Rigid Body Dynamics deals with the movement of rigid bodies that are interconnected, influenced by the external forces acting upon them. The chapter dynamics is one of the important chapters in physics for NEET. Around 8 marks are assigned for questions from this topic, and the related topic 'motion of the system of particles and rigid body' has questions worth 12 marks from it. 

• In the NEET 2021 syllabus, Rigid Body Dynamics constitute about 6.67% of the total weightage of the question paper. 

• After topics like matter, fluid mechanics, thermal expansion, etc. this topic constitutes the highest weightage in the exam.

• The expected number of questions that may be asked in NEET 2021 from the topic rigid dynamics is around 2 to 3.

Candidates need a deep understanding of the topic to develop quick problem-solving skills. Read the article to get simple definitions of all the important topics with solved examples for better understanding. 

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Moment of Inertia

The capacity of a system to resist the changes that are produces while a body undergoes rotational motion is known as the Moment of Inertia. Moment of Inertia is denoted by I.

The formulae to calculate the moment of inertia of a body is;

I=mr2

where m denotes the mass of the body or particle and r denotes the perpendicular distance from the axis from which the moment of inertia has to be calculated.

In case there are many particles for which the moment of Inertia has to be calculated;

I = miri2

Where m is the sum of the product of mass and r is the perpendicular distance to its axis.

For different objects, with the different axis of rotation about which the rotational motion occurs, the formula for calculating the moment of inertia varies.

A solid cylinder or disc rotating around its symmetry axis

• I = ½ mr2

• If it is rotating around its central diameter,

• I = ¼ mr2 + 1/12 ml2

• Where m is the mass of the particle, r is the radius and l is the length.

A hoop rotating around its symmetry axis

• I = mr2

• If it rotates about its diameter,

•  I = ½ mr2

A solid sphere

• I = 2/5 mr2

A hollow sphere or a spherical shell

• I = 2/3 mr2

A rod rotating about its centre

• I = 1/12 ml2

If it rotates around an axis at its end

• I = 1/3 ml2

Sample Question

Q: If three identical rods with mass m and length l forms an equilateral triangle, what will be the moment of inertia of the triangular frame about an axis running parallel to its one side and passing through the opposite vertex of the other two sides.

1. 5/3 ml2

2. 3/2 ml2

3. 5/4 ml2

4. 5/2 ml2

A: Option C 5/4 ml2

Solution: In a triangle ABC, where the axis PQ is parallel to AB and at C,

The rods AC and BC have the same moment of inertia as both of them are rotating around the axis PQ at their ends.

So,

IAC = IBC = 1/3 ml2 sin2 60o

ml2/3 sin2 60o = ml2/4

Moment of inertia of the rod AB about the axis PQ

IAB = m (l√3/ 2)2 = ¾ 3l

I = IAB + IAC + IBC = ¾ ml2 + ml2/4 + ml2/4 = 5/4 ml2

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Radius of Gyration

An imaginary distance between the centroid and a cross-section at which the moment of inertia to be calculated is defined as the radius of gyration. It is denoted by k when the radius of gyration comes into the equation.

I = mk2

Where I is the moment of inertia, m denotes the mass of the object and k is the radius of gyration.

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The Relation between Torque and Moment of Inertia

Torque is defined as the measure of force which causes an object to undergo rotational motion around its axis. Torque is directly proportional to the angular acceleration in an object which is in rotational motion and the moment of inertia. It is denoted by τ

T = Iα

Where I is the moment of inertia and α is the rate of change of angular velocity ie; angular acceleration.

Sample Question

Q: A uniform disc of mass M and radius R with a cord wrapped around its perimeter is supported in frictionless bearings. If the cord is subjected to a constant downward pull T what will be the angular acceleration of the disc?

1. 2T/MR

2. MR/2T

3. MR/T

4. T/MR

A: A) 2T/MR

Solution: Torque on the disc

T = TR : ( T = Iα ) ( I = ½ MR2)

α = T/R = TR/ ½ MR2

= 2TR/ MR2 = 2T/MR

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Angular Momentum

A rotating body which resists the change in its state produces an angular momentum.

For a point object, which rotates around an axis,

L = r x p

Where L is the angular momentum, r is the radius or the distance between the rotating object and its axis and p is the linear momentum.

For an extended object, which rotates around its axis,

L = I x w

Where L is the angular momentum, I is the moment of inertia created by the rotational motion and w is the angular velocity.

Sample Question

Q: If a pully of mass 2kg and radius 0.1 m rotates at a constant angular velocity of 4 rad/sec, what will be the angular momentum?

