9th Physics Federal Board Notes: Unit 4 Dynamics II | Complete Guide

9th Physics Federal Board Notes: Unit 4 Dynamics II

Complete study guide covering torque, equilibrium, friction, centripetal force, orbital velocity with solved exercises and practical applications

9th Physics Federal Board Unit 4 Dynamics II Torque and Equilibrium Centripetal Force Friction Reading Time: 25 min

🔬 Introduction to Unit 4: Dynamics II

Unit 4: Dynamics II explores the principles of rotational motion, equilibrium, friction, and circular motion. This unit builds upon the foundation of Newton's laws of motion and extends these concepts to rotational systems. You'll learn about torque, center of mass, stability, friction, centripetal force, and orbital mechanics - all essential concepts for understanding how objects move and interact in our physical world.

Multiple Choice Questions

1. A seesaw balances perfectly with two children of equal weight sitting at equal distances from the fulcrum. If one child moves closer to the fulcrum:

A. The seesaw remains balanced.

B. The seesaw tips towards the child who moved closer.

C. The seesaw tips towards the child who stayed further away.

D. The seesaw topples.

Correct Answer: C
When one child moves closer to the fulcrum, their moment arm decreases. Since torque equals force multiplied by distance (τ = F × d), the torque produced by that child decreases. The child who remains at the original distance produces more torque, causing the seesaw to tip in their direction.

2. When line of action of the applied force passes through its pivot point then moment of force acting on the body is:

A. maximum

B. minimum

C. zero

D. infinite

Correct Answer: C
When the line of action of a force passes through the pivot point, the perpendicular distance from the pivot to the line of action is zero. Since torque τ = F × d, if d = 0, then τ = 0.

3. If a body is at rest or moving with uniform rotational velocity, then torque acting on the body will be:

A. maximum

B. minimum

C. zero

D. infinite

Correct Answer: C
According to Newton's first law for rotation, if a body is at rest or rotating with uniform angular velocity, the net torque acting on it must be zero. This is analogous to translational motion where zero net force means constant velocity.

4. A body in equilibrium must not have:

A. speed

B. quantity of motion

C. velocity

D. acceleration

Correct Answer: D
A body in equilibrium has zero net force and zero net torque acting on it. According to Newton's second law (F = ma), if net force is zero, acceleration must be zero. The body can have constant velocity (including zero) but cannot be accelerating.

5. A uniformly rotating fan is said to be in:

A. static equilibrium only

B. dynamic equilibrium only

C. both in static and dynamic equilibrium

D. not in equilibrium

Correct Answer: B
A uniformly rotating fan is in dynamic equilibrium because it has constant angular velocity (zero angular acceleration), meaning net torque is zero. It is not in static equilibrium because static equilibrium requires the body to be at rest.

6. You throw a weighted fishing net into a calm lake. As the net sinks, it opens fully underwater, spreading out its mesh evenly. Compared to the moment it left your hand, where is the net's center of mass now?

A. Higher in the water column.

B. Lower in the water column.

C. At the same depth but slightly shifted horizontally.

D. Unchanged from its position when thrown.

Correct Answer: B
As the weighted fishing net opens and spreads out underwater, its mass distribution changes. The weights are designed to sink and spread the net, causing the center of mass to move downward in the water column compared to its compact form when thrown.

7. A tightrope walker is carrying a long pole while walking across a rope. The stability of the walker is affected if the pole is:

A. long and placed vertically

B. long and placed horizontally

C. short and placed vertically

D. short and placed horizontally

Correct Answer: B
A long pole held horizontally increases the moment of inertia, making it harder for the tightrope walker to rotate and fall. This provides greater stability. The horizontal position maximizes the distance of mass from the pivot point (the walker's feet), increasing rotational inertia.

8. It is more difficult to walk on a slippery surface than on a non-slippery one because of:

A. reduced friction

B. increased friction

C. high grip

D. lower weight

Correct Answer: A
Walking requires friction between your feet and the ground to push backward and propel yourself forward. On slippery surfaces, friction is reduced, making it difficult to generate the necessary force for walking without slipping.

