Mechanics

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

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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?

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:

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!

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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

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

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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,

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

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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,

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.’

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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

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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?

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

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

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.

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 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

1. 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.

Solution:

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

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

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

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

FAQ section

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)

1. Parallel Axis Theorem

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

Solution :

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

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?

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

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!

1. Rules to Remember

• 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.

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FAQ section

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

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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:

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

Final Answer: We get V_a = V_b

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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:

$v_A = 2v_B$

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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

Moderate to Difficult Examples on Pulley Problems

Question 4:

Solution:

Part (a) :

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)

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Question 5:

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

But, we need relations between Vb and Va

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

We get,

$v_A = 2v_B$

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Question 6:

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!

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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!