4.1.1 Explain Projectile Motion (CRQ – 3 Marks)

• Definition: Projectile motion is the motion of a body thrown at an angle θ with an initial velocity under the effect of gravity only.

  • The object follows a curved (parabolic) path.

  • Only gravitational force (g) acts downward.

  • Motion occurs in two dimensions (x and y).

4.1.2 (a) Derive Time of Flight (ERQ – 5 Marks)

• Initial Setup: Consider a projectile thrown with an initial velocity u at an angle θ.

Step 1: Vertical Component of Velocity

• The vertical component of velocity is given by:

  • u_y=usin⁡θ

Step 2: At Maximum Height

• At maximum height, the vertical velocity v_y is 0. Using the first equation of motion:

  • vy = uy - gt

• Setting v_y = 0 leads to:

  • 0 = u imes ext{sin}( heta) - gt

  • Rearranging gives:
    t = rac{u imes ext{sin}( heta)}{g}

  • This is the time to reach maximum height.

Step 3: Total Time of Flight

• Since the motion is symmetric:

  • T = 2t

  • T = 2 imes rac{u imes ext{sin}( heta)}{g}

4.1.2 (b) Derive Maximum Height (ERQ – 5 Marks)

• At Maximum Height: Again, compressing the understanding:

  • v_y = 0

• Using the third equation of motion we have:

  • v^2y = u^2y - 2gH

  • Substituting gives:

  • 0 = (u imes ext{sin}( heta))^2 - 2gH

  • Rearranging leads to:

  • 2gH = u^2 imes ext{sin}^2( heta)

  • Hence,

  • H = rac{u^2 imes ext{sin}^2( heta)}{2g}

4.1.2 (c) Derive Horizontal Range (ERQ – 5 Marks)

• Step 1: Horizontal Motion

  • Since there is no acceleration in the horizontal direction:

  • R = u_x imes T

• The horizontal component of velocity is:

  • u_x = u imes ext{cos}( heta)

• Thus, total time of flight is:

  • T = rac{2u imes ext{sin}( heta)}{g}

Step 2: Substitute

• Substitute values into range equation:

  • R = (u imes ext{cos}( heta)) imes rac{2u imes ext{sin}( heta)}{g}

  • Simplifying gives:

  • R = rac{2u^2 imes ext{sin}( heta) imes ext{cos}( heta)}{g}

• Using identity $2 imes ext{sin}( heta) imes ext{cos}( heta) = ext{sin}(2 heta)$ gives:

  • R = rac{u^2 imes ext{sin}(2 heta)}{g}

4.1.2 (d) Maximum Horizontal Range (CRQ – 3 Marks)

• Range Formula:

  • R = rac{u^2 imes ext{sin}(2 heta)}{g}

• The maximum value of $ ext{sin}(2 heta)$ is 1. Thus,

  • ext{sin}(2 heta) = 1

  • Hence:

  • 2 heta = 90

  • heta = 45

• Maximum Range:

  • R_{ ext{max}} = rac{u^2}{g}

4.1.3 Solve Word Problems (Concept) (CRQ – 3 Marks)

• Steps to solve projectile problems:

  1. Resolve velocity into components:

  • u_x = u imes ext{cos}( heta)

  • u_y = u imes ext{sin}( heta)

  1. Treat horizontal and vertical motions separately.

  2. Use the equations of motion for vertical and horizontal paths:

  • Vertical motion: $s = vt$

  • Horizontal motion: Apply the motion equations.

4.1.4 Effect of Air Resistance (CRQ – 3 Marks)

• With Air Resistance:

  • Velocity decreases (air opposes motion).

  • Maximum height decreases.

  • Range decreases.

  • Time of flight increases.

  • Path is no longer a perfect parabola.

4.2.1 Describe Angular Displacement, Angular Velocity & Angular Acceleration (CRQ – 3 Marks)

Angular Displacement (θ)

• Definition: Angular displacement is the angle covered by an object during rotation.

  • Measurement Units:

  • Radians (rad) - SI unit.

  • Degrees (°) and revolution (rev).

  • Defined as the angle subtended at the center of a circle.

Angular Velocity (ω)

• Definition: Angular velocity is the rate of change of angular displacement.

• Formula:

  • ext{ω} = rac{ ext{Δθ}}{ ext{Δt}}

• SI Unit: rad/s.

