Category: Physics Notes PPT

Introduction

A machine which consists of a wheel attached to a small axle so that both of them (the wheel and axle) rotate together, and in this process force is transferred from one to another: this is defined as the wheel and axle machine. The axle is supported by a thing or a bearing which allows rotation. A small force which is applied from the wheel can amplify large loads attached to the axle. This is looked as the version of the lever. A drive force is applied tangentially to the perimeter of the wheel and a load force is applied to the axle. These two are balanced around a hinge which is known as the fulcrum. 

Wheel Axle

Wheel-axel machine is a simple machine which makes tasks easier in terms of force by applying the concept of mechanical advantages. wheel and axle is assembly formed by two disks, or cylinders, of different diameters mounted together so that they can rotate together around the common axis and the thin rod that needs to be turned is called the axle and the wider object fixed to the axle, on which we are apply force is called the wheel. A tangential force applied to the periphery of the large disk can exert a larger force on a load attached to the axle thus achieving the mechanical advantage. Also the wheel and axle does not dissipate or store energy, because it has no almost no friction as well as elasticity,thus power output at the axle equals to the input by the force applied to the wheel. 

Mechanical Advantage

Machines make our work easier to do. We use several machines in our daily life to make our work easier. Mechanical  advantage is basically when we put a small amount of energy and a huge amount of work is done by that small energy. Being specific,  it’s the ratio of force that the machine produces which is the output, to the force which is put into the machine by humans which is also known as the input. The wheel and axle comes under the category of six simple machines that are lever, pulley, inclined plane, wedge and screw. We try to use the simple machines because they give us the mechanical advantage. For example if we carry weights in our arms we will definitely feel burdened than carrying on a cart which moves on wheels. The axle and wheel are two circle-like structures that rotate together for the work to be done. The force is transferred from the axle to the wheel in most of the cases. It’s the ratio of radius of the wheel to the radius of the axle. Note- the radius is the half of the diameter and the diameter goes from the middle of the circle touching both the ends of the circle.it crosses through the origin or the midpoint of the circle. Although radii is used in the calculation of the wheel and axle but we can also do it with the help of the diameter of wheel and axle, it provides us with the same result.

Since the wheel and axle system rotates around its own bearings, thus the points on the circumference, or edge, of the wheel move faster than points on the circumference, or edge, of the axle. Therefore, a force applied to the edge of the wheel must be less than the force applied to the edge of the axle, since power is the product of force and velocity .Thus if a and b be the distances from the centre of the bearing to the edges of the wheel A and the axle B and If the FA (input force) is applied to the edge of the wheel A and the force FB  at the edge of the axle B is the output, then  a/b is the ratio of the velocities of points A and B is given by , so the ratio of the output force to the input force, or mechanical advantage is given by

Work is basically the constant force that is done on a system and it is the product of the component of the force in the direction of motion times the distance through which the force acts. And the capacity to work is called energy. Work is the term that is used for the displacement done by any force in physics. In other words, we can say that work and energy are the two essential elements to understand any physical movement. 

Well, here we will discuss the work-energy theorem, limitations, and work-energy theorem examples. To complete the work, energy is needed, and hence in this theorem, we will know the relation between energy and work. Work relates to displacement, and displacement relates to kinetic energy. The work-energy theorem affirms that the work done on any object is comparable to the difference in kinetic energy of the object. So, according to the theorem statement, we can define the work-energy theorem as follows.

K_f – K_i = W

Where, 

K_f = Final kinetic energy

K_i = Initial kinetic energy

W = Net-work done on the object.

Types of Work-Energy Theorem

Various types of energy are there like potential energy, kinetic energy, thermal energy, electrical energy, chemical energy, and nuclear energy. Kinetic energy is basically the type of energy that an object or particle has as an outcome of its movement and when the work is done on an object by applying net force, the item accelerates and gains kinetic energy. So the Kinetic energy is a property of a moving object or particle that is definitely by its mass as well as its motion. Now the Work-energy theorem states that the work done on a particle by the sum of all forces acting is equal to the change in the kinetic energy of the particle, in other words, net-work done on a system is equal to the change in Kinetic energy.

