Mechanics

Maximum Force such that both blocks move together

Jun 19, 2026 PCM TUTORIALS 0 Comment

A block A (mass 4 kg) is placed on a smooth horizontal surface. Another block B (mass 2 kg) is placed on top of A. A horizontal force F is applied to block A as shown. The coefficient of static friction between the two blocks is μ = 0.25. Find the maximum value of force F such that both blocks move together without slipping.

Complete Video Solution below

Maximum Force such that both blocks move together Without Slipping: Understanding the Real Physics

One of the most common friction problems in JEE Main, JEE Advanced, and NEET involves two blocks placed one above the other. A horizontal force is applied to the lower block, and students are asked to find the maximum force that can be applied so that both blocks move together without slipping.

At first glance, this may look like a standard formula-based problem. However, the real value of this question lies in the concepts it teaches.

In fact, many students solve such problems mechanically without understanding the actual role of friction. As a result, they often make mistakes when the problem is slightly modified.

Before jumping to equations, let us first understand the physics behind this situation.

The First Question: How Does the Upper Block Move?

Suppose a force is applied to the lower block.

The lower block is directly connected to the external force, so it is obvious why it starts moving.

But what about the upper block?

No force is directly applied to it.

So why should it move at all?

This is the most important question in the entire problem.

Before writing equations, always ask yourself:

What is the force responsible for accelerating the upper block?

Once you answer this question, the entire problem becomes much easier.

Many students immediately start looking for formulas.

A better approach is to first understand the physics.

Do not start with formulas. Start with physics.

The Hero of the Story: Static Friction

The upper block moves because of static friction.

Many students think friction only opposes motion.

This is not entirely true.

A more accurate statement is:

Friction opposes relative motion or the tendency of relative motion between two surfaces.

In this problem, the lower block tries to move forward. If there were no friction, the lower block would slide out from beneath the upper block.

Static friction prevents this from happening.

It pulls the upper block forward and allows both blocks to move together.

Therefore, static friction is not acting as an enemy here.

It is actually helping the upper block move.

This is a very important idea that students should remember.

Static Friction Is Smarter Than Most Students Think

Many students believe that static friction is always equal to μN.

This is one of the most common misconceptions in mechanics.

Static friction is not a fixed force.

It adjusts itself according to the requirement of the situation.

If a small friction force is needed, static friction provides a small force.

If a larger friction force is needed, static friction increases its value.

However, it cannot increase forever.

There is a maximum limit beyond which static friction cannot go.

As long as the required friction remains below this limit, the blocks continue to move together.

The moment the required friction exceeds this limit, slipping begins.

Understanding this single idea can solve a large number of friction problems.

Why Friction Direction Confuses Students

One of the most common mistakes in friction problems is assuming the wrong direction of friction.

Students often memorize rules instead of thinking physically.

The correct approach is simple.

Ask yourself:

If friction were absent, what would happen?

Without friction, the lower block would move forward while the upper block would tend to stay where it is because of inertia.

Therefore, relative to the lower block, the upper block would appear to move backward.

Static friction opposes this tendency.

As a result, friction acts forward on the upper block.

According to Newton’s Third Law, an equal and opposite friction force acts backward on the lower block.

The direction becomes obvious once you understand the physical situation.

A Common Mistake Students Make

Many students immediately draw friction opposite to the motion of the object.

This shortcut often creates confusion.

Remember:

Friction does not oppose motion. Friction opposes relative motion.

This small distinction can completely change the solution.

Whenever you are confused about friction direction, imagine the situation without friction.

The correct direction usually becomes obvious.

A small mistake here can spoil the entire problem.

As teachers often say:

A small mistake in friction direction can make the entire solution go wrong.

Moving Together Means No Slipping

The statement “both blocks move together” is extremely important.

It tells us that there is no relative motion between the two blocks.

Since there is no slipping, kinetic friction does not come into the picture.

Only static friction is acting.

Many students immediately use friction formulas without first identifying the type of friction involved.

This is dangerous.

Always determine whether slipping is occurring or not before choosing the friction model.

As long as the two blocks move together, static friction is responsible for maintaining the common motion.

Can Static Friction Do the Job?

Now we come to the central idea of the problem.

The upper block needs a certain force to accelerate along with the lower block.

That force is supplied by static friction.

However, static friction is not unlimited.

It has a maximum possible value.

As the applied force on the system increases, the acceleration of the system increases.

A larger acceleration means the upper block requires a larger friction force.

Initially, static friction can easily provide the required force.

But as the applied force keeps increasing, a stage is reached where the required friction becomes equal to the maximum friction available.

This is the limiting situation.

At this point, static friction is working at its highest possible value.

This condition determines the maximum force that can be applied while keeping both blocks together.

What Happens Beyond This Limit?

Suppose we increase the applied force even further.

Now the upper block requires more friction than static friction can provide.

But friction cannot exceed its maximum limit.

As a result, static friction fails to maintain common motion.

The upper block can no longer keep up with the lower block.

Relative motion begins.

Slipping starts.

The moment slipping starts, static friction disappears and kinetic friction takes over.

This is why the problem asks for the maximum force.

It is asking for the boundary between two situations:

Both blocks move together.
The upper block starts slipping.

Understanding this boundary is much more important than memorizing any formula.

Why This Problem Is More Important Than It Looks

At first glance, this appears to be a simple two-block friction problem.

However, the ideas used here appear again and again in mechanics.

The same concepts are used in:

Friction problems
Wedge problems
Pulley systems
Circular motion with friction
Advanced Newton’s Laws questions

The goal is not to remember an answer.

The goal is to understand how forces interact and how friction helps maintain common motion.

Once you understand the physics behind this question, many seemingly difficult problems become much easier.

Physics Is About Understanding, Not Memorization

Many students search for shortcuts and formulas.

However, formulas are only the final result of physical reasoning.

If you understand the role of friction, the direction of friction, and the condition for slipping, you can derive the result whenever required.

That is the real goal of learning physics.

The two-block friction problem teaches us something far more valuable than a numerical answer.

It teaches us how static friction helps two surfaces maintain common motion and how slipping begins when friction reaches its limit.

Once this concept becomes clear, many friction problems become surprisingly simple.

Remember:

Do not start with formulas. Start with physics.

When the physics is clear, the mathematics becomes easy.

Final Solution

Given:
$m_A\;=\;4\;kg$

$m_B=2kg$
$\mu_s=0.25$
Maximum static friction:

f_{max\;=\;\mu_s\;m_B\;g}

System acceleration at limiting condition:

$a_{max\;=\;\mu_s\;\;g}$
$a_{max\;=\;0.25\times\;\;10}$
$a_{max}=\;2.5\;m/s^2$

For the complete system:

$F_{max}=(m_A+m_B)a_{max}$ $NF_{max}=6\times2.5=15\,N$

Answer: $F_{max}=15N$

Author: Deep Aman Sir

Founder, PCM TUTORIALS | Physics Faculty for JEE, NEET & CBSE

Blog Mechanics

Pseudo Force

Feb 18, 2025 PCM TUTORIALS 0 Comment

Pseudo Force

Pseudo Force: Understanding the “Fictitious” Force of Non-Inertial Frames

Physics is full of concepts that seem counterintuitive at first but become fascinating once you dive deeper. One such idea is the pseudo force (sometimes called fictitious force).

