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Understanding Centre of Mass
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Today, we'll discuss the centre of mass for various shapes, like spheres, cylinders, rings, and cubes. Can anyone tell me where the centre of mass of a sphere is located?
I think it's at the center of the sphere!
That's correct! The centre of mass for a uniform sphere is indeed at the geometric center. Now, what about a cylinder?
Isn't it also at the center along the axis?
Exactly! The centre of mass for a uniform cylinder lies along its central axis. What about a ring?
Would it be at the center point of the hole in the ring?
Yes, itβs at the center of the ring! Lastly, where would you expect the centre of mass of a cube to be?
It should be at the geometric center of the cube.
Correct! Great job, everyone! Remember, for uniform objects, the center of mass is usually at their geometric center.
Calculating CM Location
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Let's tackle a practical example. If we take a molecule like HCl, where chlorine is much heavier than hydrogen, how would we find the centre of mass?
We could use the masses and the distances between them to calculate the center of mass, right?
Correct! Since chlorine is around 35.5 times the mass of hydrogen and the distance is given, we can find the location of the CM using the weighted average formula. Does anyone remember the formula for CM?
Isnβt it CM = (m1*x1 + m2*x2) / (m1 + m2)?
That's right! You'll use that to find the approximate location of the center of mass of HCl. Fantastic work!
Center of Mass in Motion
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Imagine a trolley moving with a steady speed. If a child inside runs around, what happens to the centre of mass of that system?
Wouldn't the centre of mass stay the same unless the trolley's speed changes?
Exactly! The CM of the system (trolley plus child) moves at the speed of the trolley, regardless of the child's movements.
So, as long as the net external force is zero, the center of mass remains constant?
Right again! You all are doing wonderfully. This is a key concept in understanding motion within systems.
Vector Product and Area
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Letβs cover a new topic on vectors. If you were to find the area of a triangle formed by two vectors, how would you do that?
I remember we can use the vector product to find it!
Exactly! The area is half the magnitude of the vector product of the two vectors. Does anyone remember how to express that?
Yes! It's Area = 1/2 * |a Γ b|.
Fantastic! This shows how the vector product isnβt just about direction but also has practical applications in geometry.
Angular Momentum
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Shifting gears a bit, letβs talk about angular momentum. Who can tell me what it represents and how it relates to torque?
Angular momentum is the rotational equivalent of linear momentum, right?
Correct! And what happens to angular momentum when there is no external torque?
It stays constant or is conserved!
Exactly! Understanding angular momentum is crucial, especially in systems where rotational forces are at play. Youβre all picking this up very quickly!
Introduction & Overview
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Quick Overview
Standard
The Exercises section provides a series of problems ranging from simple identification of concepts to complex application-based scenarios, enabling students to test and deepen their understanding of the topics covered in this chapter.
Detailed
The Exercises section encompasses a comprehensive variety of questions and problems pertaining to the concepts discussed in the previous sections of the chapter on Systems of Particles and Rotational Motion. These exercises are structured to assist learners in applying theoretical knowledge to practical situations, checking their understanding of key principles related to center of mass, motion, torque, angular momentum, and more. The problems range from straightforward exercises that require recalling definitions to more intricate calculations necessitating the application of multiple concepts. Engaging with these exercises is vital for consolidating one's grasp and facilitating a deeper comprehension of the material covered.
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Identifying Centre of Mass
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6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Detailed Explanation
This exercise prompts students to identify the centre of mass of various shapesβsphere, cylinder, ring, and cubeβwith uniform mass density. The centre of mass refers to the point where the weighted relative position of the distributed mass sums to zero. For standard shapes, the centre of mass will be at their geometric centerβthis means: a sphere's centre of mass is at its center, a cylinder's is at its center along its axis, a ring's lies at its center, and a cube's at the center where the sides meet. The final part asks whether a body's centre of mass can lie outside its physical boundaries, to which the answer is yes, as seen in a ring where the mass is equally distributed at a distance away from the center hole.
Examples & Analogies
Imagine balancing a seesaw: if the masses on either side are equal, it balances at the midpoint, representing the centre of mass. Similarly, if you take a hula hoop and hold it up, the centre of mass is where you find balance, which is precisely at the center of the hoop, regardless of the hoop's size.