Angular Acceleration (α)

• Definition: Angular acceleration is the rate of change of angular velocity.

• Formula:

  • ext{α} = rac{ ext{Δω}}{ ext{Δt}}

• SI Unit: rad/s².

4.2.2 Derive Relationship Between Linear & Angular Quantities (ERQ – 5 Marks)

• Consider a Particle Moving in a Circular Path of Radius r:

1. Displacement Relation

• Arc length (linear displacement):

  • s = r heta

  • Thus, linear displacement depends on the radius (r) and angular displacement (θ).

2. Velocity Relation

• Differentiating with Respect to Time:

  • v = r imes ext{ω}

  • Where v is linear velocity and ω is angular velocity.

3. Acceleration Relation

• Differentiating Again:

  • a = r imes ext{α}

  • Where a is linear acceleration, and α is angular acceleration.

• Final Important Relations:

  • s = r heta

  • v = r ext{ω}

  • a = r ext{α}

• Conclusion: These show the connection between linear and rotational motion.

4.2.3 Solve Word Problems Related to Rotational Motion (CRQ – 3 Marks)

• To Solve Rotational Motion Problems:

  1. Convert degrees to radians (if needed).

  2. Use the proper formulas:

  • s = r heta

  • v = r ext{ω}

  • a = r ext{α}

  1. Apply sign conventions carefully.

  2. For small time intervals, use instantaneous values.

4.3 Centripetal Force & Centripetal Acceleration

4.3.1 Define Centripetal Force & Centripetal Acceleration (CRQ – 3 Marks)

Centripetal Force

• Definition: Centripetal force is the force that bends the straight path of a body into a circular path.

  • It is not a separate force; it can be tension, friction, gravity, etc.

  • Direction: Towards the center of the circular path.

• Formula:

  • F_c = rac{mv^2}{r}

Centripetal Acceleration

• Definition: Centripetal acceleration is the acceleration produced due to the change in direction of velocity in circular motion.

  • Exists even if speed is constant.

  • Direction: Always directed towards the center.

• Formula:

  • a_c = rac{v^2}{r}.

4.3.2 Derive Centripetal Acceleration (Uniform Speed) (ERQ – 5 Marks)

• Consider a body moving in a circle with constant speed:

Step 1: Important Concept

• Velocity is a vector quantity. Even if speed is constant, the direction keeps changing; therefore, acceleration exists.

Step 2: Similar Triangle Concept

• From the geometry of velocity vectors:

  • rac{ ext{Δv}}{v} = rac{ ext{Δs}}{r}

• But, ext{Δs} = v ext{Δt}

• Therefore:

  • ext{Δv} = rac{v^2 ext{Δt}}{r}

Step 3: Acceleration

• a = rac{ ext{Δv}}{ ext{Δt}}

• Substituting gives:

  • a = rac{v^2}{r}

• Final Result:

  • a_c = rac{v^2}{r}

• Direction: Towards center (inward).

4.3.3 Justify That Centrifuge Separates Materials (CRQ – 3 Marks)

• A centrifuge rotates at high speed, producing large centripetal acceleration.

• Effect on Particles:

  • Heavier particles experience a greater outward effect.

  • Denser particles move outward faster than lighter ones.

• Conclusion: Therefore, materials are separated based on density.

4.3.4 Explain Why Objects in Orbiting Satellites Appear Weightless (CRQ – 3 Marks)

• Weightlessness: Does not mean gravity is zero.

• A satellite and astronauts are in continuous free fall.

• Both fall with the same acceleration (g).

• Observation: No normal reaction force acts on them.

• Conclusion: Therefore, felt weight becomes zero, hence astronauts appear weightless.

4.3.5 Artificial Gravity to Counter Weightlessness (ERQ – 5 Marks)

• Weightlessness Challenges: Creates difficulties in space.

• Solution: Rotating Space Station

  • When the station rotates:

  • A person moves in a circular path.

  • Centripetal acceleration acts toward the center.

  • An outward centrifugal effect is felt.

• To Create Artificial Gravity:

  • Using the relationship:

  • rac{v^2}{r} = g

• Substituting:

  • v = r ext{ω}

• Thus,

  • r ext{ω}^2 = g

• Conclusion: This produces artificial gravity equal to Earth’s gravity. Therefore, rotation can simulate gravity in space.

