Physics Tutor in Deira Dubai

Physics Tutor in Deira Dubai by Kumar Physics Classes with ladder on inclined plane, torque concepts, toppling condition, and moment of inertia of a disc after removing a circular hole. Contact +91 9958461445.

 

Physics Tutor in Deira Dubai 

+91-9958461445

If you live in Deira Dubai and you are struggling in Physics, then you are on the right page. Many students in Deira Dubai study in CBSE, ICSE, IGCSE, IB, A-Level, AP Physics, NEET, IIT JEE Main and IIT JEE Advanced, but they face one common problem: Physics concepts are not clear.

Sometimes the school teacher is teaching very fast. Sometimes the student understands the topic in class, but when numerical questions come, everything becomes confusing. Marks do not come in tests, parents become worried, and students start feeling that Physics is very difficult.

But Physics is not difficult. Physics becomes difficult only when the basic concepts are weak.

At Kumar Physics Classes, Kumar Sir teaches Physics in a simple, friendly and concept-based way. He explains every topic from zero level and then slowly takes the student to board level and competitive exam level.

If you are living in Deira Dubai and you need a strong Physics Tutor, you can contact Kumar Sir for online Physics classes.

Physics Tutor in Deira Dubai for CBSE, ICSE, IGCSE, IB, A-Level, AP, NEET and IIT JEE

Kumar Physics Classes provides online Physics tuition for:

  • CBSE Class 11 Physics

  • CBSE Class 12 Physics

  • ICSE Physics

  • IGCSE Physics

  • IB Physics SL and HL

  • A-Level Physics

  • AP Physics 1

  • AP Physics 2

  • AP Physics C Mechanics

  • AP Physics C Electricity and Magnetism

  • NEET Physics

  • IIT JEE Main Physics

  • IIT JEE Advanced Physics

Kumar Sir focuses on:

  • Concept clarity

  • Step-by-step derivations

  • Numerical practice

  • Formula understanding

  • Doubt clearing

  • Regular assignments

  • Exam-oriented preparation

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Moment of Inertia of a Body After Removing a Circular Part

Now let us understand an important Physics concept.

Suppose we have a uniform circular disc of radius R. From this disc, a smaller circular part of radius R/2 is removed. Now we have to find the new moment of inertia of the remaining body.

This type of question is very important for CBSE, NEET, IIT JEE, AP Physics and A-Level Physics.

Basic Idea

When a part is removed from a body, we treat the removed part as negative mass.

So:

Moment of inertia of remaining body = Moment of inertia of complete body – Moment of inertia of removed part

This is the main idea.

Step 1: Moment of Inertia of Complete Disc

For a complete disc of mass M and radius R about its central axis:

I complete = 1/2 MR²

Step 2: Mass of Removed Disc

Radius of removed disc = R/2

Area of complete disc = πR²

Area of removed disc = π(R/2)²

Area of removed disc = πR²/4

So removed area is one-fourth of complete area.

If mass of complete disc is M, then mass of removed disc is:

m = M/4

Step 3: Moment of Inertia of Removed Disc About Its Own Centre

For removed small disc:

radius = R/2
mass = M/4

Moment of inertia about its own centre:

I small cm = 1/2 m(R/2)²

Now put m = M/4:

I small cm = 1/2 × M/4 × R²/4

I small cm = MR²/32

Step 4: If Removed Disc Is Concentric

If the smaller disc is removed from the centre of the bigger disc, then both centres are same.

So:

I remaining = I complete – I small

I remaining = 1/2 MR² – MR²/32

I remaining = 16MR²/32 – MR²/32

I remaining = 15MR²/32

So final answer:

I remaining = 15MR²/32

Case 2: If Small Disc Is Removed Away From Centre

If the small circular part is removed from a point away from the centre, then we must use Parallel Axis Theorem.