Centre of Mass in HCl Molecule
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6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Γ (1 Γ = 10^-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Detailed Explanation
For this exercise, we need to calculate the centre of mass (CM) of the HCl molecule using the formula for the centre of mass of a two-body system: CM = (m1x1 + m2x2) / (m1 + m2), where m is the mass of each atom, and x is the position of each atom. Given that chlorine is 35.5 times the mass of hydrogen, if we consider the mass of H to be 1 unit, then Cl will be approximately 35.5 units. With the distance between the two nuclei as 1.27 Γ , we can set a coordinate system where hydrogen is at 0 and chlorine at 1.27 Γ to find the CM position, factoring in their respective masses.
Examples & Analogies
Think of a seesaw again, but this time with different weights on each side. If you have one side heavier than the other, the balance point (or CM) shifts towards the heavier side. In the case of HCl, chlorine being significantly heavier means the 'balance point' will be closer to chlorine, illustrating how mass distribution affects the centre of mass.
Child Running on Trolley
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6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Detailed Explanation
This question explores the concept of the centre of mass (CM) in a system of multiple objects. When the child is at rest on the trolley, the system (trolley + child) moves forward with speed V. The total CM speed is unaffected by the child's movement on the trolley because the system's speed is dictated by the external motion. Whether the child runs forward or backward on the trolley, that internal movement does not change the overall speed of the CM, which remains at speed V since there are no external forces acting horizontally on the system.
Examples & Analogies
Imagine riding a train. If you walk from the back to the front, although you're moving, your speed relative to the ground and the train moves steadily forward remains unchanged. Similarly, the CM of the trolley and child system remains moving at speed V to the outside observer, regardless of the child's actions.
Area of Triangle Formed by Vectors
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6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a Γ b.
Detailed Explanation
In vector mathematics, the area of a triangle formed by two vectors can be expressed using the cross product. The area A of the triangle is given by A = 0.5 * |a Γ b|. This result comes from half the magnitude of the cross product representing the area of the parallelogram formed by the vectors. Since a triangle is half of that parallelogram, we donβt need to do any extra calculations; this property of cross products lends natural simplicity to determining areas in vector spaces.
Examples & Analogies
Picture a sailboat whose sails are positioned at an angle to catch the windβthese sails form a triangle. The stronger the wind (bigger vectors), the bigger the area of the triangle, which means the boat can sail faster (more force). The relationship established by the cross product gives a direct mathematical reflection of how effectively vectors can interact geometrically to create magnitudes like area.
Volume of Parallelepiped from Vectors
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6.5 Show that a.(b Γ c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.
Detailed Explanation
The volume of a parallelepiped, formed by three vectors, can be determined using the scalar triple product. For vectors a, b, and c, the volume V is expressed as V = |a.(b Γ c)|. This scalar triple product geometrically represents the signed volume of the parallelepiped, showing that the volume is equivalent to the product of the area of the base (the magnitude of b Γ c) and the height (the projection of vector a onto the direction perpendicular to the base).
Examples & Analogies
Think of a box (the parallelepiped) that you want to fill with balls. You can visualize how by using three different lengths (the three vectors), the orientation and lengths will determine how much space it fills. This can apply to many situations, be it stacking boxes in a warehouse or filling spaces in an apartment!
Components of Angular Momentum
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6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz.
Detailed Explanation
This exercise requires students to determine the angular momentum (l) in terms of its components. Angular momentum is given by l = r Γ p. The position vector r can be represented as (x, y, z) while momentum p can be (px, py, pz). The cross product expands into components that can be expressed as: l_x = ypz - zpy, l_y = zpx - xpz, and l_z = xpy - ypx. This establishes the angular momentum based on each of the three coordinates, helping students see the multidimensional nature of motion.
Examples & Analogies
Consider a spinning toy top. The way it remains upright and spins is due to the torque applied when spun. Each point on the top contributes to the overall motion, which can be broken down into how it moves across three axes, akin to how we break down the angular momentum in this exercise: each component demonstrates how a specific aspect of motion affects overall stability.
Angular Momentum and Torque
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6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken.
Detailed Explanation
This exercise explores the idea of angular momentum concerning two particles moving in opposite directions. The formula for angular momentum L for each particle can be determined with respect to a certain point. When calculating the angular momentum L = r Γ p for different reference points, the symmetry in the arrangement means that the total angular momentum remains unchanged. The reason underlying this is due to the conservation of angular momentum: regardless of the position from which you measure, the interplay of distances and parallel movements cancels out variations in the direction of the vectors.