As we discussed earlier, kinetic energy changes due to external forces or energies like gravity or friction. You probably remember the law of energy conservation. According to the law of conservation, energy is only changed from one form to another. Let’s study the following work-energy theorem example for better understanding. 

According to the Work energy theorem,

Work (completed by all kinds of energy or forces) = Change (Difference) in Kinetic Energy

Formula for Work-Energy Theorem

W_net = K−K₀ =ΔK

Wnet is net-work and K is final kinetic energy and K₀ is the initial kinetic energy, and lastly, ΔK is changes in kinetic energy  

Or

Wg + W_N + W_f  =K_f – K_i

Where Wg = work done by gravity

W_N = work done by a reasonable force

W_f = work done by friction

K_f = final kinetic energy

K_i = initial kinetic energy

Work Done by a Constant Force

The constant force will result in constant acceleration. Let’s take the constant acceleration as ‘a.’ 

From the motion equation, we get

v² = u² + 2as…….. equation (1)

Therefore,

2as = v² – u² ………… equation (2)

In equation 2, multiply both the sides with ‘m’ mass.

(ma).s = (mv²–mu²)²

Mass X Acceleration = Force 

F.s = (mv²–mu²)²…………….. equation (3)

Comparing the above equations (2) and (3),  we get,

Work done by force (F) = F.s

Where’s’ is the displacement of the body.

Work is Done by (Non-Uniform) Variable Force.

Now, consider the resulting equation of work. 

W = F.ds

The above equation is valid only for such conditions where the force is constant and can’t be applicable for non-constant variables. 

Work-Energy Theorem for Variable Force

In the given figure, force is constant for a little displacement, and then after the force is variable till the final position. So, you can see the change in the applied force from the graph. 

W = [int x_{f} x_{i} F(x)dx] 

The shaded portion represents the work done by force F(x).

It represents the work that is done by a variable force. A graphical approach to this would be finding the area between F(x) and x from xi to xf.

Application of Work-Energy Theorem

The application of the Work-Energy theorem is that it is very useful in analyzing situations where a rigid body should move under several forces. A rigid body cannot store the potential energy in its lattice due to its rigid structure and it can only possess kinetic energy. The work done by any force acting on a rigid body is eventually equal to the change in its kinetic energy and it is the basis of the work-energy equation for rigid bodies.

Limitations of Work-Energy Theorem

Initially, the rule came from Newton’s second law, and hence it is applicable to the particles. Objects that are like particles are considered for this rule. So, if all the object particles behave like particles, we can consider the whole object as a particle.

Work energy theorem is used to solve different types of problems but it has a limitation that it does not give complete information about the real cause of motion which is the dynamics of Newton’s second law of motion and is called the scalar form of Newton’s second law of motion. Work energy theorem also does not define the direction of velocity.

Solved Examples

Example 1. Consider a bullet that has 20g mass and a velocity of 500 m/s. It suddenly struck a tree and went to another side with 400 m/s velocities. Find the work that is done by the bullet. 

A) 900 J    B) 500 J   C) 800 J   D) 950 J

Solution:

Mass of the bullet (given), m = 20 g (= 0.02 kg). 

Beginning velocity of the bullet = 500 m/s. 

Closing velocity of the bullet = 400 m/s. 

Calculate the given bullet’s energy difference, and we can get the work done on the tree by the bullet. 

Hence, according to the Work-Energy Theorem, 

we have: 

Δ(K.E.) of the bullet = 1/2{0.02(500)2 – 0.02(400)2}

Therefore, 

Δ(K.E.) of the bullet = 900 J

Example 2. A block of mass 10 kg starts moving up the incline with 20m/s. It reaches the top and comes back to its initial position and stops. What is the work done by friction in the whole process?

