Physics | Physics & Electronics

Understanding Projectile Motion with Derivations

5 February 20264 November 2024 by Saurabh Salvi

Projectile Motion is one of the most common concepts we encounter in our daily lives. Studying it becomes important because the ‘understanding’ (not mugging up formulae) that we take here comes in handy while analyzing complex concepts and experiments.

“Formulae give you Marks, but derivations give you understanding!

Another Fun article to read to understand the above quote would be the following: What is a Machine? -3 idiots machine definition

We already have a video on our Channel about such an experiment, on our channel, where knowledge of Projectile is needed for the Analysis part :

Calculating e/m Ratio – Watch the full video on YouTube by clicking here

Table Of Contents

1. Equation of Trajectory for the Projectile
2. Finding Expression for Horizontal Range
3. Finding Time of Flight Expression
4. Finding Maximum Height Covered
Conclusion

1. Equation of Trajectory for the Projectile

We all have played ‘catch-catch,’ and we know what the motion of a ball looks like when it’s thrown. But how do you describe that curve mathematically? What’s the equation of that curve?

The figure helps to divide the projectile motion into 2 separate axes (X and Y)

Most Important Tip that fixes everything related to Projectile!

Divide this whole 2-D situation into separate 1-D problems (X and Y)

Please make sure to note down all quantities in X separately. This will be your separate problem
Note down all quantities in Y separately. This will be your other separate problem
At last, combine them to get results for 2-D motion.

Following the tip:

X direction

$u_x = u \cos \theta$

$a_x = 0$

Y direction

$u_y = u \sin \theta$

$a_y = -g$

Starting from the origin (point where the ball is thrown) till point P (x,y):

Solving in the X direction:

$S_x = u_x t + \frac{1}{2} a_x t^2$

Substitute the values as discussed above

$x = u \cos\theta \times t$ –(1)

Solving in the Y direction:

$S_y = u_y t + \frac{1}{2} a_y t^2$

$y = u \sin\theta \cdot t + \frac{1}{2} (-g) t^2$ –(2)

Substitute values of ‘t’ from equation (1) into equation (2). This gives us an equation only in terms of x and y. This is what we refer to as ‘Locus’ or Equation of Trajectory here.

$y = x \tan\theta – \frac{1}{2} \frac{g x^2}{u^2 \cos^2\theta}$

The above equation is quadratic in nature. The Projectile Trajectory is a Parabola!

2. Finding Expression for Horizontal Range

Range is basically the ‘horizontal’ distance that the ball covers (from the origin to the point where it lands). It means that we need to find R in the figure

This can be easily found since x = 0 and x = R are the two roots of the parabola.
To calculate the roots, simply put y = 0 in our Equation of Trajectory

The figure is used to derive an expression for the range

Calculations :

Substituting y = 0 and taking ‘x’ common on the RHS of the Equation of Trajectory, we get the equation as:

$0 = x \left( \tan\theta – \frac{g x^2}{2u^2 \cos^2\theta} \right)$

Hence, we get some possibilities:

1st possibility:

$x=0$

This is quite obvious since the ball was at ground level before it was just thrown

2nd possibility:

$0 = \left( \tan\theta – \frac{g x^2}{2u^2 \cos^2\theta} \right)$

On Simplifying,

$x =R = \frac{u^2 \sin 2\theta}{g}$

* In the simplifying process, you need to use the $\sin 2\theta$ trigonometric identity

3. Finding Time of Flight Expression

Let’s find out how much time the body stays in the projectile motion.

Note that :

On landing, after completing the motion, the displacement in the Y direction is zero (Pause and observe !), since it again came to the same Y-level (S_y = 0 here)
The time taken for this displacement in Y to become 0 is nothing but the Time of Flight (T)

In Y direction:

$S_y = u_y t + \frac{1}{2} a_y t^2$

As discussed, make the following substitutions: S_y = 0; t = T

$0 = u \sin\theta \cdot T + \frac{1}{2}(-g)T^2$

On simplifying,

$T = \frac{2u \sin\theta}{g}$

4. Finding Maximum Height Covered

If we look at the Equation of Continuity, in this case (simple throwing), the trajectory resembles a downward parabola since a<0

$y = x \tan\theta – \frac{1}{2} \frac{g x^2}{u^2 \cos^2\theta}$

Comparing this with $y = ax^2 + bx + c$ , we would get,

$a = -\frac{g}{2u^2 \cos^2\theta}$

$b = \tan\theta$

$c = 0$

Hence, visualizing it graphically,

The figure is used to derive an expression for the Maximum height covered by the projectile using mathematical approach

Finishing the Calculations,

On calculating (-b/2a), we get x = R/2
On calculating (-D/4a), we get:

$y = H = \frac{u^2 \sin^2\theta}{2g}$

So, this is how we get the expression for the Maximum Height reached, denoted by ‘H.’

Conclusion

So, this is how we derive the expressions for Time of Flight, Range, Maximum Height, and Equation of Trajectory. Note that – With this same approach, we can solve almost any kind of problem related to Projectile Motion. And, why is that the case?

Because whatever we discussed in this blog/article is not a trick or something, it’s a complete concept with a complete understanding. This makes us equipped to solve any kind of problem (Throwing a ball from a cliff, a projectile on an incline, or just anything for that matter). You are now in a very good position to solve problems

Stress vs Strain? – Which comes first?

5 February 20262 November 2024 by Saurabh Salvi

It’s Stress vs Strain, and you won’t want to miss the action!

‘Stress‘ and ‘Strain‘ are the most encountered terms when it comes to studying Elasticity. Though it might not have such a huge weightage in competitive exams like JEE/NEET, but trust me, ladies and gentlemen: “There’s no Mechanical Engineering without these 2 terms” – Being a student pursuing my UG degree in Mechanical, I can say this with full assurance. And if there’s no Mechanical, there are no cars, bridges, buildings, etc.