A: For an object which rotates around its own axis,

L = T x w

Here, I = ½ mr2

I = ½ 2 x 0.12

= 0.02/2 = 0.01 kg m2

L = 0.01 x 4 = 0.04 kg m2 s-1

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Relation Between Angular Momentum and Torque

Torque can be defined as the twisting force for rotational motion of an object. So, torque is the change in angular momentum of a rotating object. The variables here are angular momentum and time.

T = dL/dt

Where T is torque L is the angular momentum and t is time.

Important Formulas to Remember

• Moment of Inertia; I = mr2

• Torque; T = Iα

• Angular Momentum; L = r x p ; L= I x w

• Relation between Torque and Angular Momentum; T = dL/dt

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Previous Year Solved Sample Questions

Question: A stone is tied to a string and is rotating in a circle. If the string is cut, why does the stone travels away in a different path rather than continuing in its circular path of motion?

1. Because of the centripetal force

2. A centrifugal force exerted on the stone

3. Because of its inertia

4. A centripetal force acts on the stone

Answer: (C ) Because of its inertia

Question: When a cricket ball with M as its mass and R as its radius rotates on its axis, what will be the moment of inertia?

1. I = MR2

2. I = 2/5 MR2

3. I = ½ MR2

4. I = 2/3 MR2

Answer: A cricket ball is a hollow sphere. Hence. (D) I = 2/3 MR2

Question: A thin wire with length L and uniform mass density p forms a loop with its centre as O. What will be the moment of inertia of the loop about an axis XY as shown in the image.

1. 3pL3/8π2

2. 5pL3/8π2

3. pL3/16π2

4. pL3/8π2

Answer: (A) 3pL3/8π2

L=2πr, where r = L/ 2π

 m = pL

Moment of inertia at axis XY

IXY = 3/2 mr2

= 3/2 pL (L/2π)2 = 3 pL3/8π2

Question: A uniform cylinder with L as length and R as radius about its perpendicular bisector has I as the moment of inertia. What is the ratio of L/R when the moment of inertia is minimum.

Answer: Length of the cylinder is L

The radius of the cylinder is R

I = mR2/4 + mL2/12

= m/4 ( R2 + L2/3 ) = m/4 (V/πl + L2/3 )

Differentiate I with respect to l

dI/dl = m/4 (-V/πl2 + 2l/3)

For maxima and minima, dI/dl = 0

So, m/4 (-V/πl2 + 2l/3) = 0

V/πl2 = 2L/3

R2/l = 2L/3

L2/R2 = 3/2

√(L2/R2) = √(3/2)

L/R = √(3/2)

Hence the ratio of L/R when the moment of inertia is at the minimum will be √3/2

Question: If a particle moves with constant angular velocity in a circular motion then during the motion its

1. Both energy and momentum is conserved

2. Momentum is conserved

3. None of the above is conserved

4. Energy is conserved

Answer: (d) Its energy is conserved.

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Tips and Tricks for Rigid Body Dynamics

• While answering the questions from rigid body dynamics, always draw and name the diagrams that are said in the situation. It helps in better understanding of the described scenario, and in figuring out the apt equations.

• Most of the questions will require two or more equations from different parts of the topic to reach the final answer. So, you must be well-versed with all the equations in the topic. Leaving out even one of them may lead to completely missing out a question even if you know all the other equations required to solve the question.

• Solving questions with different difficulty levels will help you to understand the twisted ways a simple question can be asked. It also helps with quickening the pace of your problem-solving skills.

• Make flashcards of formulas and important concepts for quick references. This way it will be easy to remember the topics you studied just by referring to the flashcards.

• NCERT textbooks are a great resource when preparing for NEET that is available to everyone. Practice the exercise questions at the end of each chapter, to be thorough with your understanding of the topic.

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A Study Plan for NEET 2021 Rigid Body Dynamics

To do a final revision for NEET, candidates can go through the following preparation plan. 

• The last month before examination should mostly be dedicated to solving question papers and re-visiting the portions which you feel difficult.

• For those who wish to appear for the NEET in the coming years, don’t burden yourself with so many topics to cover in so little time, start early and take your time to familiarize yourself with the topics.

Here is a One-month Revision Plan for NEET 2021 Candidates which candidates can use or apply in the last one month or can also following during the whole preparation time. 

For the topic Rigid Body Dynamics, you can take 1 to 2 weeks of your whole preparation period. This topic is important, yet it is easier to grasp once you get the basic concepts right. During the one-month revision, a day or two will be well enough to brush up your memory on rigid body dynamics and its sub-sections.