A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity . Its kinetic energy is
1
2
mr
3
4
Official Solution
Correct Option:
(1)
Kinetic energy , and for ring Hence
02
PYQ 1988
hard
physicsID: neet-ug-
A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are
1
up the incline while ascending and down the incline while descending
2
up the incline while ascending as well as descending
3
down the incline while ascending and up the incline while descending
4
down the incline while ascending as well as descending
Official Solution
Correct Option:
(2)
mg component is always down the plane whether it is rolling up or rolling down.Therefore, for no slipping, sense of angular acceleration should also be same in both the cases. Therefore, force of friction f always act upwards.
03
PYQ 1988
medium
physicsID: neet-ug-
Let be the moment of inertia of a uniform square plate about an axis that passes through its centre and is parallel to two of its sides. is a line in the plane of the plate that passes through the centre of the plate and makes an angle with . The moment of inertia of the plate about the axis is then equal to
1
2
3
4
Official Solution
Correct Option:
(1)
In the figure, is perpendicular to is perpendicular to By symmetry,
By theorem of perpendicular axes,
Again From and , or or Hence moment of inertia about axis
04
PYQ 1989
medium
physicsID: neet-ug-
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of with respect to the initial position. The object is
1
ring
2
solid sphere
3
hollow sphere
4
disc
Official Solution
Correct Option:
(4)
Body is disc. The correct option is (d).
05
PYQ 1989
medium
physicsID: neet-ug-
A solid cylinder of mass and radius rolls without slipping down an inclined plane of length and height . What is the speed of its centre of mass when the cylinder reaches its bottom -
1
2
3
4
Official Solution
Correct Option:
(3)
06
PYQ 1990
easy
physicsID: neet-ug-
A flywheel rotating about a fixed axis has a kinetic energy of 360 J when its angular speed is 30 rads-1. The moment of inertia of the wheel about the axis of rotation is
1
0.6 kgm2
2
0.15 kgm2
3
0.8 kgm2
4
0.75 kgm2
Official Solution
Correct Option:
(3)
Flywheel rotating about a fixed axis works on the principle of conservation of angular momentum, to store rotational energy.
The kinetic energy of the flywheel = 360 J, angular speed, ω = 30 rad/s
The relation between kinetic energy, angular speed, and moment of inertiais as follows.
K.E. = ½ Iω2
Where I = moment of inertia
ω = angular speed
Putting the values in the formula, we get
360 = ½ I(30)2
Simplifying the expression.
So, the moment of inertia of the wheel is 0.8 kgm2 when its kinetic energy is 360 Joules.
Hence, the correct option is (C) is 0.8 kgm2.
07
PYQ 1991
medium
physicsID: neet-ug-
A particle of mass m = 5 is moving with a uniform speed v = 3 in the XOY plane along the line Y = X + 4. The magnitude of the angular momentum of the particle about the origin is
1
60 units
2
40 units
3
zero
4
7.5 units
Official Solution
Correct Option:
(1)
Y + X + 4 line has been shown in the figure.
Y=4, So OP=4,
The slope of the line can be obtained by comparing with the equation of line y = mx + c m = tan
If we draw a line perpendicular to this line. Length of the perpendicular = OR
Angular momentum of particle going along this line units
08
PYQ 1992
medium
physicsID: neet-ug-
The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height h form rest without sliding is
1
2
3
4
Official Solution
Correct Option:
(1)
P.E. = total K.E ,
09
PYQ 1994
medium
physicsID: neet-ug-
A solid spherical ball rolls on a table. Ratio of its rotational kinetic energy to total kinetic energy is
1
2
3
4
Official Solution
Correct Option:
(4)
Linear K.E. of ball and rotational K.E. of ball Therefore total K.E. = . And ratio of rotational K.E. and total K.E. .
10
PYQ 1996
medium
physicsID: neet-ug-
The formation of canal system in sponges is due to
1
folding of inner walls
2
gastro-vascular system
3
reproduction
4
porous walls.