9. For an object moving with terminal velocity, its acceleration:

A. Increases with time

B. decreases with time

C. is zero

D. first increases then decreases

Correct Answer: C
Terminal velocity occurs when the downward force of gravity is balanced by the upward force of air resistance. At this point, the net force is zero, so according to Newton's second law (F = ma), acceleration is zero. The object continues to fall at constant velocity.

10. The correct order of comparison for the terminal speeds of a raindrop, snowflake, and hailstone is:

A. Raindrop > Snowflake > Hailstone

B. Hailstone > Raindrop > Snowflake

C. Snowflake > Raindrop > Hailstone

D. Raindrop = Snowflake = Hailstone

Correct Answer: B
Terminal velocity depends on mass and cross-sectional area. Hailstones are dense and compact with high mass relative to their area, giving them the highest terminal velocity. Raindrops are smaller and less dense than hailstones. Snowflakes have very low density and large surface area, resulting in the lowest terminal velocity.

11. You are trying to loosen a nut using a spanner, but it is not working. In order to open the nut, you need to:

A. Insert a pipe to increase length of spanner

B. use a spanner of small length

C. use plastic and soft spanner

D. tie a rope with spanner

Correct Answer: A
Torque (τ) equals force multiplied by the perpendicular distance from the pivot (τ = F × d). Increasing the length of the spanner increases the moment arm (d), which increases torque for the same applied force, making it easier to loosen the nut.

12. The force that always changes direction of velocity and not its magnitude is called:

A. gravitational force

B. electric force

C. centripetal force

D. friction

Correct Answer: C
Centripetal force acts perpendicular to the velocity vector of an object in circular motion. This perpendicular force changes the direction of velocity but not its magnitude, resulting in uniform circular motion.

13. The reason that a car moving on a horizontal road gets thrown out of the road while taking a turn is:

A. the reaction of the ground

B. rolling friction between tyre and road

C. gravitational force

D. lack of sufficient centripetal force

Correct Answer: D
When a car takes a turn, it requires centripetal force to move in a circular path. This force is provided by friction between the tires and the road. If the car is moving too fast, or if friction is insufficient, the required centripetal force cannot be provided, causing the car to continue in a straight line (due to inertia) and potentially leave the road.

14. A car drives at steady speed around a perfectly circular track.

A. The car's acceleration is zero.

B. The net force on the car is zero.

C. Both the acceleration and net force on the car point outward.

D. Both the acceleration and net force on the car point inward.

Correct Answer: D
In uniform circular motion, even though speed is constant, velocity is constantly changing direction, so there is acceleration (centripetal acceleration) directed toward the center of the circle. According to Newton's second law (F = ma), the net force must also point toward the center, as this is the centripetal force required for circular motion.

15. A satellite of mass 'm' is revolving around the earth with an orbital speed V. If mass of the satellite is doubled, its orbital speed will become:

A. double

B. half

C. one fourth

D. remain the same

Correct Answer: D
The orbital speed of a satellite depends on the mass of the central body (Earth) and the orbital radius, but not on the mass of the satellite itself. This is derived from the equation v = √(GM/r), where M is Earth's mass, r is orbital radius, and G is the gravitational constant. The satellite's mass cancels out in the derivation.

Short Response Questions

1. Why long spanner is used to open or tight nuts of vehicle's tyre? While tightening a small nut, extra-long wrench is not suitable. Why?

🔧 Torque and Spanner Length

A long spanner is used to open or tighten vehicle tyre nuts because torque (τ) is the product of force (F) and the perpendicular distance from the pivot point (d), expressed as τ = F × d. A longer spanner increases the moment arm (d), allowing greater torque to be applied with the same amount of force, making it easier to turn stubborn nuts.

However, for smaller nuts, an extra-long wrench is not suitable because:

It may generate excessive torque that could damage the nut or bolt threads
Small nuts typically require less torque for proper tightening
Over-tightening can strip threads or deform components
Precision control is more difficult with a very long tool

2. Why door knobs are fixed at the edge of door? What will happen if the door knob is at the middle of the door?

🚪 Door Knob Placement and Torque

Door knobs are fixed at the edge of the door to maximize the moment arm (perpendicular distance from the hinge), which increases torque for a given applied force. According to the torque equation τ = F × d, with the knob at the edge, even a small force can produce sufficient torque to rotate the door easily.

