J
KE_rotational = ½Iω² = mv²/5 = 2 × 71.43/5 = 28.57 J
Total KE = 71.43 + 28.57 = 100 J = mgh ✓
Note: Translational KE is 5/7 of total, rotational KE is 2/7 of total.
Question 23: Two particles of masses 5 kg and 10 kg are separated by 0.5 m. Find: (a) the gravitational force between them, (b) the gravitational field strength at the midpoint between them. (G = 6.67 × 10⁻¹¹)
$$F = \frac{Gm_1 m_2}{r^2} = \frac{6.67 \times 10^{-11} \times 5 \times 10}{0.25} = \frac{3.335 \times 10^{-9}}{0.25} = 1.334 \times 10^{-8} \text{ N}$$
(b) Gravitational field at midpoint (r = 0.25 m from each):
Field due to 5 kg mass (toward it): g₁ = GM₁/r² = 6.67×10⁻¹¹ × 5/0.0625 = 5.336 × 10⁻⁹ N/kg
Field due to 10 kg mass (toward it): g₂ = GM₂/r² = 6.67×10⁻¹¹ × 10/0.0625 = 1.067 × 10⁻⁸ N/kg
Since the fields point in OPPOSITE directions (toward each respective mass):
g_net = g₂ − g₁ = (1.067 − 0.534) × 10⁻⁸ = 5.33 × 10⁻⁹ N/kg (toward the 10 kg mass)
Question 24: A wheel of moment of inertia 0.5 kg·m² is rotating at 20 rad/s. A frictional torque of 0.2 N·m is applied. How many revolutions does the wheel make before coming to rest?
Using ω² = ω₀² − 2αθ:
0 = (20)² − 2(0.4)θ
0 = 400 − 0.8θ
θ = 400/0.8 = 500 rad
Revolutions = 500/(2π) = 79.6 revolutions
Question 25: Prove that the escape velocity from the surface of a planet of radius R and acceleration due to gravity g is v_e = √(2gR). Hence calculate the escape velocity from a planet where g = 4 m/s² and R = 3200 km.
To escape, the kinetic energy must equal the gravitational potential energy (to reach r = ∞ where KE = 0 and PE = 0):
½mv² = GMm/R
v² = 2GM/R
Since g = GM/R², we have GM = gR²:
v² = 2gR²/R = 2gR
$$v_e = \sqrt{2gR}$$ ✓
Calculation:
$$v_e = \sqrt{2 \times 4 \times 3.2 \times 10^6} = \sqrt{2.56 \times 10^7} = 5060 \text{ m/s} = 5.06 \text{ km/s}$$
End of Challenge Questions
Dear student, you have completed all the challenge questions for Unit 2! Remember:
- Always draw diagrams for projectile and circular motion problems.
- Resolve forces into components before applying equations.
- Check units at every step — this catches many errors.
- For gravitation problems, always use the distance between CENTERS.
- Practice is the key to mastering physics. Keep solving problems!