Whenever you sit in a bus that suddenly accelerates or take a sharp turn in a car, you feel as if some mysterious force is acting on you, pushing you backward or sideways. But in reality, there is no such physical agent pushing you. That’s the world of pseudo force — a force that appears only when we observe motion from a non-inertial reference frame.

This article will explore pseudo force in detail — its definition, origin, mathematical expression, examples, and applications. By the end, you’ll understand why pseudo forces are not “real” in the strict sense, yet are extremely useful in solving problems of mechanics.

1. Reference Frames: The Foundation

Before talking about pseudo force, we need to revisit the concept of reference frames.

Reference Frame: A coordinate system from which we observe and measure the motion of objects.
Inertial Frame: A frame of reference in which Newton’s laws of motion hold true without modification. Typically, a frame at rest or moving with constant velocity relative to the “fixed stars” is considered inertial.
Non-Inertial Frame: A frame that is accelerating or rotating relative to an inertial frame. In such frames, Newton’s laws do not appear to hold unless we introduce additional forces — the pseudo forces.

Example:

Standing on the ground (which we approximate as an inertial frame for most problems), you see a ball falling vertically under gravity.
Sitting in an accelerating car (a non-inertial frame), the same ball seems to move backward, even though no physical backward force is acting on it.

2. What is a Pseudo Force?

Definition:
A pseudo force is an apparent force that arises when we describe motion from a non-inertial frame. It has no physical interaction or agent behind it. Instead, it is introduced mathematically to make Newton’s second law applicable inside a non-inertial frame.

Key Points:

It is not a “real” force — it doesn’t arise due to any physical contact or field.
It is proportional to the mass of the object.
It always acts opposite to the acceleration of the non-inertial frame with respect to the inertial frame.

3. Mathematical Expression of Pseudo Force

Suppose we have:

An inertial frame $S$ .
A non-inertial frame $S ’$ , accelerating with acceleration $a⃗0\vec{a}_0$ relative to $S$ .
An object of mass $m$ .

In inertial frame $S$ :

$ma⃗=∑F⃗realm\vec{a} = \sum \vec{F}_{\text{real}}$

But in non-inertial frame $S ’$ :
The acceleration of the object relative to $S ’$ is different, so Newton’s law does not balance unless we introduce a fictitious force:

$ma⃗′=∑F⃗real+F⃗pseudom\vec{a}’ = \sum \vec{F}_{\text{real}} + \vec{F}_{\text{pseudo}}$

Where,

$F⃗pseudo=−ma⃗0\vec{F}_{\text{pseudo}} = -m\vec{a}_0$

Thus, the pseudo force is directly proportional to the mass of the body and opposite to the acceleration of the non-inertial frame.

4. Everyday Examples of Pseudo Force

(i) The Bus Ride

When a bus suddenly accelerates forward, passengers feel a backward push.
From the ground (inertial frame), passengers tend to remain at rest due to inertia.
From the bus frame (non-inertial), it looks as if a backward force is pushing them. That is the pseudo force.

(ii) Elevator Problems

In a downward accelerating lift, a person feels lighter.
In an upward accelerating lift, the person feels heavier.
The apparent weight is explained by introducing pseudo force equal to $ma_0$ , where $a_0$ is the acceleration of the lift.

(iii) Rotating Reference Frames (Centrifugal Force)

When you take a sharp turn in a car, you feel “thrown” outward.
From the inertial frame, you are simply trying to maintain a straight-line path, but the car is turning beneath you.
From the rotating (car’s) frame, an outward pseudo force called the centrifugal force appears.

(iv) Coriolis Force

A special pseudo force observed in rotating frames, responsible for large-scale effects like trade winds, cyclones, and ocean currents on Earth.

5. Pseudo Force vs Real Force

Feature	Real Force	Pseudo Force
Origin	Arises from physical interaction (contact, field, gravity, electromagnetism, etc.)	Arises due to acceleration of reference frame
Agent	Always has a physical source	No physical agent
Newton’s Laws	Can be explained directly by Newton’s second law	Introduced to make Newton’s laws valid in non-inertial frames
Example	Gravitational force, tension, friction	Backward push in an accelerating bus, centrifugal force

6. Types of Pseudo Forces (Required for JEE/NEET level)

Linear Acceleration Pseudo Force: Appears when the frame is linearly accelerating.
- Expression: $F⃗pseudo=−ma⃗0\vec{F}_{\text{pseudo}}=-m\vec{a}_0$ .
Centrifugal Force: Appears in rotating frames, directed radially outward.
- Expression: $Fc=mω2rF_c = m\omega^2 r$ .

7. Pseudo Force in Elevators – A Classic Example

Let’s derive the apparent weight in a lift:

Actual weight = $m g$ .
Lift acceleration = $a$ .

In the elevator’s frame:

$weight=N=mg−Fpseudo\text{Apparent weight} = N = mg – F_{\text{pseudo}}$

Where pseudo force = $- ma$ .

So,

If lift accelerates upward ( $a > 0$ ):
$N = m g + ma$ . → You feel heavier.
If lift accelerates downward ( $a > 0$ ):
$N = m g - ma$ . → You feel lighter.
If $a = g$ (free fall): $N = 0$ . → Weightlessness.

8. Pseudo Force in Rotating Frames – Centrifugal and Coriolis

When a body is in a rotating frame (like a rotating merry-go-round), it seems to be “thrown” outward.

From inertial frame: The body tries to move tangentially (straight line).
From rotating frame: An outward pseudo force $F=mω2rF=m\omega^2 r$ is introduced to explain the tendency.

9. Applications of Pseudo Forces

Engineering and Design:
- Designing elevators, centrifuges, rotating machines.
- Vehicle safety, roller-coaster rides, airplane maneuvers.
Meteorology:
- Coriolis force explains trade winds, cyclones, jet streams.
Space Science:
- Artificial gravity in rotating space stations uses centrifugal pseudo force.
Daily Life:
- Balancing in buses or trains, predicting motion while turning.
Education:
- Helps students reconcile Newton’s laws with real-life observations in accelerating systems.

10. Common Misconceptions

Pseudo forces are imaginary, so they are useless.
Wrong! They are extremely useful tools to simplify analysis in non-inertial frames.
Centrifugal force pushes objects outward.
In reality, no outward force exists in the inertial frame. It is the inertia of motion that makes objects resist circular motion.
Pseudo forces are optional.
If you are working in a non-inertial frame, you must include them to apply Newton’s laws consistently.

11. Visualizing Pseudo Forces

If you are writing this for a blog, here are diagrams you can add (describe in captions):

Passenger being thrown backward in an accelerating bus.
Elevator free-fall with zero apparent weight.
Centrifugal force on a ball tied to a rotating string.
Coriolis effect deflection on Earth.

These visuals help readers connect theory with experience.

12. Pseudo Force and General Relativity – A Deeper Note

Interestingly, pseudo forces give a glimpse into Einstein’s general relativity. Gravity itself can be thought of as a pseudo force that arises because we observe motion from a non-inertial frame of curved spacetime.