If the door knob were placed in the middle:

The moment arm would be halved, reducing torque by 50% for the same applied force
Significantly more force would be required to open or close the door
The door would feel much heavier and harder to operate
People with less strength might struggle to use the door

3. If you drop a feather and a bowling ball from the same height, which one will reach the terminal velocity first? Which one of them will hit the ground first?

🍃 Terminal Velocity Comparison

The feather will reach terminal velocity first because it has a much lower terminal speed due to its large surface area relative to its mass. Air resistance quickly balances the small gravitational force acting on the feather.

The bowling ball will hit the ground first because:

It has a much higher terminal velocity due to its greater mass relative to cross-sectional area
It accelerates for a longer time before reaching terminal velocity
Even at terminal velocity, the bowling ball falls much faster than the feather's terminal velocity
In a vacuum (no air resistance), both would hit simultaneously, but in air, the bowling ball's superior mass-to-area ratio gives it the advantage

4. Why do ice skates effortlessly slide on ice, while your shoes cause skidding?

⛸️ Friction on Ice

Ice skates slide effortlessly on ice due to a combination of factors:

Pressure melting: The narrow blade of ice skates exerts high pressure, which lowers the melting point of ice, creating a thin layer of water that acts as a lubricant
Low friction surface: Ice naturally has low friction, and the skate blade minimizes contact area
Smooth blade surface: Polished metal blades reduce friction further

Shoes cause skidding because:

They have larger contact area, distributing weight and reducing pressure
Rubber or textured soles are designed for high friction on normal surfaces
On ice, this high friction material cannot grip the smooth surface effectively
The irregular contact creates unpredictable friction points leading to loss of control

5. Why is it easier to push a car on a flat road with inflated tyres than with flat tyres?

🚗 Rolling Friction and Tyre Pressure

It's easier to push a car with inflated tyres because of reduced rolling friction:

Less deformation: Inflated tyres maintain their shape and deform less when rolling, minimizing energy loss to tyre deformation
Smaller contact area: Properly inflated tyres have optimal contact with the road, reducing friction
Smoother rolling: The rigid structure of inflated tyres allows for more efficient transfer of pushing force to motion

With flat tyres:

Increased deformation creates more internal friction within the tyre material
Larger contact area with the road increases sliding friction
More energy is converted to heat rather than forward motion
The car essentially needs to be dragged rather than rolled

6. Why do you spread your feet when you have to stand in the aisle of a bumpy, crowded school bus?

🚌 Stability on Moving Vehicles

Spreading your feet in a bumpy, crowded school bus increases stability by:

Lowering center of gravity: A wider stance effectively lowers your center of mass relative to your base of support
Increasing base area: A larger support base makes it harder for external forces to move your center of mass outside this area
Better force distribution: Impacts from bumps and crowd movements are distributed over a wider area
Improved balance control: Wider stance allows for quicker weight shifting to counteract sudden movements

This stance utilizes principles of stability where an object is more stable when its center of mass is lower and its base of support is wider.

7. Why is it easier to balance a moving bicycle than a stationary one?

🚲 Bicycle Stability Dynamics

A moving bicycle is easier to balance due to several physical principles:

Gyroscopic effect: The spinning wheels act as gyroscopes, creating stability through conservation of angular momentum
Trail effect: The front wheel contact point trails behind the steering axis, creating self-correcting steering
Inertia: The bicycle's forward momentum resists changes in direction, helping maintain straight-line motion
Rider input: A moving bicycle allows for subtle steering corrections that aren't possible when stationary

When stationary, these stabilizing effects are absent, and the bicycle must be balanced entirely through rider adjustments against a very narrow base of support.

8. Why is it difficult to stand a pencil on its tip?

✏️ Stability and Center of Mass

It's difficult to stand a pencil on its tip because:

Very small base area: The tip provides an extremely narrow support base
High center of mass: Most of the pencil's mass is located far above the support point
Unstable equilibrium: Any slight disturbance moves the center of mass outside the tiny base area
Minimal restoring force: When tilted, gravity creates very little restoring torque to bring it back to vertical

For an object to be stable, its center of mass must remain above its base of support. The pencil's high center of mass and tiny base make this condition nearly impossible to maintain without continuous adjustment.

