Rotational Motion is a very basic concept to form a solid base for important topics like Simple Harmonic Motion, Circulation Motion & Laws of Motion. So, aspirants are advised not to skip it while preparing for JEE Main 2021. If you have a look at the 2020 Question Paper and you would find; One question was asked from Simple Harmonic Motion, one question was related to Circular Motion, and two questions were from Laws of Motion. Overall, the weightage of these types of questions is around 13.2%. 

• Rotational Motion and SHM question are not that easy. There are so many formulas and derivations, but you can solve them without any problem if you understand the concept. Check JEE Main Physics Syllabus
• Generally, Mass and Radius are given in questions. Every important topic in Physics is linked to other topics. JEE Main aspirants Must-have a clear understanding of Rotational Motion to Solve SHM questions.

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JEE Main Test series - Previous Year Questions and Solutions

Q.1 The linear Mass density of thin rod AB of length L varies from A to B as λ(x)=λ0 (1+x/L) where X is the distance from A. If M is the Mass of Rod then it’s moment of inertia about an axis passing through A and Perpendicular to the rod is: [ 1 marks JEE MAIN 2020] 

1. 7/8 ML2
2. 2/5 ML2
3. 3/7 ML2
4. 15/12 ML2

Solution: dm = λdx = λ0 (1+x/L)

After doing the integration of both side;

M = λ0L + λ0L2 /2L = 3λ0L/2

 λ0 = 2M/3L ………. (1.)

To solve it, let’s find out Momentum of Inertia of dx

dI = dmx2

After Integrating both side;

= λ0 [L3/3+L3/4]

= 7L3 λ0/12…………. (2.)

Now, put the value of λ0 from equation (1.) to into equation (2.)

I = 7L3 λ0/12 x 2M/3L

I = 7/18 ML2

Q.2 What will be the momentum of inertia near its axis of a hollow ice-cream cone that is open at the top. Its height is H, radius is R, and mass is M. [ 1 marks JEE MAIN 2020] 

1. MR2/2
2. M2H/
3. MR2+MH2/3
4. MR2/3

Solution:

Apply the given conditions, you will get;

Momentum of inertia of cone = Momentum of inertia of a circular disk of mass (M) and radius (R) = MR2/2

Q.3 A satellite is in an elliptical orbit around a planet P. It’s observed that the velocity of the satellite when it’s farthest from the planet is 6 times less than that when it’s closest to the planet. The ration of distances between the satellite and planet at closest and farthest points is: [ 1 mark JEE MAIN 2020]

1. 1:4
2. 1:6
3. 1:3
4. 1:2

Solution:

Apply Angular momentum conservation rule.

Li = Lf

r (min) x V (max) = R (max) x v (min)

Put the value of Vmax (6) into the above-mentioned formula...

You will get the answer 1:6

Q.4 Four-point masses, each of mass (M), are fixed at the corners of a square of slide L. The square is rotating with angular frequency w, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the picture. The angular momentum of the square about this axis is: [ 1 mark JEE MAIN 2020]

1. 4ml2w
2. 3ml2w
3. 2ml2w
4. ml2w

Solution:

I = m . l2/2 . 2 + m . (√2L)2

3ml2

Because L= Iw

L = 3ml2w

Q.5 A clock has a continuously moving second’s hand of 0.1 length. The average accelerator of the tip of the hand in meter second square is of the order of: [ 1 mark JEE MAIN 2020] 

1. 10-4
2. 10-3
3. 10-2
4. 10-1

Solution:

l = r = 0.1m

W = 2π/t = 2π/60 = 0.105 rad/sec

a = w2R

= 0.105 x 0.105 x 0.1

= 0.0011

= 1.1 x 10-3 (order of the average acceleration)

Q.5  A hoop of radius r and mass m rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the center is zero. Find out the velocity of the center of the hoop when it cases to slip? [ 1 mark JEE MAIN 2016]

1. rw0
2. rw0/2
3. rw0/3
4. rw0/5

Solution:

Conservation of angular momentum at point of contact;

mr2w0 = mvr + mr2w

= mvr + mr2 (v/r) {v = rw}

= mvr + mvr

= 2mvr

mr2w0 = 2mvr

V = w0r/2 = rw0/2

Q.7 Distance of the center of mass of a solid uniform cone from its vertex is z0. If the radius of its base r and its height is h then z0 is equal to: [ 1 mark JEE MAIN 2016]

1. h2/4r
2. 3h/4
3. 5h/8
4. 3h2/8r

Solution:

Let the density of a solid cone p.

dm = p π r2 d y

ycm = ∫ ydm / ∫ dm

= h∫0 π r2 d y p x y / 1/3 πR2hp

= 3h/ 4

Q.8 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Momentum inertia of cube about an axis passing through its center and perpendicular to one of its face is: [ 1 mark JEE MAIN 2016] 

1. 4M R2/ 3√3 π
2. 4M R2/ 9√3 π
3. M R2/ 16√2 π
4. M R2/ 32√2 π

Solution:

Diagonal of the cube = Diameter of the sphere

√3a = 2R

a = 2R / √3 

Let,

M = mass of sphere

M1 = mass of cube

M/M1 = 4/3π R3 / a3

M1 = 2M /√3  π

Momentum of inertia of cube = M1a2/6

= 2M/√3 π x (2R / √3)/6

= 4M R2/ 9√3 π

Q.9 A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed w rad/sec. about the vertical. About the point of suspension: [ 1 mark JEE MAIN 2015]

1. Angular momentum is conserved.
2. Angular momentum changes in magnitude, but not in direction.
3. Angular momentum changes in direction & Magnitude
4. Angular Momentum changes in direction but not in magnitude.

Solution:

T = mg x l sin θ

T is perpendicular to the angular momentum L of the bob.

So, the direction of L changes and magnitude remains the same.

Q.10 A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass M and radius R. If the string doesn’t slip on the cylinder, with what acceleration will the mass fall? [ 1 mark JEE MAIN 2014]

1. g
2. g/2
3. 5g/6
4. 2g/3

Solution:

a = R α

mg – T = ma [Newton’s second law]

Torque on cylinder due to tension in string about the fixed point,

T x R = I α

T x R = mR2 α = mR2 (a/R)

Because, t= ma and I =MR2

mg – ma = ma

a = g/2

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JEE Main Rotational Motion- Quick Formulas for Revision

JEE Main aspirants are advised to revise formulas that is given before their examination. It’s important to remember formulas to solve questions quickly.

1. V avg = s/t
2. W avg = θ/ 2
3. Velocity = ds/dt
4. Angular velocity = dθ/dt
5. Average acceleration = v/t
6. Average angular acceleration = s/t
7. Instantaneous acceleration = dv/dt
8. Instantaneous angular acceleration = dw/dt
9.  V = V0 + a t
10.  X- x0 = v0 +1/2 a t2
11.  V2 = v2o + 2a (x – x0)