In Einstein’s view:

Objects in free fall are actually in inertial motion (no real force).
Observers standing on Earth feel a downward “gravitational force” only because the Earth is accelerating upward relative to free-falling objects.

Thus, pseudo force in Newtonian mechanics acts as a stepping stone to understanding modern physics.

13. Summary

Pseudo force arises in non-inertial frames.
Formula: $F⃗pseudo=−ma⃗0\vec{F}_{\text{pseudo}} = -m\vec{a}_0$ .
Examples include the backward push in an accelerating bus, elevator apparent weight, centrifugal and Coriolis forces.
They are not “real” but are extremely useful in calculations.
They connect classical mechanics to deeper ideas in relativity.

14. Conclusion

Next time you feel pushed backward in a speeding bus, remember — it’s not a mysterious hidden hand but your own inertia seen from a non-inertial frame. The pseudo force is a clever tool invented by physicists to keep Newton’s laws working consistently even in accelerating systems.

Understanding pseudo force enriches our grasp of dynamics, helps us appreciate everyday experiences, and lays a foundation for advanced physic

Mechanics

Kinematics: Understanding Uniform Motion

Jun 21, 2024 PCM TUTORIALS 0 Comment

Kinematics: Understanding Uniform Motion

What is Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the causes of motion. Among the various types of motion, uniform motion is a fundamental concept that provides a foundation for understanding more complex movements. In this article, we will delve into the intricacies of uniform motion, explore its characteristics, and see its applications in our daily lives.

What is Uniform Motion?

Uniform motion refers to the movement of an object at a constant speed in a straight line along specific fixed direction. In other words, the object covers equal distances in equal intervals of time, regardless of how small these intervals might be. This type of motion is characterized by:

Constant Velocity: The speed/magnitude of velocity and direction of the object’s motion remain unchanged.
Zero Acceleration: Since the velocity is constant, there is no change in speed or direction, resulting in zero acceleration.

As , $\overrightarrow{v\;}\;=\;constan t\;$

Therefore, $\overrightarrow{a\;}=\;\frac{d\overrightarrow v}{dt}=\;0$

Mathematical Representation

Consider an object is moving along +X axis with uniform velocity ‘v’ as shown in diagram. Assume that object is at position $x=x_1$ at time $t=t_1$ and reaches at position $x=x_2$ at time $t=t_2$

We know that velocity of object is defined as rate change of position of object w.r.t. time. Mathematically,

$v=\frac{dx}{dt}\;$

$\Rightarrow dx\;=\;v\;dt\;$

$\int_{x_1}^{x_2}dx\;=\;\int_{t_1}^{t_2}v\;dt\;$

$\int_{x_1}^{x_2}dx\;=\;v\int_{t_1}^{t_2}\;dt\;$ (v= constant/uniform)

$\left|x\right|_{x_1}^{x_2}=\;v\;\left|\;t\;\right|_{t_1}^{t_2}$

$x_2\;-x_1\;=\;v\;(t_2-t_1)$

$\Delta x\;=v\;\Delta t$

$v=\frac{\Delta x}{\Delta t}$

$Hence,\;v=\frac{dx}{dt}=\frac{\Delta x}{\Delta t}=constant$

Clearly, uniform motion reflects that the object covers equal distances in equal intervals of time, regardless of how small these intervals might be.

When object/particle starts motion from origin (x=0)

When particle starts motion from origin then initial displacement of particle is zero i.e. at t= 0 , x=0 . And it moves with constant velocity v, finally reaches at P in time t. Thus, at time t = t , x = x as shown in following diagram .

We know that for uniform motion,

$\Delta x\;=v\;\Delta t$

$x-0= v(t-0)$

$x = v t$

Position-Time (x-t) graph of uniform motion starting from origin

When object/particle starts motion with some initial displacement

When particle does not starts motion from origin but from $x= x_o$ then initial displacement of particle is non-zero i.e. at $t=0 , x= x_o$ . With constant velocity v, it finally reaches at position x in time t as shown in following diagram . Thus, at time t = t , x = x

We know that for uniform motion,

$\Delta x\;=v\;\Delta t$

$x-x_o= v(t-0)$

$x = x_o+ v t$

Position-Time(x-t) graph of uniform motion of particle with non-zero initial displacement

Graphical Representation

Position-Time Graph

In a distance-time graph for uniform motion, the distance (y-axis) plotted against time (x-axis) yields a straight line with a constant slope. The slope of this line equals the velocity of the object.

Velocity-Time Graph

In a velocity-time graph for uniform motion, the velocity (y-axis) plotted against time (x-axis) is a straight horizontal line. This horizontal line indicates that the velocity remains constant over time.

Uniform Motion vs. Non-Uniform Motion

Understanding the difference between uniform and non-uniform motion is crucial. While uniform motion features a constant velocity, non-uniform motion involves a change in speed or direction, indicating acceleration. Real-world motions are often non-uniform due to varying forces acting on objects, but uniform motion serves as an idealized model to simplify analysis.

Conclusion

Uniform motion, with its simplicity and predictability, is a cornerstone concept in kinematics. It lays the groundwork for more complex analyses of motion in physics and engineering. By grasping the fundamentals of uniform motion, one can better understand the intricate dynamics of moving objects, both in theoretical contexts and real-world applications. Whether it’s a car cruising on a highway or the precise movement of a spacecraft, uniform motion continues to be a vital principle in the study of mechanics.

Mechanics

A spaceship is launched into a circular orbit close to the earth’s surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull.

Apr 13, 2024 PCM TUTORIALS 3 Comment

Additional Velocity Required to Overcome Earth's Gravitational Pull for Satellite Escape

Escape velocity is the minimum velocity an object must attain to break free from the gravitational attraction of a massive body without further propulsion. For an object on the surface of the Earth, this velocity is approximately 11.2 kilometers per second (or about 25,000 miles per hour). But for a satellite already in orbit, the calculation is little more complex.

Satellites in low Earth orbit (LEO) typically travel at speeds around 8 kilometers per second (km/s). This velocity, known as orbital velocity, is necessary to counteract the gravitational force pulling the satellite towards Earth, allowing it to maintain a stable orbit.

However, what if we seek to free a satellite from Earth’s gravitational grasp altogether? Escaping Earth’s gravitational influence requires imparting additional velocity to the satellite. Suppose we have a satellite orbiting near the Earth’s surface, with an orbital velocity of approximately 8 km/s. To break free from Earth’s gravity, an additional velocity of approximately 3.2 km/s is needed.

Example

Q.A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth =6400 km , g= 9.8 $ms^{-2}$

(A) $3.2 km/s$

(B) $11.2 km/s$

(C) $1.5 km/s$

(D) $8 km/s$

Solution

We know that , orbital speed of satellite is

$v_o=\;\sqrt{\frac{GM}r}\;$

$where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth$

$Also, r= R+ h$ where R is radius of planet/earth and h is height of satellite from surface of earth

$v_o=\;\sqrt{\frac{GM}{R+h}}\;$

Near to earth's surface, h can be neglected

So, $R\;+\;h\;\approx\;R$

$v_o=\;\sqrt{\frac{GM}{R}}\;$

$v_o=\;\sqrt{\frac{gR^2}{R}}\;$

$v_o=\;\sqrt{gR}\;$

$v_o\;=\;\sqrt{Rg}\approx\;8\;km/s\;$

As we know that the escape velocity on earth surface is

$v_e\;=\;\sqrt{2Rg}\approx\;11.2\;km/s\;$

Addition velocity required is

$v_{add}\;=\;v_e\;\;-\;v_{o\;}$

$v_{add}\;=11.2 - 8 = 3.2 km/sec$

(A) is correct option

Conclusion

This additional velocity is crucial for overcoming the gravitational potential energy barrier that binds the satellite to Earth. When the satellite reaches this escape velocity, its kinetic energy surpasses the gravitational potential energy, allowing it to break away from orbit and venture into interplanetary space.