Solution:

If we want to use the formula of work, we need the friction coefficient to calculate the frictional force. But that is not given. So let’s attempt to implement the work-energy theorem. Forces acting on the block are gravity, normal reaction, and frictional force. So,

Wf + WN + Wg = Kf − Ki

Here, WN is zero as force is always perpendicular to the displacement. Being a conservative force Wg is zero as the body returns to its initial position. Also,

Ki = 12 × 10 × 400 = 2000 J

Kf = 0

Wf = −2000 J

Hence the work done by friction is negative.

Conclusion

The article is very helpful for students to understand the work-energy theorem. It also contains some solved examples and applications of the work-energy theorem.

Heat of fusion, also called the specific latent heat of a substance, is the amount of heat energy provided to one gram of substance that changes its state from a solid to a liquid keeping pressure constant.

For example, If we want to change ice to water, a specific amount of heat is required which will be dependent upon the heat of fusion of ice and the amount of water present.

The Heat of fusion is denoted by Δl

The heat of fusion formula is given as

q = m·Δl

where

q is heat energy

m is mass

ΔHf is the heat of fusion

Example:

Find the amount of heat needed to melt 200gms of ice, if the heat of fusion of ice is 330 J/gm

Ans:

Mass of ice(m) = 200gm

Heat of fusion(Δl) = 330 J/gm

Heat needed(q) = m.Δl

= 200 x 330

=66000J or 6.6 x 104 J.

Applications of Heat of Fusion

The applications related to the heat of fusion can be seen very commonly in our daily life around us. Many of the popular household products are created using the heat of the fusion process, which can be seen in a variety of applications. The most common and easy example of use of the heat of fusion is the melting of ice into water form. The manufacturing industries are home to the many great majorities of examples of the heat of fusion that are in use. The following examples have been worked upon to make them perfect over hundreds of years and are still being worked on to be used today. The heat of fusion is a major requirement to perform the operation of coin production, glassblowing, forging metal items, and even in transforming blow moulded polymers into household products. The plastic Coke bottle from the vending machine, the change in your pocket, and the vase of glass on a fireplace mantle all have been created and went through a heat of fusion manufacturing process.

Copper and Solid Zinc, which are also the metals used in American pennies, are heated at a high temperature in a casting furnace until they attain the liquid phase using the heat of fusion method. The molten zinc and copper are then poured into a mould and are moulded into the shape of long bars once they have reached their liquid phase. The molten metal transforms from its liquid form to a solid-state throughout this process of casting, resulting in a solid bar. Heavy machinery is used which helps in flattening these long bars, which are then stamped over and are shaped into thousands of coins. A monetary system would not be easy and could not exist in the United States without the process of heat of fusion.

Concept of Heat of Fusion

The concept of heat of fusion can be understood as when the amount of energy required to melt a given quantity of a solid at its melting point temperature is measured by heat of fusion. In the other Words, when a certain mass of liquid transforms and solidifies, it also indicates the amount of energy lost during this. For example, we can see that water has a fusion heat of around 80 calories per gram. This means that it consumes 80 calories of energy to melt 1 gram of ice at the temperature of zero degrees Celsius into water. The value of heat of fusion can vary here depending upon what type of substance is being used.

We can see that the heat received by ice is equal to the heat lost by water, for example. The heat of fusion is represented by the symbol ΔHf.

Melting is the process through which a solid material transforms into a liquid. To allow the solid-state particles to break loose from one other, the melting process will require an increase in energy. Fusion heat is the source of this energy. Although the heat of fusion is not the same for all substances, it is a constant value for each type of substance.

Physics is one of the most fundamental scientific disciplines. It aims to study that portion of natural science which deals with motion, behavior, energy, and force, through space and time. 

In this particular article, we will discuss one such important topic of Physics, related to Position. The team of experts has explained the concepts in the simplest language for the students to grasp them without much difficulty. 