Official Solution
Correct Option:
(4)
The body wall of sponges encloses a large cavity, the spongocoel and in most cases also contains in its thickness numerous small cavities, the canals. Either the spongocoel or certain canals are lined by choanocytes with flagella. The ceaseless beating of flagella maintains a steady current of water through the canals in the sponge body. The current of water enters through small pores, the dermal ostia, perforating the porocytes and, after passing through various canals, enters the spongocoel, and finally leaves through a larger aperture, the osculum, or apertures, the oscula. All the cavities in a sponge body are inter- communicating, and are collectively referred to as the canal system. The current of water that flows through the canal system brings in food and oxygen, and carries away carbon dioxide and nitrogenous waste materials. Thus, the canal system helps the sponge in nutrition, respiration and excretion.
11
PYQ 1997
easy
physicsID: neet-ug-
A couple produces
1
purely translational motion
2
purely rotational motion
3
both translational and rotational motion
4
no motion
Official Solution
Correct Option:
(2)
A couple defined more than two forces operating on a body in equal and opposing directions with an interval between their lines of action.
The pair is exerting no force as a consequence. As a result, the item does not travel linearly.
As a result, the coupling just causes the item to rotate.
Discover More Topics From This Chapter: System Of Particles & Rotational Motion
12
PYQ 1999
easy
physicsID: neet-ug-
A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
1
2g/3
2
g/2
3
5g/6
4
g
Official Solution
Correct Option:
(2)
For the mass m, As we know, So, .....(i) Torque about centre of pully .....(ii) From Eqs. (i) and (ii), we get, Hence, the acceleration with the mass of a body fall is g/2.
13
PYQ 1999
medium
physicsID: neet-ug-
A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let , and be the magnitude of velocities of points P,Q and C respectively, then
1
2
3
4
Official Solution
Correct Option:
(1)
In case of pure rolling bottom most point is the instantaneous centre of zero velocity. Velocity of any point on the disc, where r is the distance of point from O.
Therefore, the correct option is (a).
14
PYQ 2000
medium
physicsID: neet-ug-
For a hollow cylinder and a solid cylinder rolling without slipping on an inclined plane, then which of these reaches earlier
1
solid cylinder
2
hollow cylinder
3
both simultaneously
4
can't say anything
Official Solution
Correct Option:
(1)
Solid sphere reaches the bottom first because for solid cylinder and for hollow cylinder . Acceleration down the inclined plane . Solid cylinder has greater acceleration, so it reaches the bottom first.
15
PYQ 2000
easy
physicsID: neet-ug-
For the adjoining diagram, a triangular lamina is shown the correct relation between , & is: ( – moment of inertia)
1
2
3
4
Official Solution
Correct Option:
(2)
For triangular lamina Longest side Smallest side Therefore
So, the correct option is (B):
16
PYQ 2000
easy
physicsID: neet-ug-
A mass is performing vertical circular motion (see figure). The average velocity of the particle is increased, then at which point the string will break :
1
A
2
B
3
C
4
D
Official Solution
Correct Option:
(2)
When a sphere is rotating in a vertical circle, it exerts the maximum outward pull when it is at the lowest point B. Therefore, tension at B is maximum = Weight So, the string breaks at point B.
17
PYQ 2001
hard
physicsID: neet-ug-
A disc is rolling the velocity of its centre of mass is Vcm then which one will be correct
1
The velocity of highest point is 2Vcm and point of contact is zero
2
The velocity of highest point is Vcm and point of contact is Vcm
3
The velocity of highest point is 2Vcm and point of contact is Vcm
4
The velocity of highest point is 2 Vcm and point of contact of contact is 2Vcm
Official Solution
Correct Option:
(1)
The correct option is(A): The velocity of highest point is 2Vcm and point of contact is zero.
18
PYQ 2001
medium
physicsID: neet-ug-
Two particles having mass M and m are moving in a circular path having radius R and r. If their time period are same then the ratio of angular velocity will be
1
2
3
1
4
Official Solution
Correct Option:
(3)
Let radius for the particle of mass M = OB = r and for the particle of mass m = OC = R Let linear velocity for a particle of mass M = v1 and for the particle of mass m = v2 Let angular velocity for the particle having mass M = and for the particle having mass m= Let Time period for the particle having mass M = T1 and for the particle having mass m = T2
Given: T1 = T2
The above equation generated is the formula for angular velocity. hence:
Therefore, the ratio of the angular velocity will be 1:1. Therefore, the correct option is (C) : 1.