For more questions on mechanics click here

Mechanics

Kepler’s law of period

Mar 7, 2024 PCM TUTORIALS 0 Comment

Kepler's law of period

Kepler's law of period for planetary motion

According to Kepler’s laws of period for planetary motion, the square of the period of revolution of as planet around the sun is directly proportional to the cube of the semi-major axis of its orbit.

Mathematically, it can be expressed as $T^2 = k a^3$ , where $T$ is the orbital period of the planet, $a$ is the semi-major axis of its orbit, and $k$ is a constant that is the same for all planets orbiting the Sun.

Example

Q.A geostationary satellite is orbiting the earth at a height of 6R above the surface of earth where R is the radius of earth. The time period of another satellite at a distance of 3.5 R from the centre of earth is

(A) $16.96 hours$

(B) $8.48 hours$

(C) $9.25 hours$

(D) $15.76 hours$

Solution

We know that , orbital speed of satellite is

$v_o=\;\sqrt{\frac{GM}r}\;$

$where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth$

$T\;=\frac{dis\tan ce\;}{speed}=\;\frac{2\pi\;r}{v_o}$

$T\;=\;\frac{2\pi\;r}{\sqrt{\displaystyle\frac{GM}r}}$

$T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}$

$T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}\\T^2=\;4\;\pi^2\;\frac{r^3}{GM}$

$\Rightarrow T^2\;\;\propto\;r^3$

This is known as Kepler's Law of period for planetary motion

$r= R+ h$ where R is radius of planet/earth and h is height of satellite from surface of earth

$\Rightarrow r_1 = R+ 6R=7R \;and \;r_2= 3.5R$

$\Rightarrow T\;\propto\;r^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{r_2}{r_1}\right)^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{3.5R}{7R}\right)^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{1}{2}\right)^\frac32$

$\frac{T_2}{T_1}=\frac1{2\sqrt2}$

$T_2 =T_1 * \frac1{2\sqrt2}$

$T_2 =24 * \frac1{2\sqrt2}$

$T_2 =12* \frac1{\sqrt2}$

$T_2 =12* 0.707= 8.48 \;hours$

(B) is correct option

Conclusion

Kepler’s third law is also known as law of period which helps us to find the time period of orbital motion for various distances from sun. It includes following applications.

Predicting Planetary Motion: Kepler’s law of periods allows astronomers to predict the orbital periods of planets based on their distances from the sun. This law has been crucial in the study of our solar system and in the discovery and characterization of exoplanets in other solar systems.
Comparing Orbits: By comparing the orbital periods and semi-major axes of different planets or moons, astronomers can gain insights into the structure and dynamics of planetary systems. For example, comparing the orbital periods of moons around a planet can provide information about their relative distances from the planet.
Verification of Kepler’s Laws: Kepler’s laws of planetary motion played a significant role in the development of Newton’s law of universal gravitation and the laws of motion. They served as a crucial test for Newtonian physics and provided evidence for the gravitational force between celestial bodies.

Mechanics

Moment of Inertia of a Rod having non-uniform, linearly increasing mass density

Mar 3, 2024 PCM TUTORIALS 0 Comment

Moment of Inertia of Rod having non-uniform, linearly increasing mass density

Moment of Inertia

Moment of inertia is of great importance when we come to discuss the rotational dynamics. The the word “inertia” refers to resistance against any change in state of object (mass). Specifically, the moment of inertia is measure of a body’s resistance to changes in its rotation. The greater the value of moment of inertia, the more is difficulty it is to cause change in its rotation.

The moment of inertia is often denoted as “I” and it is analogous to mass (m) of the object. In linear motion, mass quantifies an object’s resistance to linear acceleration (change in velocity) in response to a force. Similarly, in rotational motion, the moment of inertia quantifies an object’s resistance to angular acceleration (change in angular velocity) in response to a torque. Moreover, both are scalar quantities.

Factors affecting Moment of Inertia

Here are the main factors affecting the moment of inertia:

Mass Distribution
Shape of the Object
Axis of Rotation
Size and Dimensions
Symmetry

Infact, we can can generalize the above all factor into one key factor – the mass distribution of body about axis of rotation which incorporates he definition of moment of inertia.

$I= m_1 r_1^{2} + m_2 r_2^{2}+ …… m_n r_n^{2}$ . Lets consider following example to grasp the concept of moment of inertia.

Example - On Moment of Inertia of Rod

The density of a rod AB increases linearly from A to B. Its mid-point is O and centre of mass is at C. Four axes pass through A, B, O and C , all perpendicular to the length of rod. The moments of inertia of the rod about these axes are $I_{A}$ , $I_{B}$ , $I_{O}$ and $I_{C}$ respectively. Then

(A) $I_{A}> I_{B}$

(B) $I_{C}< I_{B}$

(C) $I_{O}> I_{C}$

(D) All of these

Solution

The mass distribution of body about axis of rotation determine the value of moment of inertia.

$I= m_1 r_1^{2} + m_2 r_2^{2}+ …… m_n r_n^{2}$ .

If the overall mass is distributed nearer to axis of rotation then M.O.I. of the object(rigid body) will be smaller. And if mass is distributed away from axis of rotation it results into larger M.O.I.

Let us first locate the center of mass (C.O.M.) of the rod by using formula:

x_{com}=\;\frac{\int dm\;x}{\int dm}=\frac{\int dm\;x}M

$\\x_{com}=\;\frac{\int_0^Ldm\;x}{\int_0^Ldm}=\frac{\int_0^Ldm\;x}M$ ….(1)

Given that ,the linear mass density of the rod varies linearly along the length as $\lambda\;=\lambda_o\;x$

Mass of the small element dx, $dm= \lambda\ dx$

dm= \lambda_o \;x dx

By integrating on both sides, we can calculate total mass of the rod

M= \int_0^L\;dm\;=\int_0^L\;\lambda_o\;x\;dx

M=\lambda_o\;\int_0^L\;\;x\;dx

$M=\lambda_o\;\left|\frac{x^2}2\right|_0^L=\;\frac12\lambda_o\;L^2$ ….(2)

Now, we have to calculate , $\int_0^Ldm\;x\;$

\int_0^Ldm\;x\;=\;\int_0^L\;\lambda_o x\;dx\;x\;

\Rightarrow\int_0^Ldm\;x\;=\;\;\lambda_o\int_0^L\;x^2\;dx\;

\Rightarrow\;\int_0^Ldm\;x\;=\;\;\lambda_o\;\left|\frac{x^3}3\right|_0^L=\;\;\frac{\;\lambda_o\;L^3}3