Examples & Analogies
Imagine two ice skaters moving in opposite directions, forming a tug-of-war at a center point. No matter if you measure their speeds from the rink's edge or from the center, their combined angular effect will be constantβthey spin about this central point uniformly, reflecting how angular momentum holds irrespective of the observational standpoint.
Moments of Forces on a Rigid Body
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6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angles made by the strings with the vertical are 36.9 Β° and 53.1 Β° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.
Detailed Explanation
This exercise involves understanding how weight distribution affects stability and balance. Given the system is in equilibrium, the moments about any point must sum to zero, leading us to calculate the torques from both support strings acting on the bar and analyzing their respective distances from the centre of gravity. By balancing the torques provided by the forces from the strings against the weight of the bar, we can set up equations to find the distance of the center of gravity, d, from the left end.
Examples & Analogies
Think of balancing a broom vertically on your palm. If someone pushes the broom in one direction, your body naturally compensates by shifting the other way to keep it stable. Similarly, each weight in this exercise affects how the bar balances, demonstrating how forces affect its centre of gravity.
Force Exerted on Car Wheels
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6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Detailed Explanation
In this exercise, students must consider how the weight and center of mass of the car affect the forces exerted on the wheels when it is at rest. You calculate the distribution of weight across the wheels based on the position of the center of gravity relative to the axles and how gravitational force is distributed. This is done using torque equilibrium principles to balance the system, leading to the calculations of forces on each wheel.
Examples & Analogies
Picture balancing backpack straps both on your shoulders evenly when filled equally: if one shoulder has a heavier load, that shoulder feels more strain (force). Similarly, when the vehicle's center of gravity shifts, so does the force on each wheel, directly reflecting how weight is managed across the system.
Torque on Cylinder and Sphere
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6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Detailed Explanation
This exercise compares the rates of angular acceleration you would find in different shaped objects under the same torque conditions. Since the moment of inertia differs between the hollow cylinder and the solid sphere, the same applied torque results in different angular accelerations. The sphere will have a smaller moment of inertia, leading to a higher angular speed compared to the hollow cylinder as inertia resists changes in motion.
Examples & Analogies
Think of a rolling ball versus a hoop down a slope: even though they start at the same point and roll due to gravity, the tightened axis of the ball allows it to speed ahead more than the hoop, which is more resistant to acceleration due to its structure.
Kinetic Energy of Rotating Cylinder
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6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Detailed Explanation
This exercise requires calculating the kinetic energy associated with the rotation of the cylinder using the formula for rotational kinetic energy, K = 1/2 I Ο^2, where I is the moment of inertia computed as I = 1/2 m r^2 for the cylinder, and Ο is the angular speed given. The angular momentum would be calculated using L = I Ο as well, leading to both energetic measures of the system.
Examples & Analogies
Imagine pedaling a stationary bike: when you begin to pedal faster (increasing angular speed), you require more energy and feel it in your legs (kinetic energy). Similarly, the faster the cylinder spins, the more energy it generates and momentum it builds, akin to the bike speeding up upon greater effort.
Child in Rotating Turntable
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6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the childβs new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Detailed Explanation
In this exercise, the child is effectively changing their moment of inertia and, with conservation of angular momentum, the angular speed must adjust inversely. As the child retracts arms and reduces their moment of inertia, consistency in rotational speed (momentum) requires an increase in angular speed. For part b, the kinetic energy formula will show how more compact configurations lead to greater speeds and thus energy levels, reflecting real-world scenarios in conservation laws.
Examples & Analogies
Think of an ice skater spinning: when they pull their arms in close to their body, they spin faster. The same process occurs with the child on the turntable: as they fold their arms, the rotation speeds up to maintain equilibrium and momentum, increasing the kinetic energy observed.
Rope Wound Round Hollow Cylinder
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6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Detailed Explanation
This exercise requires calculating both angular and linear acceleration provided given external force acting on a cylinder. Using Ο = F * r to find the torque created about the center of the cylinder and applying the moment of inertia for the hollow cylinder, insights into dynamics also reveal a relationship between linear and angular acceleration through the radius and radians, giving a fuller understanding of rotational motion.
Examples & Analogies
Imagine tugging a rope tied to a bicycle wheel. When you pull, the wheel rotates in response to that force applied at the edge. This example directly reflects how forces lead to accelerations consistent in both dimensions: pulling creates angular shift at the wheel, exemplified in the calculations here.