9. Why does a car skid when taking a curve too fast?

🔄 Centripetal Force and Skidding

A car skids when taking a curve too fast because of insufficient centripetal force:

Centripetal force requirement: For circular motion, F_c = mv²/r, where m is mass, v is speed, and r is curve radius
Friction limitation: The maximum centripetal force available is μN, where μ is the coefficient of friction and N is the normal force
Speed squared dependence: Required centripetal force increases with the square of speed
Inertia: When centripetal force is insufficient, the car continues in a straight line due to inertia

When the required centripetal force exceeds the maximum frictional force available (μmg), the tires lose grip and the car skids outward from the curve.

10. If a ball is swung on the end of a string in a vertical circle, at what point is the string most likely to break? Why?

🎾 Tension in Vertical Circular Motion

The string is most likely to break at the bottom of the swing because tension is greatest at this point due to two additive factors:

Centripetal force requirement: T_centripetal = mv²/r
Weight of the ball: At the bottom, tension must support both the centripetal force AND the weight of the ball

The total tension at the bottom is: T_bottom = mg + mv²/r

At the top of the swing, tension is actually reduced because gravity helps provide some of the centripetal force: T_top = mv²/r - mg

This makes the bottom position the point of maximum stress on the string.

11. Why are geostationary satellites placed at a specific height above the Earth?

🛰️ Geostationary Orbit Physics

Geostationary satellites are placed at a specific height (approximately 35,786 km above Earth's surface) because at this altitude:

Orbital period matches Earth's rotation: The satellite completes one orbit in exactly 24 hours
Stationary relative to Earth: The satellite appears fixed in the sky from Earth's surface
Equatorial orbit requirement: Must orbit directly above the equator to maintain fixed position

This specific height is determined by setting the centripetal force equal to gravitational attraction:

mv²/r = GMm/r²

And since v = 2πr/T, where T = 24 hours, we can solve for r to find the specific orbital radius required for geostationary orbit.

Long Response Questions

1. Define like and unlike parallel forces with examples.

Type of Force	Definition	Examples	Characteristics
Like Parallel Forces	Forces that are parallel to each other and act in the same direction	Two people pushing a car from behind Multiple books stacked on a shelf Weights placed on a beam	Same direction of application Can be added directly Produce translational motion
Unlike Parallel Forces	Forces that are parallel to each other but act in opposite directions	Tug-of-war between two teams See-saw with children on both ends Opening a door by pushing and pulling	Opposite directions of application Can create rotational effects May produce zero net force but non-zero torque

2. Define moment of force or torque. Also write its mathematical form and units.

⚙️ Torque (Moment of Force)

Definition: Torque is the turning effect of a force about a pivot point. It measures how effectively a force causes rotation.

Mathematical Form: τ = F × d × sinθ

Where:

τ = Torque (Greek letter tau)
F = Applied force
d = Distance from pivot to point of force application
θ = Angle between force vector and lever arm

For maximum torque, the force should be applied perpendicular to the lever arm (θ = 90°), simplifying to:

τ = F × d

Units:

SI Unit: Newton-meter (Nm)
British Unit: Pound-foot (lb-ft)
Dimensional Formula: [ML²T⁻²]

Direction: Torque is a vector quantity. The right-hand rule determines direction: curl fingers in direction of rotation, thumb points in torque direction.

3. Define center of mass.

🎯 Center of Mass

Definition: The center of mass of a body or system of particles is the point where the entire mass of the body or system can be considered to be concentrated for translational motion analysis.

Mathematical Definition: For a system of particles, the center of mass position vector R is given by:

R = (m₁r₁ + m₂r₂ + ... + mₙrₙ) / (m₁ + m₂ + ... + mₙ)

Where mᵢ are individual masses and rᵢ are their position vectors.

For continuous bodies: The center of mass is found by integration over the entire volume.

Key Properties:

The center of mass may or may not coincide with actual mass in the body
For symmetrical, uniform density objects, center of mass is at geometric center
External forces applied at the center of mass produce pure translation without rotation
The motion of the center of mass follows Newton's second law: F_net = Ma_cm

4. Define center of gravity.

⚖️ Center of Gravity

Definition: The center of gravity of a body is the point where the total weight of the body can be considered to act.