In this article, students will be able to learn the following concepts: 

Position Formulas – Introduction 

In order to know the motion of an object, it is important to understand its position. To describe the position, we need to take a point of reference. Often, we make use of the Earth as a point of reference to identify the position of an object. The point to note here is that we should take the reference which is generally not in a stationary position, and thus, is in motion. Follow the full article to understand the concept well. 

What is the Position of an Object?

The true position of an object is its location, which is determined by the basic dimensions. The true position helps to control the variation of a specific feature from its desired position.

For example, the screw holes on a cylinder head cover must match the screw holes on the engine casing in size as well as position. In case they don’t, the two parts will not mate, and the lubrication oil for the valve assembly will leak out, defeating the purpose of the cover. The cylinder cover is useless, and a different piece must be used.

The Formula for the Position is Represented as: 

Case 1: 

[Delta x = x_2 – x_1],

Where x1 is the first position of the body, 

x2 is the second position after undergoing displacement,

And  Δx is the rate of change in the displacement. 

 

Case 2: 

If the body changes its position after time t, the rate of change in position at any moment of time t, x(t), is articulated as

 

[x(t) = frac{1}{2} alpha t^2 + v_0t + X_0]

 

Where x(t) is the position of the body with time t,

x0 is the initial position of the body,

v0 is the initial velocity of the body, and 

α is the acceleration the body possesses.

 

Example:

A boy who has an initial velocity of 2 m/s had already covered a distance of 10 m. If it has a constant acceleration of 2 m/s2, find the position of the boy at the end of 5s.

Ans:

Position of the boy = [x(t) = frac{1}{2} alpha t^2 + v_0t + X_0]

=½ x 2 x 25 + 2 x 5 + 10  = 45m

Tips for Preparation for IIT JEE     

The exam is competitive because lakhs of students appear every year, but with the right guidance and organized preparation, even you can clear this exam. Some of the tips that our experts give to the aspirants are as follows: 

• Knowing the syllabus well is the first step, get a printout of the syllabus and try to remember all the topics and subtopics mentioned in the syllabus.

• Deciding on your study material; you should not confuse your mind by referring to too many sources for a single topic. Identify your source and stick to one source for one topic. You can check out the study materials provided on the website. 

• Have a clear idea about the marks allocation to every topic. We have already done that for you. Go through these before you start off your preparation.

• Read the topics from the notes on your own once, and watch the related videos to grasp the concept better.

• Simultaneously, start practising the past year’s JEE papers and other sample papers. Compare and learn from the solution of these papers prepared by our subject experts. 

• Keep practising and keep revising the notes, make your own short notes for quicker revision. 

• Interact with the mentors to get regular feedback, support, and guidance to keep you on track. 

Nobody can stop you if you keep following the process with sincerity and dedication. 

Average velocity is the expression of total displacements possessed by a body in time ‘t’. So, if a body travels various displacements viz: s₁, s₂, s₃,……,n in a given time ‘t’, then finding average velocity becomes easy by using the following average velocity equation:

In SI systems, the unit for the average velocity is the same as that of the velocity, i.e., m/s. In this article, we will derive the formula to find average velocity and also the average speed equation.

Average Speed Formula Derivation

Let’s suppose that a particle travels varying distances viz: s₁, s₂, s₃, s₄,…., sn in their respective velocities viz: v₁, v₂, v_3,….,v_n in the same direction, so we know that the time is equivalent to the division of distance by the speed, for this scenario, we have the following formula:

Total time taken ‘t’  = [frac{s_1}{v_1}] + [frac{s_2}{v_2}] + [frac{s_3}{v_3}] + ….. + [frac{s_n}{v_n}]

We know that the speed = distance/time, so here the formula for the average speed goes:

Average speed v_av = [frac{s_1 + s_2 + …..}{(frac{s_1}{v_1} + frac{s_2}{v_2} + ……)}] 

Now, if we say the varying distances are the same and the body travels with two varying speeds v₁ and v₂ over two equal distances, i.e., s₁ = s₂, the formula becomes the following:

= [frac{2s}{(frac{s}{v_1} + frac{s}{v_2} + ……)}]

= [frac{2v_1 v_2}{(frac{1}{v_1} + frac{1}{v_2})}]

vav = [frac{2v_1 v_2}{(v_1 + v_2)}]

This expression clearly describes us with an idea that average speed is the equivalent of the harmonic mean of individual speeds.