Let distance between centre of big disc and centre of small removed disc = d

Moment of inertia of removed disc about centre of big disc:

I removed about big centre = I small cm + md²

Now:

I small cm = MR²/32

m = M/4

So:

I removed about big centre = MR²/32 + (M/4)d²

Therefore:

I remaining = I complete – I removed

I remaining = 1/2 MR² – [MR²/32 + (M/4)d²]

I remaining = 15MR²/32 – (M/4)d²

This is the general formula.

Special Case: If Small Disc of Radius R/2 Touches the Outer Boundary

If the removed disc of radius R/2 touches the outer boundary internally, then distance between centres is:

d = R/2

Now:

I remaining = 15MR²/32 – (M/4)(R/2)²

I remaining = 15MR²/32 – MR²/16

I remaining = 15MR²/32 – 2MR²/32

I remaining = 13MR²/32

So final answer:

I remaining = 13MR²/32

Kumar Sir Style Explanation

देखो बेटा, जब भी किसी body में से कोई part काट दिया जाता है, तो हमें डरना नहीं है। Simple rule याद रखो:

Complete body का moment of inertia लो.
Removed part का moment of inertia निकालो.
Removed part को minus कर दो.

अगर removed part का centre same axis पर है, तो direct minus कर दो.

अगर removed part side में है, तो पहले Parallel Axis Theorem लगाओ:

I = Icm + md²

फिर उसको complete body से subtract करो.

यही पूरी कहानी है.

Important Formula Summary

Complete disc:

I = 1/2 MR²

Removed small disc radius R/2:

m = M/4

I small cm = MR²/32

If removed from centre:

I remaining = 15MR²/32

If removed at distance d from centre:

I remaining = 15MR²/32 – (M/4)d²

If removed disc touches outer boundary:

d = R/2

I remaining = 13MR²/32

Why This Concept Is Important

This concept is important because it combines:

  • Moment of inertia

  • Area-mass relation

  • Parallel Axis Theorem

  • Composite bodies

  • Negative mass method

  • Rotational dynamics

Many students make mistakes because they directly subtract formulas without checking the axis. Axis is the most important thing in moment of inertia.

Always remember:

Moment of inertia depends on axis.

Why Students in Deira Dubai Should Learn Physics from Kumar Sir

Students in Deira Dubai often study in good schools, but still they may not get personal attention. Physics needs one-to-one explanation. If the student does not understand vectors, force, torque, centre of mass, moment of inertia, electrostatics, current electricity, optics or modern physics, then marks will not improve only by reading books.

Kumar Sir teaches in a way that students understand:

  • What is the meaning of formula?

  • Where does the formula come from?

  • How to apply formula in numerical?

  • What is the common mistake?

  • How to write answer in exam?

  • How to solve tricky NEET and JEE questions?

This is why Kumar Physics Classes is helpful for students living in Deira Dubai.

Online Physics Classes from Delhi to Deira Dubai

You may be sitting in Deira Dubai, and Kumar Sir may be sitting in Delhi, but online Physics classes make learning very easy.

The student can attend the class from home. There is no travel time, no traffic problem and no location issue.

You only need:

  • Laptop or tablet

  • Good internet connection

  • Notebook

  • Pen

  • Regular practice habit

Kumar Sir teaches live, explains concepts, solves numerical questions, gives assignments and clears doubts.

Online classes are very useful for Dubai students because timings can be adjusted according to UAE time.

Why Choose Kumar Physics Classes?

Students choose Kumar Physics Classes because:

  • 30+ years teaching experience

  • Strong command over Physics

  • Simple English and Hindi explanation

  • CBSE, ICSE, IGCSE, IB, A-Level, AP, NEET and IIT JEE support

  • Concept-based teaching

  • Numerical problem solving

  • Step-by-step derivations

  • Doubt clearing

  • Online classes for Dubai students

  • Flexible UAE timing

  • Personal attention

  • Regular practice and tests

Kumar Sir believes that Physics becomes easy when the concept becomes clear.