Through this article, we are going to find out which one of the two comes first – It’s going to be Stress vs Strain?

Table Of Contents

1. Introduction to Stress:
2. Introduction to Strain:
3. Final Decision: Stress vs Strain | Who wins the Match?

1. Introduction to Stress:

Stress is defined as the Internal Restoring Force acting per unit area.

Now, What is this Internal Restoring Force? Let’s understand the process to know what happens inside the material.

The atoms inside the solid are arranged in a spring-ball system. So, when a load (external force) is applied, it disturbs the equilibrium state by deforming the springs. This deformation is responsible for the Internal restoring force, and we call it restoring because it tends to bring the system back to its equilibrium.

The following flow-chart explains the process :

Explanation of Flowchart (Expand This)

First, the external force is applied. As we have already discussed, the spring ball system is an analogy for the arrangement of atoms in bodies. This external force causes a disturbance in this system. As we all know, springs are elastic in nature, and hence, they give birth to a restoring force in order to restore back to equilibrium. This indirectly induces stress in the body, as stress is defined as Restoring Force per unit area

The flowchart shows the insider view of what happens inside the body when an external force is applied. This actually decides who's the winner in Stress vs Strain — Fig. Flowchart

2. Introduction to Strain:

Strain is defined as ‘Change in Dimensions / Original Dimensions.’

$Strain = \frac{Change}{Original}$

Again, there are types of strain :

Longitudinal Strain – Change happens in the length
Shear Strain – There is a shift that leads to an angle change
Volumetric Strain – Change happens in the Volume

Representation of what happens during longitudinal strain — Fig. Longitudinal Strain (expansion)

Representation of what happens during shear strain — Fig. Shear Strain (measured as angle)

Representation of what happens during volumetric strain — Fig. Volumetric Strain (compression)

Important Note :
A careful observation of the Flowchart above would tell that: There is ‘Strain’ coming into the picture at the second step, since on applying load, there is a deformation happening. This is exactly what we discuss in Strain.

The flowchart now reveals the secret and decides who's the actual winner in the Stress vs Strain match

Explanation of the Flowchart (Expand This)

Here, we are actually revealing the secret – The disturbance that the external force caused inside the atom arrangement was nothing but the strain that came into existence, and then you get the restoring force.

3. Final Decision: Stress vs Strain | Who wins the Match?

Now, it’s very much clear from the above discussion that: it’s the Strain which comes first! All of this is because of the Definition of Stress. Most of the time, we just memorize the formula of Stress as ‘Force/Area’, which is not complete.

The Complete Answer is: Stress = Restoring Force/Area

What is stress in a material?

Stress is the internal resistance developed within a material when an external force acts on it. It is defined as force per unit area. Basically, it represents how strongly the material tries to oppose the applied load.

What is strain and how is it different from stress?

Strain is the deformation or change in dimension that occurs due to applied stress. It is the ratio of the change in length to the original length. Stress depends on external force and area, while strain describes the actual deformation the material undergoes.

Why is strain a dimensionless quantity?

For this, refer to the definition in the second question. Strain is calculated by dividing the change in length by the original length. Since both quantities have the same units, they cancel out. This makes strain a pure ratio with no units.

Can there be no stress while strain is non-zero? (Conceptual)

Yes, this is very much possible. This situation occurs during free expansion or contraction, such as when a material changes length due to temperature but is not externally restricted.

Perpendicular Axis Theorem + Mixed Problem

5 February 202616 October 2024 by Saurabh Salvi

What is the Perpendicular Axis Theorem, and where do we even use it?

In the last article, we learned about a very useful theorem called the ‘Parallel Axis Theorem‘. But as well know, it can be applied only when the 2 axes under consideration are parallel to each other.

But what if we want to know MOI about an axis that is not in the plane? Perpendicular Axis Theorem comes to our rescue

Table Of Contents

1. Perpendicular Axis Theorem
2. Examples based on Perpendicular Axis Theorem
3. Practice Question – Solve it yourself!
FAQ section

1. Perpendicular Axis Theorem

The figure explains the perpendicular axis theorem

Conditions:

Applicable to only planar 2-D bodies
3 axes to be considered
2 axes in the plane of the body, and 3rd should be perpendicular to both
All 3 axes need to be concurrent (all of them should pass through the same point)

Descriptive Statement:

‘The moment of inertia of the planar body about an axis perpendicular to the plane is equal to the sum of moment of inertia of two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.’

(Observe that all the points have been covered in the ‘Conditions’ section above)

Mathematical Expression :

$I_z = I_x + I_y$

2. Examples based on Perpendicular Axis Theorem

Question: Find the moment of inertia Ip passing through the center of mass C of the square plate having mass M and side length L.

The figure related to the question is given.

Solution:

Would there be any difference in the answer you get in Figure 1 and Figure 2 ? (Remember, it’s a square plate)

In Figure 1, the square plate rotates about the vertical axis passing through its center of mass C — Figure 1

In Figure 2, the square plate rotates about the horizontal axis passing through its center of mass C — Figure 2

The answer is: NO! This is because of the beautiful symmetry that this square plate holds about the axis shown in both cases. There is the same mass distribution about the axis in both cases. That’s the reason you can’t really make out the difference!