19
PYQ 2002
medium
physicsID: neet-ug-
A disc is rotating with angular speed . If a child sits on it, what is conserved :-
1
linear momentum
2
angular momentum
3
kinetic energy
4
potential energy
Official Solution
Correct Option:
(2)
When a child sits on a rotating disc, no external torque is introduced. Hence the angular momentum of the system is conserved. But the moment of inertia of the system will increase and as a result, the angular speed of the disc will decrease to maintain constant angular momentum. angular momentum = moment of inertia angular velocity]
20
PYQ 2002
medium
physicsID: neet-ug-
A circular disc is to be made by using iron and aluminium so that it acquired maximum moment of inertia about geometrical axis. It is possible with :-
1
aluminium at interior and iron surround to it
2
iron at interior and aluminium surround to it
3
using iron and aluminium layers in alternate order
4
sheet of iron is used at both external surface and aluminium sheet as internal layers
Official Solution
Correct Option:
(1)
A circular disc may be divided into a large number of circular rings. Moment of inertia of the disc will be the summation of the moments of inertia of these rings about the geometrical axis. Now, moment of inertia of a circular ring about its geometrical axis is , where M is the mass and R is the radius of the ring. Since the density (mass per unit volume) for iron is more than that of aluminium, the proposed rings made of iron should be placed at a higher radius to get more value of . Hence to get maximum moment of inertia for the circular disc, aluminium should be placed at interior and iron at the outside.
21
PYQ 2002
medium
physicsID: neet-ug-
A rod of length is 3 m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at :-
1
1.5 m
2
2 m
3
2.5 m
4
3.0 m
Official Solution
Correct Option:
(2)
Here where is a constant mass of small element of length is
22
PYQ 2002
medium
physicsID: neet-ug-
A solid sphere of radius R is placed on smooth horizontal surface. A horizontal force F is applied at height h from the lowest point. For the maximum acceleration of centre of mass, which is correct?
1
h = R
2
h = 2R
3
h = 0
4
no relation between h and R
Official Solution
Correct Option:
(4)
Since there is no friction at the contact surface (smooth horizontal surface) there will be no rolling. Hence, the acceleration of the centre of mass of the sphere will be independent of the position of the applied force F. Therefore, there is no relation between h and R.
23
PYQ 2003
easy
physicsID: neet-ug-
A thin circular ring M and radius ‘r’ is rotating about its axis with a constant angular velocity ω. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be -
1
2
3
4
Official Solution
Correct Option:
(2)
Use law of conservation of angular momentum.
So, the correct option is (B):
24
PYQ 2003
easy
physicsID: neet-ug-
A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom-
1
2
3
4
Official Solution
Correct Option:
(3)
25
PYQ 2003
medium
physicsID: neet-ug-
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be:
1
2
3
4
Official Solution
Correct Option:
(3)
Moment of Inertia We know that,
Translational Kinetic Energy =
Rotational Kinetic Energy = =
Total Energy = Translational Kinetic Energy + Rotational Kinetic Energy
Total Energy = +
Total Energy =
Required fraction =
=
=
=
So, the correct option is (C):
26
PYQ 2004
medium
physicsID: neet-ug-
A wheel having moment of inertia 2 kg–m2 about its vertical axis, rotates at the rate of 60 rpm about the axis. The torque which can stop the wheel's rotation in one minute would be
1
N-m
2
N-m
3
N-m
4
N-m
Official Solution
Correct Option:
(2)
The torque required to stop the wheel's rotation in one minute is approximately N-m
27
PYQ 2004
medium
physicsID: neet-ug-
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is
1
2 : 1
2
5 : 6
3
2 : 3
4
1 : 2
Official Solution
Correct Option:
(2)
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is 2 : 1.
28
PYQ 2004
medium
physicsID: neet-ug-
Consider a system of two particles having masses and . If the particle of mass is pushed towards the mass centre of particles through a distance ' ', by what distance would be particle of mass move so as to keep the mass centre of particles at the original position ?