$\Rightarrow\;\int_0^Ldm\;x\; =\;\;\frac{\;\lambda_o\;L^3}3$ ….(3)

From(1),(2)and (3)

x_{com}=\frac{\int_0^Ldm\;x}M\;=\frac{\frac{\;\lambda_o\;L^3}3}{\frac{\;\lambda_o\;L^2}2}\;

\Rightarrow\;x_{com}=\frac23L

According to Parallel axis theorem ,

$I_{parallel\;\;}=\;I_{COM}\;+\;M_{total\;}\;d^2$

$I_{O\;\;}=\;I_{C}\;+\;M \;(OC)^2$

$I_{B\;\;}=\;I_{C}\;+\;M \;(BC)^2$

$I_{A\;\;}=\;I_{C}\;+\;M \;(AC)^2$

AC = 2L/3 , BC = L/3, OC= L/6

As, AC > BC> OC

Hence, $I_A> I_B>I_O> I_C$

(D) is correct option

Conclusion-

In case of variable mass density total mass can be calculated by integration. And apply formula $x_{com}=\;\frac{\int dm\;x}{\int dm}$ to find position/coordinate of center of mass. Understanding of parallel axis theorem is key concept which relates all moment of inertias about any parallel axis with axis passing through centre of mass. Distribution of mass away from axis of rotation leads to greater value of moment of inertia about that axis.

For more problems on Moment of Inertia of Rod, click here

Mechanics

A ball of mass m moving at a sped v makes a head on collision with an identical ball at rest. If the kinetic energy of the balls after collision is [latex]\frac34th[/latex] of the original. Then the coefficient of restitution is

Feb 27, 2024 PCM TUTORIALS 0 Comment

Head on collision : Important case of partial elastic collision

Q.A ball of mass m moving at a speed v makes a head on collision with an identical ball at rest. If the kinetic energy of the balls after collision is $\frac34th$ of the original. Then the coefficient of restitution is

(A) $e=\frac1{\sqrt2}$

(B) $e=\frac12$

(C) $e=\frac{\sqrt2}3$

(D) $e=\frac13$

Solution

Key concepts to solve the following problem:

Head-on collision
Coefficient of restitution
Conservation of linear momentum

Given that, mass of each ball $= m$

According to law of conservation of momentum

$m_1 u_1 +m_2 u_2 = m_1 v_1 + m_2 v_2$

$m v +m *0 = m v_1 + m v_2$

$m v = m (v_1 + v_2)$

$v = (v_1 + v_2)$

$v_1 + v_2 = v$ .....(1)

Given that , $K.E._{net,f\;}=\frac34K.E._{net,i\;}$

Therefore, $\frac12\;m_1\;v_1^2\;+\;\frac12\;m_2\;v_2^2\;=\;\frac34\left[\frac12\;m_1\;u_1^2\;+\;\frac12\;m_2\;u_2^2\;\right]$

$\frac12\;m\;\left(\frac{v(1-e)}2\right)^2\;+\;\frac12\;m\;\left(\frac{v(1+e)}2\right)^2\;=\;\frac34\left[\frac12\;m\;v^2\;+\;\frac12\;m\;{\ast\;0}^2\;\right]$

$\frac12\;\;\frac{v^2\;\left(1+e\right)^2}4\;+\;\frac12\;\;\frac{v^2\;\left(1-e\right)^2}4\;=\;\frac34\left(\frac12\;v^2\;\right)$

$\Rightarrow\left(1+e\right)^2+\left(1-e\right)^2=3$

$\Rightarrow1+\;e^2+\;2e\;+\;1\;+\;e^2-2e\;= 3$

$\Rightarrow\;2\;e^2=\;1$

$\Rightarrow\;e^2=\;\frac12$

$\Rightarrow\;e=\;\frac1{\sqrt2}$

(A) is correct option

The formula of coefficient of restitution is :

$e=\;\frac{vel. of sepration}{vel. of approach}$

$e=\;\frac{v_2-v_1}{u_1-u_2}$

$e=\;\frac{v_2-v_1}{v-0}$

$v_2- v_1= ev$ ....(2)

On adding (1) and (2), we get

$2 v_2 = v +ev$

$v_2=\;\frac{v(1+e)}2$

On subtracting (2) from (1), we get

$2 v_1 = v - ev$

$v_1=\;\frac{v(1-e)}2$

Before jumping into the concept of head-on collisions partial elastic collision, we have to grasp a basic understanding of lines of impact and motion. Secondly, the knowledge of coefficient of restitution is of equal importance.

Line of Impact: The line along which the colliding objects make contact during the collision.

Line of Motion: The line along which the objects are moving.

Collisions can be classified into two main types based on the orientation of the line of impact and the line of motion of the colliding objects. These types are:

Head on Collision :
- In a head on collision, the line of impact is along the line of motion of the colliding objects. The objects approach each other directly, and the impact occurs along the same straight line. This type of collision is often analyzed in the context of one-dimensional motion for simplicity.
Oblique Collision :
- In an oblique collision, the line of impact is not aligned with the line of motion of the colliding objects. The objects approach each other at an angle, resulting in an impact that is not directly along the line of motion. Oblique collisions are more complex to analyze compared to head-on collisions, as they involve vector components and require consideration of two-dimensional motion.

Coefficient of Restitution:

The coefficient of restitution (e) is a crucial factor in understanding the nature of a collision. It is defined as the ratio of the final relative velocity of separation to the initial relative velocity of approach. Mathematically, e is expressed as:

e=\;\frac{relative\;velocity\;of\;separation}{relative\;velocity\;of\;appraoch\;}

The coefficient of restitution can take values between 0 and 1, inclusively, where:

represents a perfectly inelastic collision, where the colliding objects stick together after the collision. Here, the loss in net Kinetic energy is maximum after collision. And It is not necessary that there is always a complete loss of kinetic energy

denotes a partially elastic collision, where some of kinetic energy is lost.

signifies a perfectly elastic collision, where the colliding objects bounce off each other without any loss of kinetic energy.

Conclusion:

Problem solving becomes easy if we have basic knowledge of types of collisions. Understanding the coefficient of restitution is crucial in analyzing the nature of collisions and predicting the post-collision velocities of the objects involved. Except for perfectly inelastic collisions, the colliding objects get separated after the collision, undergoing either some loss or no loss in net kinetic energy, depending on the nature of the collision. Moreover, the law of conservation is applicable in all types of collisions, whether elastic, inelastic, or partially elastic

Visit page Mechanics for more problems

Mechanics

What is the maximum compression in the spring, if the lower block is shifted to rightward with acceleration a. (Given that all surfaces are smooth as shown).

Feb 25, 2024 PCM TUTORIALS 3 Comment

Maximum compression in the spring

Q.What is the maximum compression in the spring, if the lower block is shifted to rightward with acceleration a. (Given that all surfaces are smooth as shown).

(A) $\frac{ma}k$

(B) $\frac{2ma}k$

(C) $\frac{mak}2$

(D) $\frac{ma}{2k}$

Solution

Let us assume that the upper block of mass 'm' displaces by distance 'x' towards leftward with velocity 'v' (w.r.t to lower block) when lower block is shifted towards rightward with acceleration a.