Step-1:

Remember, we need 3 axes: 2 in plane and 1 perpendicular to them and concurrent. In this, we already got 2 planar axes (Combine figure 1 and figure 2). These will be our Ix and Iy. Hence, I_x = I_y = I_p. We get

We combine figures 1 and 2, and draw both axes in a single figure

Step-2:

Now, we also know the standard MOI for a square plate about an axis passing through C and perpendicular to the plane of the square plate. i.e. ML²/6. This is our I_z

Square plate with central point C, symmetry axes marked, rotation axis out of the plane, and ML^2/6 shown toward point C.

Step-3:

Applying the perpendicular axis theorem,

$I_z = I_x + I_y$

$\frac{ML^2}{6} = I_p + I_p$

$\frac{ML^2}{12} = I_p$

3. Practice Question – Solve it yourself!

Now that we have learnt about the Parallel and Perpendicular Axis theorem, we are in a good state to apply this to a problem that requires both these theorems. (This itself is a good hint)

Question: Find the moment of inertia Ip for the uniform disc of mass M and radius R

Solid disk of mass M and radius R with a vertical dashed axis tangent to its left edge, showing rotation about that axis with moment of inertia as Ip

FAQ section

What does the Perpendicular Axis Theorem state?

The theorem states that for a planar (2D) lamina, the moment of inertia about an axis perpendicular to its plane (Iz) is equal to the sum of its moments of inertia about two perpendicular axes lying in its plane and passing through the same point:
Iz = Ix + Iy

Why is the Perpendicular Axis Theorem applicable only in a plane?

Because the theorem assumes all mass lies in a single plane. Only then can the perpendicular axis moment (Iz) be expressed purely in terms of in-plane mass distribution. In 3D bodies, mass exists off the plane, so the simple relation Iz = Ix + Iy no longer holds.

What is the physical meaning of adding Ix and Iy to get Iz?

Ix and Iy measure resistance to rotation within the plane. When you rotate the lamina about a perpendicular axis (z-axis), the mass resists rotation in both x and y directions simultaneously. Therefore, both contributions add up, giving a larger overall moment of inertia Iz.

Does the theorem work if the axes do not intersect at the same point?

No. All three axes (x, y, and z) must intersect at the same point for the theorem to hold. If axes pass through different points, the mass distribution relative to each axis changes, and the relationship Iz = Ix + Iy is no longer valid without applying the parallel axis theorem first.

Parallel Axis Theorem (in detail)

5 February 202616 October 2024 by Saurabh Salvi

To calculate the moment of inertia about an unknown axis, we often take the help of 2 Theorems, namely :

Parallel Axis Theorem
Perpendicular Axis Theorem

There’s one thing common to both, which is that you need to know at least one moment of inertia about an axis. This will act as a reference for you while calculating the unknown moment of inertia (MOI)

Table Of Contents

1. Parallel Axis Theorem
2. Example
3. Application- Where do we use parallel axis theorem?
FAQ section

1. Parallel Axis Theorem

conditions for applying parallel axis theorem — Figure 1

1.1 Conditions to apply

Applicable to all types of bodies
The axis through the center of mass (COM) should be parallel to the axis about which the MOI is to be found out

Please Note:
There are infinite axes passing through the center of mass C. Don’t just choose any axis passing through C. Choose only that axis passing through C that is parallel to the required axis.

1.2 Theorem

The mathematical equation for this theorem can be given as :

$I_p = I_{com} + Mh^2$

Where,

I_p is the MOI about the required axis

I_com is the MOI about the axis passing through the center of mass (COM)

h is the distance between the parallel axes

M is the mass of the body

1.3 Important Observation

$I_p = I_{com} + Mh^2$

There’s an important thing to note here – The term Mh² is a positive term. Hence, we can say that among all the parallel axes in the plane (shown in Figure 1), the moment of inertia of the body about the axis passing through the COM is the least. We also know the expression for torque:

$\tau = I\alpha$

Therefore, we can say that, for rotations in a given plane, choosing an axis through the center of mass gives the greatest angular acceleration for a given torque

2. Example

Question: Find the moment of inertia of the rod about the axis passing through P. The rod has mass M and length L

Uniform rod of mass M and length L fixed at one end with a dashed vertical axis at the support; Ip about that axis is to be found.

Solution :

Step-1: Choose an axis parallel to the required one, and it must pass through the COM of the body

Uniform rod of mass M and length L with a dashed axis at the center indicating Icom = ML^2/12, the pivot is at the left end at a distance L/2

The axis passing through the center of mass of this uniform rod can be given as ML²/12. This makes the distance ‘h’ between the axis about which we have to find the moment of inertia and the axis passing through the center of mass to be L/2

Step-2: Apply the Parallel Axis Theorem

And we are already aware of the standard MOI about the axis passing through the COM for the rod
From Step-1, we know h = L/2

$I_p = I_{com} + Mh^2$

$I_p= \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2$

$I_p = \frac{ML^2}{3}$

3. Application- Where do we use parallel axis theorem?

RC airplane with arrows labeling the elevator, rudder, and ailerons. — Control Surfaces on RC Aircraft

If you have ever played with RC Airplanes or even have seen an actual aircraft, you must observed that an aircraft has many control surfaces like ailerons, rudder, and elevator. These are essential to properly maneuver the aircraft as required by the pilot

But, if you observe closely, these control surfaces are hinged about an axis that does not pass through their centre of mass. Here is exactly where you need the Parallel Axis Theorem as an aid to complete your calculations.

If you are interested, you can explore more design details in the RC airplane series

FAQ section

Why do we need the Parallel Axis Theorem in the first place?

The moment of inertia of most bodies is easiest to calculate about their centre of mass. However, in real life, objects often rotate about other axes (like hinges, supports, rods, or edges). The parallel axis theorem helps us find the moment of inertia about any such shifted axis without redoing the full derivation.

Can the Parallel Axis Theorem be used for any shape?