Relationship to Center of Mass: In a uniform gravitational field, the center of gravity coincides with the center of mass. However, in non-uniform gravitational fields (very large objects or near massive bodies), they may differ.

Determination: The center of gravity can be found experimentally by suspending an object from different points and finding where the vertical lines intersect.

Importance in Stability:

An object is stable if its center of gravity lies above its base of support
Lower center of gravity increases stability
Wider base of support increases stability

Practical Applications:

Vehicle design for stability
Architecture and structural engineering
Athlete positioning in sports
Furniture and appliance design

5. How can you find the center of gravity of irregular shaped object?

📐 Finding Center of Gravity of Irregular Objects

For irregularly shaped objects, the center of gravity can be determined experimentally using these methods:

1. Plumb Line Method:

Suspend the object freely from a point
Hang a plumb line (weighted string) from the same point
Mark the vertical line along the plumb line on the object
Repeat the process suspending from a different point
The intersection of the two lines gives the center of gravity

2. Balancing Method:

Place the object on a narrow edge or point
Adjust until the object balances
Draw a vertical line through the balance point
Repeat with different orientations
The intersection point is the center of gravity

3. Mathematical Method (for composite objects):

Divide the object into regular shapes with known centers of mass
Calculate the weighted average position using: x_cg = (m₁x₁ + m₂x₂ + ...)/(m₁ + m₂ + ...)
Similarly for y and z coordinates if needed

Important Note: For objects with uniform density, the center of gravity coincides with the centroid (geometric center) of the object.

6. Define equilibrium. State and explain the principle of moments.

⚖️ Equilibrium and Principle of Moments

Equilibrium Definition: A body is in equilibrium when the net force and net torque acting on it are both zero. This means:

No translational acceleration (F_net = 0)
No rotational acceleration (τ_net = 0)

Types of Equilibrium:

Static Equilibrium: Body is at rest (v = 0, ω = 0)
Dynamic Equilibrium: Body moves with constant velocity (v = constant, ω = constant)

Principle of Moments (Law of the Lever):

"For a body in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point."

Mathematical Expression:

Στ_clockwise = Στ_{anticlockwise}

Or equivalently:

Στ = 0 (about any point)

Application Example - Seesaw:

For a seesaw to balance: F₁ × d₁ = F₂ × d₂

Where F₁ and F₂ are the weights of children, and d₁ and d₂ are their distances from the pivot.

Conditions for Complete Equilibrium:

ΣF_x = 0 (No net force in x-direction)
ΣF_y = 0 (No net force in y-direction)
Στ = 0 (No net torque about any axis)

7. How can you improve the stability of a body?

🏗️ Improving Stability

Stability refers to an object's ability to return to its equilibrium position after being disturbed. Stability can be improved through several methods:

1. Lower the Center of Gravity:

Add weight to the base of the object
Design objects with heavy bases and lighter tops
Examples: Racing cars with low profiles, wide-based furniture

2. Increase the Base Area:

Widen the support base in the direction of potential tipping
Use tripods or multiple support points
Examples: Camera tripods, wide-legged chairs, pyramid shapes

3. Proper Mass Distribution:

Distribute mass symmetrically about the center
Avoid top-heavy designs
Examples: Well-balanced vehicles, evenly loaded ships

4. Use of Counterweights:

Add weights to counteract tipping forces
Examples: Crane counterweights, sailing boat keels

5. Proper Alignment of Forces:

Ensure forces act through the center of mass when possible
Avoid applying forces that create large turning moments

Practical Applications:

Transportation: Wide wheelbase in vehicles, low center of mass in sports cars
Furniture: Wide-based chairs and tables
Structures: Pyramidal shapes in architecture
Sports: Athletic stances with wide bases and low centers of gravity

8. Define force of friction. Also write down the methods of reducing friction.

🔄 Force of Friction

Definition: Friction is the force that opposes the relative motion or tendency of relative motion between two surfaces in contact.