Contact Kumar Physics Classes

For online Physics classes in Deira Dubai, contact:

Kumar Physics Classes
Website: kumarphysicsclasses.com
Phone / WhatsApp: +91-9958461445
Email: kumarsirphysics@gmail.com

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25 Conceptual Questions on Torque, Constant Torque, Ladder and Toppling

Kumar Sir Style

1. What is torque?

Answer:
Torque is the turning effect of force. जब force किसी body को rotate करने की कोशिश करता है, उसे torque कहते हैं.

Formula:
τ = rF sinθ

2. Why does torque depend on perpendicular distance?

Answer:
Because turning effect depends on how far the force is applied from the axis. जितनी ज्यादा perpendicular distance, उतना ज्यादा torque.

Formula:
τ = Force × perpendicular distance

3. When is torque maximum?

Answer:
Torque is maximum when force is applied perpendicular to the radius.

If θ = 90°, then:

τ = rF

4. When is torque zero?

Answer:
Torque is zero when force passes through the axis of rotation.

Because perpendicular distance becomes zero.

5. What happens when torque is constant?

Answer:
If torque is constant and moment of inertia is constant, angular acceleration is also constant.

Formula:
τ = Iα

So:

α = τ / I

6. What is the relation between torque and angular momentum?

Answer:
Torque is the rate of change of angular momentum.

Formula:
τ = dL/dt

If torque is constant, angular momentum changes uniformly.

7. If net torque is zero, what happens?

Answer:
If net torque is zero, angular momentum remains conserved.

τnet = 0

So:

L = constant

8. Why is it easier to open a door from the handle?

Answer:
Because handle is far from the hinge. More distance means more torque. अगर hinge के पास push करेंगे तो door मुश्किल से खुलेगा.

9. Why does a long spanner open a nut easily?

Answer:
Long spanner gives larger perpendicular distance, so torque becomes larger for the same force.

10. What is toppling?

Answer:
Toppling means a body starts rotating about one edge and falls down.

This happens when line of action of weight passes outside the base.

11. What is the condition for toppling?

Answer:
A body topples when the vertical line through centre of mass falls outside the base of support.

12. Why does a tall object topple more easily?

Answer:
Because its centre of mass is high. A small tilt can make the line of action of weight go outside the base.

13. Why is a wide-base object more stable?

Answer:
Because the line of action of weight remains inside the base for a larger tilt angle.

14. Why does a ladder against a wall need friction?

Answer:
Friction prevents slipping. Without friction, the ladder may slide down because forces cannot balance properly.

15. What forces act on a ladder leaning against a wall?

Answer:
Forces are:

  • Weight of ladder downward

  • Normal reaction from ground upward

  • Friction at ground

  • Normal reaction from wall

  • Friction at wall, if wall is rough

16. If wall is smooth, what force acts from wall?

Answer:
Only normal reaction acts from the wall. No friction acts if wall is smooth.

17. Why do we take torque about one end of ladder?

Answer:
Because forces passing through that point produce zero torque. इससे calculation आसान हो जाती है.

18. Why does ladder slip more easily on a smooth floor?

Answer:
Because friction at floor is small. Without enough friction, the lower end moves outward and ladder falls.

19. What happens when a person climbs higher on the ladder?

Answer:
The torque due to weight increases, so chances of slipping increase.

20. Why is ladder more stable when it makes a larger angle with ground?

Answer:
When ladder is more vertical, horizontal torque effect becomes smaller, so it is more stable.

21. What is equilibrium in rotational motion?

Answer:
For rotational equilibrium:

Net torque = 0

For complete equilibrium:

Net force = 0
Net torque = 0

22. Can a body have zero net force but non-zero torque?

Answer:
Yes. A couple can produce pure rotation with zero net force but non-zero torque.

23. Can a body have non-zero force but zero torque?

Answer:
Yes. If force passes through centre of mass or axis, torque can be zero.

24. Why does a book topple when pushed from top?

Answer:
Because force applied at top creates large torque about the bottom edge.

25. Why is torque important in Physics?

Answer:
Torque explains rotation, toppling, ladder equilibrium, opening doors, spanners, wheels, pulleys and many real-life situations. It is very important for CBSE, NEET, JEE, AP Physics and A-Level Physics.

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