Average Speed Formula in Physics Derivation

If we say that a man travels to his favorite hilly area with varying speeds viz: v₁, v₂, v₃,…., and so on during time intervals viz: t₁, t₂, t₃,…, and so on, respectively. 

So, the distance travelled will be: v₁t₁ + v₂t₂ + v₃t₃

Time taken in total is t₁ + t₂ + t₃ +…..

So, the average speed by which he reached his journey is given as:

vav = [frac{v_1t_1 + v_2t_2 + v_3t_3}{t_1 + t_2 + t_3+…..}] 

If we suppose that the time taken in each interval is the same, i.e., t₁ = t₂ = t₃ =,…..= t_n, then we get the new equation in the following way:

vav = [frac{v_1 + v_2 + v_3…}{n}] 

We notice that the average speed is the arithmetic mean of the individual speeds.

Unit of Average Speed

Here, if you determine the unit of the average speed, it comes out in the following manner:

= [frac{2(m/s)(m/s)}{(m/s + m/s)}]

Canceling the common terms, we get the unit as m/s again.

Dimensional Formula of Average Speed

The dimensional formula for the average speed is as follows:

= [frac{2(LT^{-1})(LT^{-1})}{(LT^{-1}+LT^{-1})}]

The dimensional formula for the average speed comes out as LT[^{-1}].

Finding Average Speed Solved

Let’s suppose that you and your friend decide to go on a bike ride to a different state, i.e., Kerala. 

Your bike is a duke and your friend’s bike is BMW. Your bike can travel at a speed of 300 kmph, while your friend’s at the speed of 500 kmph, now, the average speed formula Physics says that if the two speeds are given then we can determine the average speed, and that is given by:

v_av = [frac{2v_1v_2}{v_1+v_2}] 

= [frac{2 ast 300 ast 500}{300 + 500}] = 375 kmph

So, the average speed comes out as 375 kmph, so this was another formula or you can say the short trick to determine the average speed. 

Now, we have something very interesting besides speed (a scalar quantity) and that is velocity, so do you know what is velocity?

Average Velocity Equation

A velocity is something that is the division of the displacement of an object by time. In simple terms, it is the displacement possessed by an object in a unit of time. we can determine the average velocity of the object’s journey.

If we know the initial velocity and the final velocity of an object, we can determine the average velocity as:

v_av = [frac {(u + v)}{2}]

Here,

u = initial velocity of the object and v is its final velocity. Both of these are measured in m/s and their mean is the average velocity, i.e., vav. The average velocity is also measured in m/s and its dimensional formula is LT[^{-1}] .

Another equation for the average velocity is:

         =  (Final position  – initial position)/(end time – starting time)

If we have to calculate the velocity at an instant, then the formula turns to the instantaneous velocity formula. 

Finding Average Velocity

Example:  A boy walks with an initial velocity of 40 m/s and he reaches his destination at 90 m/s. Calculate its average velocity. 

We are given the following data:

Initial Velocity u = 40 m/s

Final velocity v = 80 m/s

The formula to find average velocity is as follows:

vav= [frac {(u + v)}{2}]

So, our answer comes out as:

Average velocity vav = [frac {(40 + 90)}{2}] = 65 m/s.

 

Conclusion

We have seen various formulas for calculating the average speed and average velocity for different situations. The average speed formula helps in determining the average value of speed if the body is varying continuously for the given time intervals. velocity can be positive or negative depending on the direction we are choosing as positive while speed always remains positive.