Types of Friction:

Static Friction: Opposes the initiation of motion (F_s ≤ μ_sN)
Kinetic Friction: Opposes ongoing motion (F_k = μ_kN)
Rolling Friction: Opposes rolling motion (typically much smaller than sliding friction)

Methods of Reducing Friction:

1. Use of Lubricants:

Oil, grease, or other substances between surfaces
Creates a slippery layer that separates surfaces
Examples: Engine oil, bicycle chain lubricant

2. Smoothing Surfaces:

Polishing or machining surfaces to reduce roughness
Examples: Polished marble floors, machined metal parts

3. Use of Ball Bearings or Rollers:

Replace sliding friction with rolling friction
Rolling friction is typically much smaller
Examples: Bicycle wheels, drawer slides

4. Streamlining:

Reducing air or fluid resistance
Important for high-speed vehicles
Examples: Aerodynamic car designs, teardrop shapes

5. Use of Low-Friction Materials:

Teflon, graphite, or other materials with naturally low friction
Examples: Non-stick cookware, graphite lubricants

6. Reducing Normal Force:

Since F_friction = μN, reducing normal force reduces friction
Examples: Lighter objects experience less friction

9. Why rolling friction is less than sliding friction?

🎯 Rolling vs. Sliding Friction

Rolling friction is significantly less than sliding friction due to fundamental differences in how the surfaces interact:

Mechanism of Sliding Friction:

Surfaces slide against each other
Microscopic asperities (bumps) on both surfaces interlock and must be broken
Continuous breaking and reforming of molecular bonds
Substantial energy loss to heat and surface deformation

Mechanism of Rolling Friction:

Surfaces roll over each other without sliding
Minimal interlocking of surface asperities
Primary resistance comes from deformation of surfaces
Energy loss mainly due to hysteresis in material deformation

Key Differences:

Aspect	Sliding Friction	Rolling Friction
Surface Interaction	Continuous sliding and interlocking	Point or line contact with rolling
Energy Loss	High due to surface damage and heat	Low, mainly from deformation
Coefficient Values	Typically 0.1 to 1.0	Typically 0.001 to 0.01
Practical Examples	Dragging a box, brakes engaged	Wheels on roads, ball bearings

Mathematical Comparison:

For the same normal force N:

Sliding friction: F_slide = μ_sN (μ_s ≈ 0.1-1.0)

Rolling friction: F_roll = μ_rN (μ_r ≈ 0.001-0.01)

Since μ_r << μ_s, rolling friction is much smaller.

10. Explain the dynamics of an object falling through a viscous fluid and derive the expression for terminal velocity.

💧 Terminal Velocity in Viscous Fluids

When an object falls through a viscous fluid (like air or water), it experiences several forces:

Forces Acting:

Weight (W): Downward force = mg
Buoyant Force (F_b): Upward force = ρ_fluidVg
Drag Force (F_d): Upward force that increases with velocity

Drag Force Formula: For small spheres at low speeds (Stokes' Law):

F_d = 6πηrv

Where:

η = coefficient of viscosity
r = radius of sphere
v = velocity of object

Net Force Equation:

F_net = mg - ρ_fluidVg - 6πηrv

Terminal Velocity Condition: Terminal velocity occurs when net force = 0, so acceleration = 0.

mg - ρ_fluidVg - 6πηrv_t = 0

Solving for terminal velocity v_t:

6πηrv_t = mg - ρ_fluidVg

6πηrv_t = Vg(ρ_object - ρ_fluid)

For a sphere, V = (4/3)πr³, so:

6πηrv_t = (4/3)πr³g(ρ_object - ρ_fluid)

v_t = [2r²g(ρ_object - ρ_fluid)] / (9η)

Key Observations:

Larger objects have higher terminal velocities
Denser objects have higher terminal velocities
Higher viscosity fluids result in lower terminal velocities
In vacuum (no fluid), terminal velocity doesn't exist - object continues accelerating

11. Define centripetal force. Derive its mathematical form.

🔄 Centripetal Force

Definition: Centripetal force is the net force acting toward the center of a circular path that keeps an object moving in circular motion.

Key Characteristics:

Always directed toward the center of the circle
Perpendicular to the velocity vector
Changes the direction of velocity but not its magnitude
Results in centripetal acceleration

Derivation of Centripetal Force:

Consider an object moving with constant speed v in a circle of radius r.