We can solve this problem by two methods (1) Force Method (2) Energy Method from non-inertial frame of reference (Lower plank/block) .

Force Method

Acc. to Newton's 2nd law of motion,

$F_{net}=F_{pseu}-F_{spring}$

$F_{net}=ma- kx$

$m a_1 =ma- kx$

$m \frac{\operatorname dv}{\operatorname dt}=ma\;-\;kx$

$mv\frac{\operatorname dv}{\operatorname dx}=ma\;-\;kx$

$mv\;dv=\left(ma\;-\;kx\right)\;dx$

Integrating both sides

$m \int_{v_i}^{v_f}\;vdv=\int_{x_i}^{x_f}\left(ma\;-\;kx\right)\;dx$

$m\int_{0}^{0}\;vdv=\int_{0}^{x_o}\left(ma\;-\;kx\right)\;dx$

$m\left[\frac{v^2}2\right]_0^0=\;ma\;\left[x\right]_0^{x_0}-k\left[\frac{x^2}2\right]_0^{x_o}$

$m(0-0)=\;ma\;\left(x_o-0\right)-\;\frac12k\;\left(x_o^2-0\right)$

$0=\;ma x_o - \;\frac12k\ x_o^2$

$ma x_o = \frac12k\ x_o^2$

$2ma= k x_{o}$

$x_o=\frac{2ma}k$

(B) is correct option

let $a_1$ be the acceleration of upper block w.r.t the lower block

FBD of upper block from lower block reference frame

Boundary conditions for integration:

Initially, the upper block is at rest when x=0, therefore $v_{i}=0$

When the spring has maximum compression $x_{o}$ , again the block becomes stationary w.r.t. lower block.

So, $v_{f}=0$ , $when x=x_{o}$

Alternate Method

We can also solve this problem by Energy Method from non-inertial frame of reference (Lower plank/block) . We will apply Work-Energy theorem from lower block frame of reference.

Energy Method

We can apply Work-Energy theorem for initial and final positions of upper block w.r.t. lower block

Since, at initial position the upper block is at rest, therefore $v_{i}=0$

When the spring has maximum compression $x_{o}$ , again the block becomes stationary w.r.t. lower block. So, also $v_{f}=0$

Acc. to Work-Energy Theorem,

$W.D_{net \;by\;all\;forces}\;=\triangle K.E.\;$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;v_f^2-\frac12m\;v_i^2$

$W.D._{pseudo}\;\;+\;W.D._{spring}\;=\;\frac12m\;0^2-\frac12m\;0^2$

$\Rightarrow\;ma\;x_o-\;\frac12k\;x_o^2=0$

$\Rightarrow\;ma\;x_o=\;\frac12k\;x_o^2$

$\Rightarrow\;ma\;=\;\frac12k\;x_o$

$\Rightarrow\;x_o=\frac{2ma}k$

(B) is correct option

For more mechanics problems visit our question bank page

We must know that the work done by conservative forces is

$\;W.D._{cons}\;=-\triangle U$

$\;\Rightarrow\;W.D._{spring}\;=-\triangle U$

$\;\Rightarrow W.D._{spring}\;=-\;\frac12k\;x_{o}^2$

And, $W.D._{pseu}={\overrightarrow{F\;}}_{pseu}.\overrightarrow{x_o}$

$\Rightarrow W.D._{pseu}=F_{pseu}\;x_{o\;}\cos0^o$

$\Rightarrow W.D._{pseu}=ma\;x_{o\;}$

Key concept to solve above problem

Frame of reference
Pseudo force
Work-Energy Theorem
Condition for Maximum compression in the spring

Frame of reference

In mechanics , the frame of reference is directly tied to the perspective of an observer. A frame of reference is essentially a set of coordinate axes and a set of observational rules that allow an observer to quantify and describe the motion of objects. The observer’s point of view or choice of reference frame influences how the physical phenomena are described and analyzed. All of us know that the motion is a relative concept. So, frame of reference which observer chooses directly affect the observation.

For example, if you are driving to towards Taj Mahal and after some time a fellow passenger ask when will Taj Mahal come whereas he knows you are moving towards destination!
It does indeed reflect a perspective from the car’s frame of reference. The passenger is implicitly treating the car as the stationary reference point, making it appear as though the Taj Mahal is approaching them. From the frame of reference inside the car, the surroundings outside might seem to be moving, creating the perception that the destination (Taj Mahal) is approaching. Meanwhile, from an external frame of reference, such as someone on the ground, it would be clear that the car is moving towards the stationary Taj Mahal.

Let’s consider the other (following) example which helps you to grasp the concept.

If you are inside a car moving at a constant speed and you throw a ball straight up, from your perspective inside the car, the ball will appear to go straight up and down. However, an observer outside the car, seeing the entire motion, would notice that the ball follows a parabolic trajectory due to the combined motion of the car and the ball. There are mainly two type of frame of references

(1) Inertial Frame of Reference:

An inertial frame of reference is often preferred when studying the laws of mechanics. In an inertial frame, an object either remains at rest or moves with a constant velocity unless acted upon by an external force.
Newton’s laws of motion are formulated with respect to inertial frames.

(2) Non-Inertial Frames:

In non-inertial frames (accelerating or rotating frames), additional forces called fictitious forces may appear. These forces are not “real” forces but are necessary for describing motion accurately from the perspective of an observer in a non-inertial frame.
Common example include centrifugal

Pseudo force

A pseudo force is an apparent or fictitious force introduced in non-inertial reference frames to account for observed accelerations, allowing the application of Newton’s laws of motion as if the frame were inertial.

Work Energy Theorem

The work-energy theorem states that the work done on an object by all forces (conservative, non conservative, pseudo force etc. ) is equal to the change in its kinetic energy. In equation form, it can be expressed as:

$W = Δ K E$

Condition for maximum compression in the spring

When a block is attached to a spring and compressed to its maximum, the block will momentarily come to rest with respect to the reference frame to which the spring is attached., assuming no external forces act on the block after it reaches its maximum compression.

When the spring is compressed to its maximum, the potential energy stored in the spring is at its maximum, and this energy is then converted into kinetic energy as the block is released. At the point of maximum compression, the block’s velocity becomes zero before it starts moving in the opposite direction due to the restoring force of the spring. The momentary pause at maximum compression in the spring, is indeed a point where the block stops with respect to the frame to which the spring is attached.

Mechanics

For given system of two blocks, find the maximum value of force F which can be applied on the system as shown in figure so that both blocks move together. [Given coefficient of static friction between both blocks = 0.3]

Feb 23, 2024 PCM TUTORIALS 0 Comment

static friction to move two blocks together for IIT-JEE and NEET

How Static Friction helps two blocks to move together !