4.4.1 Explain Moment of Inertia with SI Unit & Dimension (CRQ – 3 Marks)

Definition

• Moment of inertia refers to the resistance of a body to rotational motion.

• It is also known as rotational inertia.

• In linear motion, inertia depends on mass (m). In rotational motion, inertia depends on moment of inertia (I).

Formula (Single Particle)

• I = mr^2

• Where:

  • $m$ = mass

  • $r$ = distance from axis of rotation.

• It depends on:

  • Mass.

  • Distribution of mass from axis.

• SI Unit: kg imes m^2.

• Dimensional Formula:

  • [ML^2].

4.4.2 Compare Moment of Inertia of Rod, Ring, Disc & Sphere (CRQ – 3 Marks)

• Comparison:

  • Smallest Moment of Inertia: Rod.

  • Largest Moment of Inertia: Ring.

  • Reason:

  • Ring has all mass farthest from the axis, hence having maximum moment of inertia.

4.4.3 Derive Relationship Between Torque, Moment of Inertia & Angular Acceleration (ERQ – 5 Marks)

Step 1: Torque Definition

• Torque is the turning effect of force, defined as:

  • au = rF ext{sin}( heta)

• For maximum torque (when θ = 90°):

  • au = rF.

Step 2: Newton’s Second Law (Linear Form)

• F = ma.

Step 3: Substitute in Torque Equation

• $ au = r(ma)

• Where:

  • a = r ext{α}

• Thus:

  • au = r(mr ext{α}).

Step 4: Since I = mr²

• Therefore, au = I ext{α}.

• Final Result:

  • Torque = Moment of inertia × Angular acceleration.

  • This is Newton’s second law in rotational motion.

4.4.4 Solve Word Problems Related to These Relations (CRQ – 3 Marks)

• To Solve Rotational Problems:

  1. Use: au = I ext{α}.

  2. Calculate I using the standard formulas.

  3. Use the correct axis of rotation.

  4. For maximum torque, maintain an angle of 90°.

  5. Check units to be consistent (kg·m²).

4.5 Angular Momentum

4.5.1 Explain Angular Momentum with SI Unit & Dimension (CRQ – 3 Marks)

Definition

• Angular momentum (L) is the quantity of rotational motion contained in a body, representing the rotational version of linear momentum.

• Formula:

  • L = I ext{ω}

  • Alternatively:

  • L = mvr

• Dependencies:

  • Mass.

  • Velocity.

  • Distance from the axis.

• SI Unit: kg imes m^2/s.

• Dimensional Formula:

  • [ML^2T^{-1}].

4.5.2 Solve Word Problems Related to Angular Momentum (CRQ – 3 Marks)

• To Solve Angular Momentum Problems:

  1. Use: L = I ext{ω} or L = mvr.

  2. Identify whether the motion is circular.

  3. Use the correct radius.

  4. Apply conservation law if no external torque is present.

4.5.3 Explain Law of Conservation of Angular Momentum (ERQ – 5 Marks)

Statement:

• If no external torque acts on a system, the total angular momentum remains constant.

Condition:

• τ_{ ext{ext}} = 0

• Mathematical Form:

  • L = I ext{ω}

  • If no external torque acts:

  • I1 ext{ω}1 = I2 ext{ω}2

• Important Concept:

  • If moment of inertia decreases, angular velocity increases.

  • If moment of inertia increases, angular velocity decreases.

  • The product remains constant.

4.5.4 Applications of Conservation of Angular Momentum (ERQ – 5 Marks)

(a) Flywheel

• Function: Stores rotational energy.

• Characteristics:

  • Large moment of inertia.

  • Maintains steady angular velocity.

• Advantages: Used to store rotational energy efficiently.

(b) Gyroscope (Navigation System)

• Application: Utilizes conservation of angular momentum.

• Key Feature: A spinning wheel resists changes in orientation.

• Use: Helps in aircraft and ship navigation.

(c) Ice Skater

• Effect: When a skater:

  • Arms out → I large, ext{ω} small.

  • Arms closed → I small, ext{ω} increases.

• Conclusion: Since I ext{ω} = ext{constant}, the skater spins faster when arms are pulled inward.