1
2
3
d
4
Official Solution
Correct Option:
(2)
from (1) and (2) we get
29
PYQ 2004
medium
physicsID: neet-ug-
A round disc of moment of inertia about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia rotating with an angular velocity about the same axis. The final angular velocity of the combination of discs is
1
2
3
4
Official Solution
Correct Option:
(3)
Applying conservation of angular momentum.
.
30
PYQ 2005
medium
physicsID: neet-ug-
A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle .The frictional force
1
converts translational energy to rotational energy
2
dissipates energy as heat
3
decreases the rotational motion
4
decreases the rotational and translational motion
Official Solution
Correct Option:
(1)
We have, mg sinA−f=ma ________________________ (1) in pure rolling = = angular acceleration Now,τ=I I = moment of inertia f r = I f r = I f =Iar2 ________________________________ (2) Substituting (2)in (1)
we have mgsinA−I =ma
here I of solid cylinder=
mgsinA− =ma
a= Now from (2) we have f = f = which should be less than or equal to μN otherwise body will slip (where u is coefficient of friction and N is normal acting on cylinder which is equal to mg cos A)
mgcosA
putting value of a
mgcosA tanA 3μ
Therefore, the correct option is (A): converts translational energy to rotational energy
31
PYQ 2005
medium
physicsID: neet-ug-
The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:
1
2
3
4
Official Solution
Correct Option:
(4)
Moment of inertia of the disc around an axis perpendicular to its plane and passing through its centre of mass ICM = MR2 The disc's moment of inertia about an axis perpendicular to its plane and running across its diameter may be calculated using the parallel axis theorem by using the formula I = ICM + Md2.
Therefore, I = MR2+MR2 = .
Correct option is (D):
32
PYQ 2005
medium
physicsID: neet-ug-
Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio:
1
1:4
2
:1
3
4:1
4
1:
Official Solution
Correct Option:
(4)
KE= I 2 =
KE1 = KE2
=
=
=
1:
Therefore, the correct option is (D): 1:
33
PYQ 2005
medium
physicsID: neet-ug-
A stone tied to the end of a string of 1m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44s, what is the magnitude and direction of acceleration of the stone?
1
ms-2 and direction along the radius towards the centre
2
2 ms-2 and direction along the radius away from centre
3
ms-2 and direction along the radius towards the centre
4
4 ms-2 and direction along the tangent to the circle
Official Solution
Correct Option:
(3)
Given, String length is 1 m. In 44 seconds, a stone makes 22 rotations. Stone thus completes revolutions in 1 second, making frequency = sec 1 = Hz. We are aware that angular speed = 2 frequency w = 2 = rad/sec. Now, acceleration equals a=2r, where r is the radius or string length. Acceleration = m/s2
Therefore, the correct option is (C): ms-2 and direction along the radius towards the centre
34
PYQ 2006
medium
physicsID: neet-ug-
The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is :
1
MR2
2
MR2
3
MR2
4
MR2
Official Solution
Correct Option:
(3)
The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is = where M is the mass of disc and R its radius. According to theorem of parallel axis, moment of inertia of circular disc about an axis touching the disc at its diameter and normal to the disc is = +
= +
=
Therefore, the correct option is (C):
35
PYQ 2007
medium
physicsID: neet-ug-
A wheel has angular acceleration of and an initial angular speed of . In a time of it has rotated through an angle (in radian) of
1
10
2
12
3
4
4
6
Official Solution
Correct Option:
(1)
36
PYQ 2007
medium
physicsID: neet-ug-
A uniform rod of length and mass is free to rotate about point The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about is , the initial angular acceleration of the rod will be:-
1
2
3
4
Official Solution
Correct Option:
(3)
37
PYQ 2008
medium
physicsID: neet-ug-
A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is . The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is
1
2
3
4
Official Solution
Correct Option:
(4)
Total mass = M, total length = L Moment of inertia of OA about O = Moment of inertia of OB about O. total
38
PYQ 2009
medium
physicsID: neet-ug-
If is the force acting on a particle having position vector and be the torque of this force about the origin, then
1
and
2
and
3
and
4
and
Official Solution
Correct Option:
(2)
is perpendicular to and .