Yes. The theorem works for all rigid bodies, independent of shape or symmetry, as long as the new axis is parallel to the centre-of-mass axis.

Is the theorem valid if the two axes are not exactly parallel?

No. Remember that the shift must be perpendicular to the axis and must maintain parallelism. If the new axis is not parallel to the original one, the moment of inertia cannot be found using this theorem, and in that case, you need to carry out the whole derivation.

Does the moment of inertia always become larger when the axis is shifted?

Yes. When you move the axis away from the centre of mass, the body’s mass is effectively farther from that new axis. This increases the resistance to rotation, so the moment of inertia always becomes larger compared to the value at the centre of mass.

Writing the Coefficient of Restitution Formula

5 February 202629 August 2024 by Saurabh Salvi

Topic Under Chapter – Center of Mass and Collisions

Coefficient of Restitution, also called COR, is one of the important concepts to be taken into consideration when it comes to designing sports equipment like badminton racquets, tennis racquets, and several types of balls, like baseball, basketball, and cricket ball, etc.

This is mainly because of the fact that these sports like basketball, badminton, tennis, etc. involve collisions, which makes terms like collision energy and rebound energy come into the picture. To relate how good the bounce will be, we have a ratio known as the ‘Coefficient of Restitution’. As simple as that!

Table Of Contents

1. Rules to Remember
2. How to use the expression of coefficient of restitution?
FAQ section

1. Rules to Remember

There are 2 cases :

Before collision (Deals with the Approach of the bodies)
After Collision (Deals with the Separation of the bodies)

For Before Collision (V_app): The velocity component that supports the approach, i.e., that component that feels as if bodies should approach each other, is positive.

For After Collision (V_sep): The velocity component that supports the separation, i.e., that component that feels as if bodies should get separated from each other, is positive.

2. How to use the expression of coefficient of restitution?

We have 2 bodies (body 1 and body 2) undergoing a collision. We will be trying to write the Coefficient of Restitution for each case.

Case 1:

Writing Velocity of approach (V_app) first :

Both u₁ and u₂ want the approach of the bodies to happen. So both will be positive. Hence, V_app is ‘u₁+u₂.’

Writing Velocity of separation (V_sep) :

v₁ and v₂ both want separation to happen. So, +v₁ and +v₂. Hence V_sep will be ‘v₁ + v₂‘

The expression for case 1 can be written as:

$e = \frac{V_{\text{sep}}}{V_{\text{app}}} = \frac{v_1 + v_2}{u_1 + u_2}$

Case 2:

Writing Velocity of approach (V_app) first :

u₁ wants an approach, but u₂ doesn’t want that. So, +u₁ but -u₂. Hence, V_app is ‘u₁-u₂.’

Writing Velocity of separation (V_sep):

v₁ wants to separate but v₂ doesn’t want that. So, +v₁ but -v₂. Hence V_sep is ‘v₁-v₂‘

The expression for case 2 can be written as:

$e = \frac{V_{\text{sep}}}{V_{\text{app}}} = \frac{v_1 – v_2}{u_1 – u_2}$

Case 3:

Writing Velocity of approach (V_app) first :

Both u₁ and u₂ want the approach of the bodies to happen. So both will be positive. Hence, V_app is ‘u₁+u₂.’

Writing Velocity of separation (V_sep):

v₂ wants to separate but v₁ doesn’t want that. So, -v₁ but +v₂. Hence V_sep is ‘-v₁+v₂‘

The expression for case 3 can be written as:

$e = \frac{V_{\text{sep}}}{V_{\text{app}}} = \frac{-v_1 + v_2}{u_1 + u_2}$

Case 4:

Writing Velocity of approach (V_app) first :

u₁ wants approach but u₂ doesn’t want that. So, +u₁ but -u2. Hence, V_app is ‘u₁-u₂.’

Writing Velocity of separation (V_sep):

v₁ and v₂ both want separation to happen. So, +v₁ and +v₂. Hence V_sep will be ‘v₁ + v₂‘

The expression for case 4 can be written as:

$e = \frac{V_{\text{sep}}}{V_{\text{app}}} = \frac{v_1 + v_2}{u_1 – u_2}$

Important to Note:

The value of the coefficient of restitution actually depends on how materials react or deform during collision, i.e., whether they store the energy elastically or lose it in the form of heat. Hence, the materials that come in contact with each other during collision also play a very important role

We can assess the value based on the stress-strain curves of those materials.

FAQ section

What is the Coefficient of Restitution?

Coefficient of Restitution or COR is the Ratio of the Rebound speed (velocity of separation) to the collision speed (velocity of approach). It helps in determining the intensity/type of collision that can take place.

How to decide whether a collision is elastic or inelastic based on the coefficient of restitution?

If the Coefficient of Restitution (COR) is 1, the collision is elastic; i.e., no kinetic energy is lost. If the COR is <1, it implies a partially elastic collision; if the COR is 0, the collision is called ‘perfectly inelastic.’

What is Total Internal Reflection (TIR)?

5 February 202624 August 2024 by Saurabh Salvi

Introduction:

Total Internal Reflection, also known as TIR, is one of the useful phenomenon which has applications in a lot of areas. One such example is of Optical Fiber Cables. In this Article, we will be discussing TIR: What is it? Under what conditions does it happen? What is the difference between ‘normal reflection’ and ‘total internal reflection’?

Table Of Contents

1. What is meant by reflection?
2. How does Total Internal Reflection happen?
3. Difference between Normal and Total Internal Reflection:
A Question from JEE Advanced Previous Year Papers:
FAQ section

1. What is meant by reflection?

In simple words, Reflection is nothing but the bouncing back of light into the same medium once it hits a polished hard surface (usually a mirror).

Incident and reflected rays meeting a surface with the normal shown between them.