Q.For given system of two blocks, find the maximum value of force F which can be applied on the system as shown in figure so that both blocks move together. [Given coefficient of static friction between both blocks = 0.3]

(A) 3N

(B) 6N

(C) 9N

(D) 12N

Solution

Free body diagram of block of upper block is

We can see that the block 1 Kg (upper block) is moved by friction force. If block moves together (without slipping) it means the frictional force is static friction. Therefore,

$f_{s,max}=\;\mu_sN$

$f_{s,max}=\;\mu_s mg$

$f_{s,max}=\;\ 0.3 * 1 * 10 = 3N$

Thus, maximum acceleration that can be produced in upper block without slipping is

$a_{max}=\frac{f_{s,max}}m=\;\frac31=3ms^{-2}$

For system of blocks , to move together the maxmium force can be

$F_{max}=\;M_{sys,net\;}\ast\;\;a_{max}$

$F_{max}=(1+2) * 3= 9N$

(C) is correct option

In the above example, we can notice clearly that lower block is acted upon by applied (in forward direction) force which causes the motion in lower block [see free body diagram given below].But, the question arises how the upper block is moving in forward direction despite being no direct force on it? What force is there that moves the upper block? The obvious answer is, the friction, which drives the upper block in same direction. Many students might think that how friction can cause motion, as they feel the friction only opposes the motion. Infact , it is a only a misconception that many student grasp due to incomplete knowledge of concept of friction.
Friction necessarily opposes relative motion (sliding) between two contact surfaces, and also some times supports the motion of one object (w.r.t. other frame) while opposing relative motion of one surface over other. When there is no actual relative motion (sliding) between two contact surfaces but exists only the tendency of sliding one surface over the other, then, the frictional force acting between the contact surfaces is static friction. The static friction, as the name says, acts to keep the surface in contact (stationary w.r.t. each other) by stopping the sliding motion between them. To move both blocks together as in the above example, static friction provides the acceleration to upper block while opposing the lower block at the same time so that there is no sliding between them. The acceleration will be maximum when maximum static friction (limiting friction) comes into play. The limiting friction can be calculated by the formula
$f_{s,max}=\;\mu_sN$

Using the maximum value of acceleration, the maximum force value can be easily found out. We have,

$F_{max}=\;M_{sys,net\;}\ast\;\;a_{max}$

F.B.D. of lower block

F.B.D. of upper block

Conclusion-

Friction plays an essential role in motion despite being a dissipative force. It generally dissipates energy; however, sometimes it utilizes a portion of it to provide kinetic energy to other systems or its own subsystems. In the present case, friction provides kinetic energy to the upper block and performs positive work. Of course, the overall work done by friction is negative. A free-body diagram is necessary for a clear understanding of all forces.

The direction of friction is simply determined by Newton’s third law of motion. It is generally taken in the opposite direction to the applied force on one body (the lower block in this case), i.e., in the backward direction. The direction of friction acting on the other body (the upper block in this case), according to Newton’s third law, will be in the forward direction, just opposite to that of the lower block.

Friction is a contact force. Static friction comes into action until the applied force becomes sufficient to initiate sliding between the contact surfaces. There are also limiting values for static friction because, up to a certain maximum, it tends to prevent actual relative motion, and eventually, sliding begins beyond the limiting friction. Thus, static friction disappears, and kinetic friction comes into play.

For more problems on static friction visit page Mechanics

Mechanics

A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . What are its translational and rotational kinetic energies respectively?

Feb 23, 2024 PCM TUTORIALS 2 Comment

Rolling of a solid disc on horizontal

Rolling , specifically a pure rolling is a classic example of combined translational and rotational motion. While talking about the rolling of a solid disc over horizontal ground, actually we are discussing a simple case of rolling. Pure rolling refers rolling without slipping. Thus, there is no presence of kinetic friction between point of contact and horizontal surface. Kinetic friction keeps acting until pure rolling is achieved. Static friction may or may not be present in the motion, it depends on on factor whether a force is present throughout the motion or not.
For permanent pure rolling(without slipping), we have conditions i.e. $v=\;R\omega$ and $a=\;R\alpha$
These are relations between angular speed and linear speed (angular acceleration and tangential acceleration) defines the situation of rolling without slipping.

Rolling of a solid disc on horizontal - A Case of Pure rolling

Here, we are considering an example which incorporates the concept of rolling of a solid disc on horizontal – a case of Pure rolling and its both kinetic energies- translational and rotational. Also, we can use these conditions (of rolling) to relate both kinetics energies of combined translational and rotational motion.

Example - A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 150 J . Its translational and rotational kinetic energies respectively are

(A) 50 J , 100 J

(B)100 J , 50J

(C) 75 J, 75J

(D) 125J , 25J

Solution

As we know that, for pure rolling(without slipping) $v=\;R\omega$

$\frac{K.E._R}{K.E_T}=\frac{{\displaystyle\frac12}I\;\omega^2}{{\displaystyle\frac12}m\;v^2}$

$\frac{K.E._R}{K.E_T}=\;\frac{\displaystyle\frac12\left(mK^2\right)\;\omega^2}{\displaystyle\frac12m\;v^2}$

$\frac{K.E._R}{K.E_T} =\frac{\displaystyle K^2\;\omega^2}{\displaystyle v^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle{(R\omega)}^2}=\frac{\displaystyle K^2\;\omega^2}{\displaystyle R^2\;\omega^2}$

$\frac{K.E._R}{K.E_T}=\frac{\displaystyle K^2\;}{\displaystyle R^2\;}$ .......(1)

Now, $I=\;\frac12m\;R^2=m\;K^2\; where\; K \;is \;radius\; of\; Gyration$

$\Rightarrow\frac{K^2}{R^2}=\frac12$ .......(2)

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac{K^2}{R^2}=\frac12$ [from (1) and (2)]

$\Rightarrow\;\frac{K.E._R}{K.E_T}=\frac12$

$\Rightarrow\ K.E_T=\;\left(\frac2{1+2}\right)\;K.E._{total}$

$\Rightarrow\ K.E_T=\;\left(\frac2{3}\right)\;*150$

$\Rightarrow\ K.E_T= 2* 50 = 100J$

Similarly, $K.E_R=\;\left(\frac1{1+2}\right)\;K.E._{total}$

$\Rightarrow K.E_R=\;\left(\frac1{1+2}\right)\;*150$

$\Rightarrow K.E_R=\;\left(\frac1{3}\right)\;*150$

$\Rightarrow K.E_R=1*50=50J$

(B) is correct option

Conclusion: -

From above discussion , we can easily understand that the basic knowledge of moment of inertia, kinetic energies and condition of rolling (without slipping) helps us to solve problems based on such concepts. For more question visit the page Physics-XI Mechanics

Category Mechanics

Complete Video Solution below

Maximum Force such that both blocks move together Without Slipping: Understanding the Real Physics

The First Question: How Does the Upper Block Move?

The Hero of the Story: Static Friction

Static Friction Is Smarter Than Most Students Think

Why Friction Direction Confuses Students

A Common Mistake Students Make

Moving Together Means No Slipping

Can Static Friction Do the Job?

What Happens Beyond This Limit?