39
PYQ 2009
medium
physicsID: neet-ug-
Four identical thin rods each of mass M and length , form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is
1
2
3
4
Official Solution
Correct Option:
(4)
the parallel axis theorem
40
PYQ 2010
medium
physicsID: neet-ug-
(1) Centre of gravity of a body is the point at which the weight of the body acts. (2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius. (3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its centre of gravity. (4) The radius of gyration of any body rotating about an axis is the length of the perpendicular drawn from the centre of gravity of the body to the axis. Which one of the following pairs of statements is correct?
1
(1) and (4)
2
(1) and (2)
3
(2) and (3)
4
(3) and (4)
Official Solution
Correct Option:
(1)
Statements(1) and (4) are true.
41
PYQ 2010
easy
physicsID: neet-ug-
A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
1
Both together only when angle of inclination of plane is
2
Both together
3
Hollow cylinder
4
Solid cylinder
Official Solution
Correct Option:
(4)
Time taken to reach the bottom of inclined plane.
Here, l is length of incline plane For solid cylinder For hollow cylinder Hence, solid cylinder will reach the bottom first. `
42
PYQ 2010
medium
physicsID: neet-ug-
A circular disk of moment of inertia it is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed . Another disk of moment of inertia is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed . The energy lost by the initially rotating disc to friction is
1
2
3
4
Official Solution
Correct Option:
(4)
As no external torque is applied to the system, the angular momentum of the system remains conserved.
According to given problem,
or Initial energy, Final energy, Substituting the value of from equation (i) in equation (iii), we get Final energy,
Loss of energy, (Using (ii) and (iv))
43
PYQ 2011
medium
physicsID: neet-ug-
A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of , the kinetic energy of the mass will
1
decrease by a factor of 2
2
remain constant
3
increase by a factor of 2
4
increase by a factor of 4
Official Solution
Correct Option:
(4)
From angular momentum conservation about centre. constant
K.E. is increased by a factor of
44
PYQ 2012
easy
physicsID: neet-ug-
A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is . It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of relative to the ground. Time taken by the man to complete one revolution is
1
2
3
4
Official Solution
Correct Option:
(3)
As the system is initially at rest, therefore, initial angular momentum According to the principle of conservation of angular momentum, final angular momentum, Angular momentum = Angular momentum of man is in opposite direction of platform. i.e or Angular velocity of man relative to platform is
Time taken by the man to complete one revolution is
45
PYQ 2012
medium
physicsID: neet-ug-
is an equilateral triangle with as its centre. and represent three forces acting along the sides and respectively. If the total torque about O is zero then the magnitude of is
1
2
3
4
Official Solution
Correct Option:
(1)
The correct option is(A): .
Taking anticlockwise torque
(where perpendicular distance of the centre from each side)
46
PYQ 2012
medium
physicsID: neet-ug-
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
1
a line perpendicular to the plane of rotation
2
the line making an angle of to the plane of rotation
3
the radius
4
the tangent to the orbit
Official Solution
Correct Option:
(1)
When a mass is rotating in a plane about a fixed point its angular momentum is directed along a line perpendicular to the plane of rotation.
47
PYQ 2012
easy
physicsID: neet-ug-
A solid cylinder of mass is rolling on a horizontal surface with velocity . It collides with a horizontal spring of force constant . The maximum compression produced in the spring will be
1
0.5 m
2
0.6 m
3
0.7 m
4
0.2 m
Official Solution
Correct Option:
(2)
Putting (For solid cylinder), we get
48
PYQ 2013
hard
physicsID: neet-ug-
A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is:
1
2
3
4
Official Solution
Correct Option:
(2)
So, the correct option is (B):
49
PYQ 2014
medium
physicsID: neet-ug-
A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis . A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions is
1
25 N
2
50 N
3
78.5 N
4
157 N
Official Solution
Correct Option:
(4)
Here, mass of the cylinder, M = 50 kg Radius of the cylinder, R = 0.5 m Angular acceleration,
Torque, Moment of inertia of the solid cylinder about its axis, Angular acceleration of the cylinder
50
PYQ 2015
medium
physicsID: neet-ug-
A mass m moves in a circle on a smooth horizontal plane with velocity at a radius The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius The final value of the kinetic energy is
1
2
3
4
Official Solution
Correct Option:
(1)
The correct answer is A: According to law of conservation of angular momentum
or (Using (i))
51
PYQ 2015
medium
physicsID: neet-ug-
A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance a from A. The normal reaction on A is
1
2
3
4
Official Solution
Correct Option:
(2)
Given situation is shown in figure.