2. How does Total Internal Reflection happen?

When a light travels from one medium to another, it bends. But whether it bends toward the normal or away from the normal depends on the fact from which medium to which medium it is going.

Rarer to Denser – bends towards the normal
Denser to Rarer – bends away from the normal

Ray entering a denser medium from a rarer medium, bending toward the normal.

Ray entering a rarer medium from a denser medium, bending away from the normal.

Now, this specific case of Total Internal Reflection (TIR) happens when the light travels from a denser medium to a rarer medium.

Rays from a denser to rarer medium showing critical angle, grazing ray, and total internal reflection.

From the figure, as we keep increasing the angle of incidence, the angle of refraction also increases until the critical angle is reached.

Critical Angle :

For each medium interface, we have a special angle defined, known as the critical angle. In simple words, it is the angle of incidence at which the refracted ray grazes along the medium-separating interface (the angle of refraction becomes 90 deg).

By Snell’s law,

$\mu_1 \sin\theta_c = \mu_2 \sin 90^\circ$

$\theta_c = \sin^{-1}\left(\frac{\mu_2}{\mu_1}\right)$

$\theta_c = \sin^{-1}\left(\frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}}\right)$

The ray from a denser medium approaches the interface at a critical angle and, after refraction, grazes along the interface

TIR Condition :

When the angle of incidence goes beyond this critical angle, the ray gets ‘reflected back’ into the ‘same medium’. This phenomenon is called ‘Total Internal Reflection.’

3. Difference between Normal and Total Internal Reflection:

3.1 Normal Reflection:

The intensity of the incident ray is much greater than the intensity of the reflected ray. This is because during normal reflection, a part of the light gets absorbed by the material through which it hits, and some of it gets transmitted further.

Diagram showing incident, reflected, absorbed, and transmitted light with a pie chart indicating their percentages

3.2 Total Internal Reflection:

In this case, the incident ray intensity is retained 100% by the reflected ray. This is the major difference between Normal Reflection and Total Internal Reflection

Ray in a denser medium hitting the boundary at an angle greater than the critical angle, producing 100% reflection.

A Question from JEE Advanced Previous Year Papers:

Question:

A light ray travelling in a glass medium is incident on the glass-air interface at an angle of incidence. The reflected (R) and transmitted (T) intensities, both as functions of theta, are plotted. The correct sketch is :

Four graphs comparing transmitted and reflected light intensities versus incident angle for different cases.

Answer : (C) option

At an angle of incidence = 0°: Most of the light (not 100%) is transmitted.
At an angle of incidence > critical angle: 100 % of light is reflected, and hence 0% transmission of light

FAQ section

What is Total Internal Reflection (TIR)?

TIR is the phenomenon in which the incident light travelling from a denser medium to a rarer medium gets reflected back into the denser medium, retaining its full intensity (100%).

What is a grazing ray?

We call a refracted ray a grazing ray when it passes along the interface separating two media.

What is meant by critical angle in TIR?

For each medium interface, we have a special angle defined, known as the critical angle. In simple words, it is the angle of incidence at which the refracted ray grazes along the medium-separating interface (the angle of refraction becomes 90 deg).

Charge Induction in Metals and Non-metals

5 February 202621 July 2024 by Saurabh Salvi

Topic under the Chapter: Electrostatics

Table Of Contents

1. Short Introduction
2. Induction in Metal and Non-metals
FAQ section:

1. Short Introduction

To understand Charge Induction, we need to first know about ‘free electrons’. These free electrons behave very similarly to the gas atoms in a container. Both of them are under continuous random motion throughout the space given to them.

Charge induction forms the working principle behind devices such as capacitors. The understanding that you would obtain by reading this concept would help you see how metals and non-metals behave in practical circuits.

Free electrons are nothing but some loosely bonded valence electrons that come out of the atom very easily, just as the loosely stitched button comes out of the shirt very easily!

Note that :

A neutral body has an equal number of positive charges and negative charges. The presence of free electrons doesn’t disturb the neutrality of the body, as free electrons are also negative charges, but the only difference is that they are free to move inside the body

2. Induction in Metal and Non-metals

When we talk of bodies, we classify them as –

Metals – The ones in which there are a lot of free electrons
Non-metals – electrons are bound to the Atom tightly (the atom loves them!)

2.1 Charge Induction in Metals

Now suppose you have a metal conductor placed in a region. And you bring a positive charge ‘+q’ in that region ‘externally’.

Important Note :

excess of electrons implies negative charge
A deficiency of electrons implies a positive charge

external charge 'q' attracts the free electrons inside the metal to accumulate on one side of the metal body, hence creating negative and positive charge separation

A flowchart explaining the process of charge induction in the case of metals

Text version for Flowchart:

Positive charges exert force on the negatively charged free electrons. Due to this attraction that the electrons feel, they get displaced towards the external positive charge. This causes one region to get an excess of electrons, i.e. negative charge, while the other has a deficiency of electrons, i.e., a positive charge. This is how charge separation happens in a conductor

These charges are ‘induced’ on the conductor due to the external charge. This phenomenon of separation of charges in a body by some external factor is called ‘Charge Induction’

2.2 Charge Induction in Non-Metals

As discussed, the basic difference between non-metal and metals is the absence of free electrons in case of non-metals.

Setup – Let’s consider the same condition. An non-metallic body has been placed in a region. Now, we bring a positive charge in vicinity of this body.

Important Note:

Atom is made up of a positively charged nucleus and an negatively charged electron cloud surrounding it. In a neutral, undisturbed atom, the negative center and the positive center, both, coincide.