Why This Problem Is More Important Than It Looks

Physics Is About Understanding, Not Memorization

Final Solution

Author: Deep Aman Sir

Founder, PCM TUTORIALS | Physics Faculty for JEE, NEET & CBSE

Pseudo Force

Pseudo Force: Understanding the “Fictitious” Force of Non-Inertial Frames

1. Reference Frames: The Foundation

2. What is a Pseudo Force?

3. Mathematical Expression of Pseudo Force

4. Everyday Examples of Pseudo Force

(i) The Bus Ride

(ii) Elevator Problems

(iii) Rotating Reference Frames (Centrifugal Force)

(iv) Coriolis Force

5. Pseudo Force vs Real Force

6. Types of Pseudo Forces (Required for JEE/NEET level)

7. Pseudo Force in Elevators – A Classic Example

8. Pseudo Force in Rotating Frames – Centrifugal and Coriolis

9. Applications of Pseudo Forces

10. Common Misconceptions

11. Visualizing Pseudo Forces

12. Pseudo Force and General Relativity – A Deeper Note

13. Summary

14. Conclusion

Kinematics: Understanding Uniform Motion

What is Kinematics

What is Uniform Motion?

As , \overrightarrow{v\;}\;=\;constan t\;

Therefore, \overrightarrow{a\;}=\;\frac{d\overrightarrow v}{dt}=\;0

Mathematical Representation

v=\frac{dx}{dt}\;

\Rightarrow dx\;=\;v\;dt\;

\int_{x_1}^{x_2}dx\;=\;\int_{t_1}^{t_2}v\;dt\;

\int_{x_1}^{x_2}dx\;=\;v\int_{t_1}^{t_2}\;dt\; (v= constant/uniform)

\left|x\right|_{x_1}^{x_2}=\;v\;\left|\;t\;\right|_{t_1}^{t_2}

x_2\;-x_1\;=\;v\;(t_2-t_1)

\Delta x\;=v\;\Delta t

v=\frac{\Delta x}{\Delta t}

Hence,\;v=\frac{dx}{dt}=\frac{\Delta x}{\Delta t}=constant

When object/particle starts motion from origin (x=0)

\Delta x\;=v\;\Delta t

x-0= v(t-0)

x = v t

Position-Time (x-t) graph of uniform motion starting from origin

When object/particle starts motion with some initial displacement

\Delta x\;=v\;\Delta t

x-x_o= v(t-0)

x = x_o+ v t

Position-Time(x-t) graph of uniform motion of particle with non-zero initial displacement

Graphical Representation

Position-Time Graph

Velocity-Time Graph

Uniform Motion vs. Non-Uniform Motion

Conclusion

Additional Velocity Required to Overcome Earth's Gravitational Pull for Satellite Escape

Example

Q.A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth =6400 km , g= 9.8 ms^{-2}

(A) 3.2 km/s

(B) 11.2 km/s

(C) 1.5 km/s

(D) 8 km/s

Solution

We know that , orbital speed of satellite is

v_o=\;\sqrt{\frac{GM}r}\;

where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth

Also, r= R+ h where R is radius of planet/earth and h is height of satellite from surface of earth

v_o=\;\sqrt{\frac{GM}{R+h}}\;

Near to earth's surface, h can be neglected

As , $\overrightarrow{v\;}\;=\;constan t\;$

Therefore, $\overrightarrow{a\;}=\;\frac{d\overrightarrow v}{dt}=\;0$

$v=\frac{dx}{dt}\;$

$\Rightarrow dx\;=\;v\;dt\;$

$\int_{x_1}^{x_2}dx\;=\;\int_{t_1}^{t_2}v\;dt\;$

$\int_{x_1}^{x_2}dx\;=\;v\int_{t_1}^{t_2}\;dt\;$ (v= constant/uniform)

$\left|x\right|_{x_1}^{x_2}=\;v\;\left|\;t\;\right|_{t_1}^{t_2}$

$x_2\;-x_1\;=\;v\;(t_2-t_1)$

$\Delta x\;=v\;\Delta t$

$v=\frac{\Delta x}{\Delta t}$

$Hence,\;v=\frac{dx}{dt}=\frac{\Delta x}{\Delta t}=constant$

$\Delta x\;=v\;\Delta t$

$x-0= v(t-0)$

$x = v t$

$\Delta x\;=v\;\Delta t$

$x-x_o= v(t-0)$

$x = x_o+ v t$

Q.A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth =6400 km , g= 9.8 $ms^{-2}$

(A) $3.2 km/s$

(B) $11.2 km/s$

(C) $1.5 km/s$

(D) $8 km/s$

$v_o=\;\sqrt{\frac{GM}r}\;$

$where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth$

$Also, r= R+ h$ where R is radius of planet/earth and h is height of satellite from surface of earth

$v_o=\;\sqrt{\frac{GM}{R+h}}\;$

So, $R\;+\;h\;\approx\;R$

$v_o=\;\sqrt{\frac{GM}{R}}\;$

$v_o=\;\sqrt{\frac{gR^2}{R}}\;$

$v_o=\;\sqrt{gR}\;$

$v_o\;=\;\sqrt{Rg}\approx\;8\;km/s\;$

$v_e\;=\;\sqrt{2Rg}\approx\;11.2\;km/s\;$

$v_{add}\;=\;v_e\;\;-\;v_{o\;}$

$v_{add}\;=11.2 - 8 = 3.2 km/sec$

(A) $16.96 hours$

(B) $8.48 hours$

(C) $9.25 hours$

(D) $15.76 hours$

$v_o=\;\sqrt{\frac{GM}r}\;$

$where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth$

$T\;=\frac{dis\tan ce\;}{speed}=\;\frac{2\pi\;r}{v_o}$

$T\;=\;\frac{2\pi\;r}{\sqrt{\displaystyle\frac{GM}r}}$

$T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}$

$T\;=\;2\pi\;\sqrt{\frac{r^3}{GM}}\\T^2=\;4\;\pi^2\;\frac{r^3}{GM}$

$\Rightarrow T^2\;\;\propto\;r^3$

$r= R+ h$ where R is radius of planet/earth and h is height of satellite from surface of earth

$\Rightarrow r_1 = R+ 6R=7R \;and \;r_2= 3.5R$

$\Rightarrow T\;\propto\;r^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{r_2}{r_1}\right)^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{3.5R}{7R}\right)^\frac32$

$\frac{T_2}{T_1}=\;\left(\frac{1}{2}\right)^\frac32$

$\frac{T_2}{T_1}=\frac1{2\sqrt2}$

$T_2 =T_1 * \frac1{2\sqrt2}$

$T_2 =24 * \frac1{2\sqrt2}$

$T_2 =12* \frac1{\sqrt2}$

$T_2 =12* 0.707= 8.48 \;hours$

The density of a rod AB increases linearly from A to B. Its mid-point is O and centre of mass is at C. Four axes pass through A, B, O and C , all perpendicular to the length of rod. The moments of inertia of the rod about these axes are $I_{A}$ , $I_{B}$ , $I_{O}$ and $I_{C}$ respectively. Then

(A) $I_{A}> I_{B}$

(B) $I_{C}< I_{B}$

(C) $I_{O}> I_{C}$

Now, we have to calculate , $\int_0^Ldm\;x\;$

$I_{parallel\;\;}=\;I_{COM}\;+\;M_{total\;}\;d^2$

$I_{O\;\;}=\;I_{C}\;+\;M \;(OC)^2$

$I_{B\;\;}=\;I_{C}\;+\;M \;(BC)^2$

$I_{A\;\;}=\;I_{C}\;+\;M \;(AC)^2$