Step 1: Centripetal Acceleration

From geometry of circular motion, the centripetal acceleration is:

a_c = v²/r

Step 2: Apply Newton's Second Law

According to Newton's second law: F = ma

Therefore, the centripetal force is:

F_c = ma_c = m(v²/r)

Alternative Forms:

Since v = ωr, where ω is angular velocity:

F_c = m(ω²r²/r) = mω²r

Since ω = 2π/T, where T is period:

F_c = m(4π²r/T²)

Since frequency f = 1/T:

F_c = 4π²mrf²

Important Notes:

Centripetal force is not a new kind of force - it's the name for the net radial force
Different forces can provide centripetal force (tension, gravity, friction, etc.)
Without centripetal force, objects would move in straight lines due to inertia

12. Write down the sources of centripetal force in the following cases: (i) A car taking a turn on a road (ii) Planet revolving around the sun (iii) Electron revolving around nucleus

🌐 Sources of Centripetal Force

(i) Car Taking a Turn on a Road

Source: Friction between tires and road surface

Explanation: When a car turns, the friction between the tires and road provides the inward force necessary for circular motion. Without sufficient friction, the car would continue in a straight line due to inertia.

Banked Curves: On banked curves, a component of the normal force also contributes to the centripetal force, reducing reliance on friction alone.

(ii) Planet Revolving Around the Sun

Source: Gravitational force between the planet and Sun

Explanation: According to Newton's Law of Universal Gravitation, F = GMm/r², where M is Sun's mass, m is planet's mass, and r is orbital radius. This gravitational attraction provides the necessary centripetal force for orbital motion.

Mathematical Relationship: GMm/r² = mv²/r, which simplifies to v = √(GM/r)

(iii) Electron Revolving Around Nucleus

Source: Electrostatic force of attraction between electron (-) and nucleus (+)

Explanation: According to Coulomb's Law, F = kq₁q₂/r², where q₁ and q₂ are charges, r is orbital radius, and k is Coulomb's constant. This electrostatic force provides the centripetal force for the electron's motion.

Quantum Consideration: In quantum mechanics, electrons don't follow classical circular orbits but exist in probability clouds. However, the electrostatic attraction still provides the centripetal force for their motion.

Case	Source of Centripetal Force	Force Equation
Car turning	Friction	F_friction = μN
Planet orbiting Sun	Gravitational force	F_gravity = GMm/r²
Electron orbiting nucleus	Electrostatic force	F_electric = kq₁q₂/r²

13. Define orbital velocity. Derive its mathematical form.

🛰️ Orbital Velocity

Definition: Orbital velocity is the minimum velocity required for a satellite to maintain a stable orbit around a celestial body.

Key Points:

Depends on the mass of the central body and orbital radius
Independent of the mass of the satellite
For circular orbits, orbital velocity is constant in magnitude but changing in direction

Derivation of Orbital Velocity:

Consider a satellite of mass m orbiting a planet of mass M at distance r from the planet's center.

Step 1: Identify Forces

The gravitational force provides the necessary centripetal force for circular motion:

F_gravity = F_centripetal

Step 2: Write Force Equations

Gravitational force: F_g = GMm/r²

Centripetal force: F_c = mv²/r

Step 3: Equate the Forces

GMm/r² = mv²/r

Step 4: Solve for Orbital Velocity (v)

GMm/r² = mv²/r

GM/r = v²

v = √(GM/r)

For Earth's Surface Orbit:

At Earth's surface, r ≈ R_Earth = 6.4 × 10⁶ m

M_Earth = 6.0 × 10²⁴ kg

G = 6.67 × 10⁻¹¹ Nm²/kg²

v = √[(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.4 × 10⁶)]

v ≈ 7.9 km/s (First cosmic velocity)

Special Cases:

Low Earth Orbit: v ≈ 7.8 km/s at ~200-2000 km altitude
Geostationary Orbit: v ≈ 3.1 km/s at 35,786 km altitude
Moon's Orbit: v ≈ 1.0 km/s around Earth

Important Note: The orbital velocity formula v = √(GM/r) shows that satellites in lower orbits move faster than those in higher orbits.

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9th Physics Federal Board Notes Unit 4: Dynamics II - Complete Solved Exercises