In non-metals, there are no free electrons, but instead, we can visualize an electron cloud. The positive external charge distorts the negatively charged clouds — (Cloud represents negatively charged electron cloud)

Dipoles get created due to the influence of an external charge in the case of non-metals

A flowchart to explain the charge separation process in the case of non-metals

Text version of Flowchart:

Positive charges exert force on the negatively charged electron cloud in the atom.

Due to this attraction, the negative center gets slightly displaced towards the external positive charge.

This causes separation of the negative and positive centers of the atom. We now call this atom a dipole.

All the dipoles tend to align themselves such that the negative side is closer to the external positive charge.

This alignment in dipoles in a non-conducting body due to an external charge is called induction in non-conducting bodies or ‘Polarization’. The separation between positive and negative charges is very very small. So, usually, we ignore it in our problem solving, etc.

Special Note :
This is why a Charged body always attracts a Neutral body. We can also conclude that :
“Attractive force on neutral conducting body will be more due to any external charge”

FAQ section:

What is meant by charge induction?

In presence of an external electric field set up by an charge placed outside a body; charge separation happens within the body. This phenomenon is called charge induction.

What is Polarization in Electrostatics?

The formation of dipoles within the atoms of non-metal due to the external electric field set up by the charge placed outside the body is called Polarization

Conversion of Galvanometer to Ammeter & Voltmeter

6 February 202625 May 2024 by Saurabh Salvi

Table Of Contents

1. What is a Galvanometer?
2. Conversion to Ammeter
3. Conversion to Voltmeter
4. Examples
- Conclusion:

1. What is a Galvanometer?

A galvanometer is a deflection-type meter used to measure current. The needle in the Galvanometer deflects when a current passes through it, and the deflection is proportional to the current.

There are two types of Galvanometers :

Uni-Directional
Bi-Directional

Unidirectional: In this case, the markings on the dial start at 0 and extend to the maximum range. It has a red terminal, which indicates it must be connected to a high-potential source, and the other is a black terminal for a lower-potential source.

Bi-Directional: The dial’s zero is in the center, and the maximum is on either side. So, from the direction of deflection, we get to know the direction of current in the conductor, and the amount of deflection gives us the magnitude.

So, the basic difference between the two is that Unidirectional can give information only about magnitude, while Bi-directional can tell direction as well as magnitude.

The inner setup of a galvanometer has something known as Coil Resistance ‘G,‘ and at maximum deflection, the safe current which flows through the galvanometer is ‘i_g‘. The symbol for a galvanometer is :

The figure shows the maximum deflection current flowing through galvanometer having coil resistance G

Before proceeding, you can cover Series and Parallel Combination

2. Conversion to Ammeter

The purpose of an Ammeter is also to measure current, but the range for current measurement is much higher.

How to Convert? – Just add a resistor with very small resistance (Shunt ‘S’) in parallel to the Galvanometer

The schematic representation of an ammeter made with the help of a galvanometer

What happens because of this?

Now suppose, ‘I‘ (I > i_g) is the current flowing in the conductor. Since S and G are connected in parallel, ‘I‘ will be divided into ‘i_g‘ and ‘I-i_g‘.
The shunt resistance S, being very small in magnitude, will attract a lot of current (since current always prefers the least resistance path). The shunt resistance S is the reason why we are able to supply a larger current than ig.
This helps us to measure a larger current, resulting in an increase in the range of the galvanometer

How to calculate this ‘I‘?

G and S are in parallel combination

$(I – i_g)\times S = i_g\times G$

From this, find the value of ‘I’. This will give us the maximum safe current allowed into this designed Ammeter

3. Conversion to Voltmeter

The Voltmeter is used specifically to measure the potential difference across the given terminals.

How to Convert? – Add a very high ‘Load’ resistance R in series to the Galvanometer

schematic representation of a voltmeter made with the help of a galvanometer

What happens because of this?

Before adding R in series,

$\Delta V = (i_g \times G)$

After adding R in series,

$\Delta V = (i_g \times G) + (i_g \times R)$

The term $(i_g \times R)$ is responsible for the increase in the range of the voltmeter. This enables us to measure larger values. And hence, we need ‘R’ as large as possible.

4. Examples

Question 1: What shunt resistance is required to make a 1.00 mA, 20 Ω galvanometer into an ammeter with a range of 0 to 50.0 mA?

Solution :

Given:

Maximum current (I) = 50 mA
i_g = 1 mA
G = 20

Solution:

$(I – i_g)\times S = i_g \times G$

By substituting, we get:

$S = 0.408\,\Omega$

Question 2: How can we make a galvanometer with $G = 20\,\Omega$ and $i_g = 1.0\,\text{mA}$ into a voltmeter with a maximum range of 10 V?

Solution :

Given:

maximum voltage to be measured $\Delta V = 10\,\text{V}$
$i_g = 1\,\text{mA}$
$G = 20\,\Omega$

Solution:

For converting a galvanometer to a Voltmeter,

$\Delta V = (i_g \times G) + (i_g \times R)$

By substituting, we get:

$R = 9980\,\Omega$

Conclusion:

So, we have learnt about the Galvanometer and how we can use it as an Ammeter and a Voltmeter.

This topic is important not only from a practical point of view but also theory exam point of view
And other than marks, it’s always good to know about our electrical instruments!

All the Best

Cathode Ray Deflection Experiment – Calculating e/m Ratio

6 February 202613 April 2024 by Saurabh Salvi

Calculation of the charge-to-mass or e/m ratio is of great importance when it comes to the subject of Modern Physics. Understanding the procedure behind this experiment is equivalent to revising the following topics as well :

Moving charge in a magnetic field
Behaviour of Charge in Electric Field
Projectile Motion

Table Of Contents

1. Importance of e/m Ratio
2. Setup of the Experiment
3. Procedure
4. Final Data
Conclusion

1. Importance of e/m Ratio

Ok, so you have a value called e/m ratio! But why is it important to calculate this value? Are there any applications of it?