Power Required by Engine
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6.14 To maintain a rotor at a uniform angular speed of 200 rad/s, an engine needs to transmit a torque of 180 N m. What is the power required by the engine?
Detailed Explanation
This exercise applies the relationship between power, torque, and angular velocity, defined through the equation P = Ο * Ο. Given that the engine maintains a constant speed, multiplying the torque provided by the angular speed yields the power output required to counteract any losses in energy to keep the rotor moving smoothly at uniform speeds.
Examples & Analogies
Picture running a fan: the continuous turning requires consistent energy provided by the motor, relating directly to both the torque exerted and the speed of rotation. The needed power balances consistently between work done against resistance in continuous motion.
Finding CG of Irregular Body
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6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Detailed Explanation
To find the new centre of gravity (CG) of the modified disk, consider using the method of sections where the CG initially lies at the center of the original disk. Cutting out the smaller section will displace the CG. Using the known positions and their respective masses, one can find the new CG by turning to the mass-weighted average of the components to derive the resulting balance considering the void created.
Examples & Analogies
Think of baking a cookie and cutting out a shape from its center. The initial mass of the cookie shapes the cavity where it was cut out, leading to new weight distribution. Likewise, when designing shapes, the balance shifts and must be recalculated to find new centers, represented precisely in this problem.
Balancing Forces on a Nailed Rod
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6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Detailed Explanation
Involving torque equilibrium, this exercise requires analyzing the vertical forces acting on the metre stick as the masses are added. The balance point is adjusted when the coins are added. By equating the torques acting around the center pivot due to both the coins and the stick itself, one determines the stickβs mass based on the balancing conditions of the evenly distributed weights.
Examples & Analogies
Think of balancing a plank on your finger with weights on one side. Just as adding weights will determine how far you need to adjust to keep balance, the adjustments in the conditions here illustrate how torque is recalculated based on added mass.
Angular Momentum of Particle
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6.17 The oxygen molecule has a mass of 5.30 Γ 10^-26 kg and a moment of inertia of 1.94 Γ 10^-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two-thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
Detailed Explanation
This exercise involves the concept of angular momentum and its relation to both translational and rotational motion. To find the average angular velocity expressed through the relationship of kinetic energy, one calculates the energies involved and extracts the rotational influence. By defining kinetic energy relations and substituting angular momentum into angular speed formulas, a deeper understanding of molecular dynamics is established.
Examples & Analogies
Think of a whirling dancer who spins while moving about a stage. Their speed (translational movement) and the speed of their spin (angular change) are connected, affecting how quickly they move across both directionsβa representation mirrored in how the oxygen molecules change with energy states.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Centre of Mass: The balance point where an objectβs mass is evenly distributed.
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Torque: The rotational force that causes an object to rotate around an axis.
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Angular Momentum: The measure of how much motion an object has when it rotates.
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Rigid Body: A solid object with fixed distances between its parts, demonstrating no deformation.
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Moment of Inertia: A measure of how difficult it is to change the rotational motion of an object.
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Angular Velocity: The speed of an objectβs rotation around an axis.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example 1: Calculating center of mass for a molecule like HCl considering its mass distribution.
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Example 2: A child running on a trolley demonstrating how system momentum is conserved.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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In a sphere, around the core, the mass does align, it's the center for sure!
π Fascinating Stories
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Imagine two kids on a seesaw, each on opposite ends forming a balance. They Know that where they sit, their weight needs to balance out!
π§ Other Memory Gems
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MOT β Mass, Orientation (shape), Torque β for determining when to use angular momentum.
π― Super Acronyms
CMOT - Centre of Mass, Orientation of Torque
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
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Term: Centre of Mass
Definition:
The point in a body or system of bodies at which the whole mass can be considered to be concentrated.
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Term: Torque
Definition:
A measure of the force that produces or tends to produce rotation or torsion of a body.
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Term: Angular Momentum
Definition:
The quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity.
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Term: Rigid Body
Definition:
An idealization of a body in which deformation is neglected and the distance between any two particles of the body remains unchanged.
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Term: Moment of Inertia
Definition:
The resistance of an object to changes in its rotational motion.
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Term: Angular Velocity
Definition:
The rate of change of angular displacement, usually measured in radians per second.