(Image to be Updated soon)

 

The orbital velocity of a body is the velocity at which it revolves around the other body. The objects that travel around the Earth in uniform circular motion are said to be in an orbit. The velocity of this orbit depends on the distance between the object under consideration and the center of the earth. This velocity is usually associated with artificial satellites because they can revolve around any particular planet. The Orbital Velocity Formula is used to calculate the orbital velocity of an object if its mass and radius are known.

 

The moving body has a tendency to move in a straight path because of its inertia. The gravitational force, on the other hand, tends to drag it down. The orbital route, which is elliptical or circular in shape, is a balance of gravity and inertia. The velocity required to achieve a balance between gravity’s pull on the body and the inertia of the body’s motion is known as orbital velocity. The orbital velocity of a satellite orbiting around the Earth is determined by its altitude above the Earth. The faster the required orbital velocity, the closer it is to the Earth.

 

At lower altitudes, a satellite collides with traces of Earth’s atmosphere, causing drag. This drag causes the satellite’s orbit to degrade, eventually causing it to fall back into the atmosphere and burn up.

 

What Is Orbital Velocity?

A natural or artificial satellite’s orbital velocity is the velocity required to keep it in orbit. The moving body’s inertia causes it to move in a straight line, while gravitational force pulls it down. The elliptical or circular orbital path thus shows a balance between gravity and inertia. If the muzzle velocity of a cannon shot from a mountaintop is increased, the projectile will go further. The bullet will never fall to the ground if the velocity is great enough. The Earth’s surface can be imagined curving away from the projectile, or satellite, at the same rate that it falls toward it. The higher the orbital velocity for a given height or distance, the more massive the body is at the center of attraction. If air resistance could be ignored near the Earth’s surface, orbital velocity would be around eight kilometers (five miles) per second. The gravitational force is weaker the farther a satellite is from the center of attraction, and the less velocity it needs to stay in orbit. See also the definition of escape velocity.

 

Orbital Velocity Equation

The equation of the orbital velocity is given by:

 

[V_{orbit} = sqrt{frac{GM}{R}}]

 

In the above equation, G stands for Gravitational Constant, M stands for the mass of the body at the center and R is the radius of the orbit.

 

The Orbital Velocity Equation is used to find the orbital velocity of a planet if its mass M and radius R are known. 

 

The unit used to express Orbital Velocity is meter per second (m/s).

 

Derive An Expression for Orbital Velocity

To derive the orbital velocity, we first need to know about the gravitational force and the centripetal force. It is important to know about the gravitational force because it is this force that allows orbiting to occur. This gravitational force is exerted by a central body on the orbiting body to keep it in its orbit. Centripetal force is also important because it is this force that is responsible for circular motion.

For the derivation of the formula, let us take a satellite of mass m which revolves around the Earth in a circular orbit of radius r at a height h from the surface of the Earth. Let M and R be the mass and radius of the Earth respectively, then r=R+h.

 

To resolve the satellite, a centripetal force [frac{mv_{0}^{2}}{r}] is needed which is provided by the gravitational force of the G[tfrac{Mm}{r^{2}}] acting between the satellite and the Earth.

Thus, equating the two equations, we got

[frac{mv_{0}^{2}}{r} = Gtfrac{Mm}{r^{2}}]  

[v_{0}^{2}= frac{GM}{r}= frac{GM}{R+h}]

Simplifying the above equation, we get

[v_{0}= sqrt{frac{GM}{R+h}}]…………….(1)

Also,GM = g[R^{2}], where g is the acceleration due to gravity

Therefore,

[v_{0}= sqrt{frac{gR^{2}}{R+h}}]

Simplifying the above equation, we get

[v_{0}= Rsqrt{frac{g}{R+h}}]

Let g’ be the acceleration due to gravity at a height h from the surface

[g^{‘}= frac{GM}{(R+h)^{2}}]

 

Conclusion

Orbital speed is a very important and rewarding chapter in physics. It helps with the understanding of orbital concepts in physics. With thorough research, you can achieve very good results in this area.