In simple words, the charge-to-mass ratio of a charge helps us to predict the behaviour of the particle under electric and magnetic fields. This ability to predict the particle’s behaviour enables us to have an idea of adjusting the setup in order to have so-and-so outcomes.

We can see its application in:

Electron Microscopes:

Electron Microscopes are known for their ability to magnify the images to a very high resolution & this is done with the help of a beam of electrons
Knowing the e/m ratio enables the scientists to control the movement of electrons and, as a result, they produce the required resolution of the image

Particle Accelerometers:

These are used to accelerate the charged particles.
By knowing the charge-to-mass ratios, we can actually control the trajectories of the particles.

2. Setup of the Experiment

The Setup mainly consists of the following things :

Filament F
Battery V
Pump
2 parallel plates, across which another battery has been connected
Current-carrying coil wire (not shown in setup)
Screen S

Electron beam entering crossed electric and magnetic fields in a velocity selector tube — Fig. Setup of the Experiment

Purpose:

Filament F: The filament is inclusive of that battery (not V) shown in the above figure. The flowchart below explains the working of filament

Flowchart explaining the working of the filament

Text Version for the above flowchart (Expand This)

The battery heats the filament
This causes the electrons to get knocked off from the atoms due to the energy given (i.e., thermionic emission)
These electrons form a cloud near the filament once they come out from the atom (since they don’t know what to do next!)

Voltage V : The plate attached to the positive terminal is used as an anode to attract the electron cloud. This is done to make the electrons accelerate. Each electron has different energies when it comes out from atoms. And when they are accelerated due to a potential difference of V, then they all end up having a different set of velocities.

Pump: To create a vacuum inside the tube
The parallel plates kept facing each other + battery setup, is used to create a uniform electric field E in the region between the two plates. Direction will be from the positive plate to the negative plate
Current-carrying coil wire: This is done to set up a steady magnetic field B (going into the plane)
Screen S: Whenever an electron strikes the screen S, it creates a spot on the screen, which helps us to detect and hence analyze the trajectory/path taken by the electron.

Note that: The E and B vectors are perpendicular to each other

3. Procedure

Step-1 :

As discussed, the anode attracts the electron cloud, which makes them accelerate towards the screen S. But well before they reach the screen, the electrons are made to pass through a region R where, for now, only the electric field is applied (B is turned off).

The current setup for Step-1 looks like:

Velocity selector showing Region R where uniform electric and magnetic fields act on the electron beam

As the electrons pass through ‘Region R‘, they undergo deflection ‘y’ due to the electric field and follow a trajectory as shown (green). We zoom into Region R to get a better understanding of what’s happening.

Zoomed picture of Region R :

Electron moving horizontally with upward deflection in a uniform downward electric field

We need the expression for deflection ‘y’.

Important: Note that the deflection is going to be measured from the axis

Some of the Projectile comes into the picture now!

Initial velocity in X direction (u_x) = v
Initial velocity in Y direction (u_y) = 0
Acceleration in X direction (a_x) = 0
Acceleration in Y direction (a_y) = $\frac{eE}{m}$
Displacement in X direction (S_x) = L
Displacement in X direction (S_y)= y

Breaking down the motion in X and Y axes separately

$S_{x} = u_{x} t$

L = vt

$S_{y} = \frac{1}{2} a_{y} t^{2}$

$y = \frac{1}{2}\left(\frac{eE}{m}\right) t^{2}$

Solving for ‘y’, we get:

$y = \frac{eE L^{2}}{2 m v^{2}}$

This ‘y’ is measured during the experiment

Step-2:

Now, we aim to find the velocity ‘v’ of the electron. Recall that the ‘y’ is the deflection –> BUT Deflection from which path? The answer is ‘the axis’. We need to find the deflection caused in the electron’s trajectory due to the electric field E, because otherwise, in the absence of E, it would just follow the straight path along the axis.

To get the speed (v) of the electrons that go undeflected, we introduce B now in addition to E to make zero deflection. This is equivalent to saying that none of the fields were present in region R

Force diagram showing velocity along +x, electric field along −y, and magnetic field into the page.

We have to balance the forces (to get zero deflection). Remember, it’s a negative charge.

Force due to electric field:

$\vec{F}_{E} = q \vec{E}$

Force on a moving charge due to a magnetic field:

$\vec{F}_{B} = q (\vec{v} \times \vec{B})$

The force due to the electric field and the force due to the magnetic field are acting on the electron in the opposite direction

We then adjust the values of E and B until the magnitudes of the forces are the same. This allows us to build a ‘velocity selector.‘

Velocity Selector :

As discussed in Section 2 of this article, all the electrons come with a different set of velocities. But, for continuing our experiment, we need only the electrons of a specific velocity to be focused on. So, how exactly distinguish those electrons?

We want the electrons to go undeflected. For that,

$\therefore \; F_{E} = F_{B}$

$eE = evB$

$v = \frac{E}{B}$

We can clearly see the relation between E and B with velocity. This means that controlling the values of E and B allows us to select the electrons that have their velocities as E/B. The electrons possessing this specific velocity will go through the region undeflected, and hence we can separate them.

Step-3:

Calculating the e/m ratio with the expressions and equations we got till now :

Substituting v = E/B in the expression of y obtained in step-1, we get:

$\frac{e}{m} = \frac{2yE}{B^{2} L^{2}}$

4. Final Data

Thomson got the value for e/m to be $1.7 \times 10^{11}\ \text{C/kg}$

Today’s accepted e/m value: $1.758820174 \times 10^{11}\ \text{C/kg}$

We can see a very good match between the two values

Conclusion

This completes our Cathode ray deflection experiment performed by Sir J.J. Thomson for calculating the e/m ratio
The article or the whole experiment procedure itself has a lot of concepts involved in it, which makes it even more important to understand, both as an Experiment as well as a good multi-concept level problem

Keep Learning!

How to Solve Pulley Block Problems? | Trick Explained

8 February 20267 January 2024 by Saurabh Salvi

Pulley Block Problems are very common in high school physics and mainly come under Newton’s Laws of Motion (NLM) chapter. These concepts are frequently asked in exams such as JEE Mains and even JEE Advanced.

Time becomes an important factor in such competitive examinations, especially JEE Mains. So, it’s always better to have some short tricks to be faster and save time. But, still, I would like to emphasize the fact that concept>>tricks

Table Of Contents

Rules to apply the Trick:
- How to use this trick in solving pulley block problems?
Easy Examples based on Pulley Problems
Moderate to Difficult Examples on Pulley Problems
Conclusion :

Rules to apply the Trick:

Remember the rules :

Only one string at a time
On the string –> ‘minus’ sign (Recall the -ve charge sign on ions from chemical bonding because that’s how I kept it in my mind)
Away from the string –> ‘plus’ sign

How to use this trick in solving pulley block problems?

Steps:

Mark the points on the string such that whole string is covered (starting to end)
For pulleys, mark for the points- where the string first comes in contact with the pulley (point 2) and where the string leaves the contact with the pulley (point 3)
Now start from one end ( say point 1). The relation will be as follows (-Va + 0 + 0 +V_b)

Why?

-V_a because V_a is going on the string

+0 because point 2 is connected to the pulley and the pulley is at rest

+0 point 3 is connected to pulley and pulley is at rest

+V_b because point 4 is connected to block B, and V_b is going away from the string

Finally equate everything to zero i.e. (-V_a + 0 + 0 +V_b = 0)

Final Answer: We get V_a = V_b

Easy Examples based on Pulley Problems

Q.1 Write down the constraint relation between velocities of block A and block B by referring the image of setup given below

Solution :

Constraint relation can be written as : (Applying trick, starting from block A)

$– v_A + 0 + 0 + v_B + v_B + 0 + 0 + 0 = 0$

Shown below is the breakdown of what each term in the above expression represents:

On simplifying, we get:

$v_A = 2v_B$

Q.2 Write down the constraint relation between velocities of block A and block B by referring the image of setup alongside.

Solution:

We start from block A (Just move along the string one-by-one) :

$– v_A + 0 + 0 + v_B + v_B + 0 = 0$

$v_A = 2v_B$

Q.3 Write down the constraint relation between velocities of block A and block B by referring the image of setup alongside.

Solution : Let’s start from block A

We start from block A (Just move along the string one-by-one) :

$– v_A + 0 + 0 + v_B + v_B + 0 = 0$

$v_A = 2v_B$

Moderate to Difficult Examples on Pulley Problems

Question 4:

The figure alongside shows a system of four pulleys with two masses, A and B. Find, at an instant:

(a) Speed of block A when the block B is going up at 1 m/s and pulley Y is going up at 2 m/s.

(b) Acceleration of block A if block B is going up at 3 m/s² and pulley Y is going down at 1 m/s².

Solution:

Part (a) :

(Lets start from the point near the rigid support)

The constraint relation can be written as:

$0 – 1 – 1 + 2 + 2 + v_A + v_A + 0 + 0 – 2 = 0$

$2v_A = 0$

$\therefore \; v_A = 0$

Part (b) : Again we start from the point attached to the rigid end

(Note that the derivative of velocity wrt time is acceleration. So once you get the velocity constraint relation, just taking its derivative wrt time will give you the constraint relation for accelerations)

The constraint relation can be written as:

$0 – 3 – 3 – 1 – 1 + a_A + a_A + 0 + 0 + 1 = 0$

$2a_A = 7$

$\therefore \; a_A = 3.5 \,\text{m/s}^2$

Question 5:

Write the constraint relations between block A and block B in the setup shown in the image alongside

Solution:

Recall that this trick can only be applied to one string at a time!

So, we take velocity of pulley X as Vx as shown in figure below and we start applying trick from block A end

The constraint relation can be written as:

$v_A + 0 + 0 + 0 + 0 + 0 + 0 + 0 + v_X + v_X = 0$

$2v_X = -v_A$

But, we need relations between Vb and Va

The constraint relation can be written as: (Starting from block B)

$-v_B + 0 + 0 – v_X = 0$

$v_X = -v_B$

On Substituting v_X value to get the relation between the velocity of A and the velocity of B,

We get,

$v_A = 2v_B$

Question 6:

Find the speed of block B when the wedges A and C are moving toward each other with speed v and the strings connected to block make an angle θ with the vertical, as shown in the figure alongside

Solution:

As we discussed, only the velocity component along the string are to be considered

Starting from the green colored star (*) mark

Same approach, mark all the points on the string first

The constraint relation can be written as: (Starting from block B)

$0 + v – v\sin\theta – v_B \cos\theta = 0$

$v_B = \frac{v(1 – \sin\theta)}{\cos\theta}$

But, we know that the value of V_B (RHS) from the above expression is always going to be positive since,

$-1 \le \sin\theta \le 1$

So, the assumed direction of V_B (upwards) is also correct!

Conclusion :

So, over here, we end this article. Do practice some more string pulley constraint problems using this trick. But also don’t miss out on the actual procedure. Tricks should always be your secondary option !!

We will keep coming up with more such articles to help you in your preparation !! Thank you for your time.

Have a good day!