Normal reaction on A Normal reaction on B Weight of the rod In vertical equilibrium,
Torque balance about centre of mass of the rod,
Putting value of from equation (i)
52
PYQ 2016
medium
physicsID: neet-ug-
A solid sphere of mass and radius is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation Will be
1
2
3
4
Official Solution
Correct Option:
(2)
Energy and Moment of Inertia Calculation
The relationship between the kinetic energy ( ), moment of inertia ( ), and angular velocity ( ) is given by:
Now, considering the ratio of the kinetic energies of a sphere and a cylinder, we can write:
Substitute the moments of inertia for a solid sphere and a solid cylinder. The moment of inertia of a solid sphere is:
And the moment of inertia of a solid cylinder is:
Now substitute these into the energy ratio equation:
Simplifying the equation, we get:
Now, simplify the fractions and powers:
Finally, multiplying the terms gives:
Thus, the ratio of the kinetic energies is:
This means that the kinetic energy of the sphere is th of that of the cylinder under the given conditions.
53
PYQ 2016
medium
physicsID: neet-ug-
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?
1
2
3
4
Official Solution
Correct Option:
(1)
Moment of Inertia Calculation for a Disc
The moment of inertia of a full disc is given by:
When a portion of the disc is removed, the mass is reduced. The mass removed is proportional to the area removed. Since the area of the removed portion is one-quarter of the total area, the mass removed is one-quarter of the total mass:
The moment of inertia of the removed portion can be calculated considering the perpendicular axis theorem. The formula for the moment of inertia of the removed portion is the sum of two parts:
Simplifying, we get:
The moment of inertia of the remaining disc is the total moment of inertia minus the moment of inertia of the removed portion:
Substituting the values:
Conclusion:
The moment of inertia of the remaining disc after the removal of a portion is .
54
PYQ 2016
easy
physicsID: neet-ug-
A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of . Its net acceleration in at the end of 2.0 s is approximately :
1
7
2
6
3
3
4
8
Official Solution
Correct Option:
(4)
Particle Acceleration Calculation
A particle at the periphery of a rotating object will have both radial and tangential acceleration. Let's calculate both accelerations step by step.
1. Tangential Acceleration:
The tangential acceleration is given by:
2. Angular Velocity:
The angular velocity is calculated using the formula:
Substitute the values:
3. Centripetal (Radial) Acceleration:
The centripetal acceleration is given by:
4. Total Acceleration:
The total acceleration is the vector sum of tangential and centripetal accelerations, calculated as:
Conclusion:
The total acceleration of the particle is approximately .
55
PYQ 2016
medium
physicsID: neet-ug-
A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?
1
Sphere
2
Both reach at the same time
3
Depends on their masses
4
Disk
Official Solution
Correct Option:
(1)
Comparison of Acceleration for Disc and Sphere
The formula for acceleration ( ) of an object rolling down an incline is given by:
Where: - is the acceleration due to gravity, - is the angle of the incline, - is the radius of gyration of the object, - is the radius of the object.
For a Disc
For a disc, the value of is:
For a Sphere
For a sphere, the value of is:
Comparison of Acceleration
From the formula for acceleration, we see that the value of for the sphere is smaller than for the disc. Therefore, the acceleration of the sphere is greater than the acceleration of the disc:
Conclusion:
Since the acceleration of the sphere is greater than the acceleration of the disc, the sphere reaches the bottom of the incline first:
56
PYQ 2016
medium
physicsID: neet-ug-
A light rod of length has two masses and attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is -
1
2
3
4
Official Solution
Correct Option:
(1)
Moment of Inertia Calculation
We are given a system with two masses, and , and their respective distances from a pivot point. We need to calculate the total moment of inertia of the system.
Step 1: Relating the Distances
The problem starts with the equation that relates the two distances and for the two masses:
Where: - is the distance of mass , - is the distance of mass , - is the total distance between the two masses.
By rearranging the equation, we get:
Solving for , we get:
Step 2: Calculating
We can find by subtracting from the total distance :
Step 3: Total Moment of Inertia
The total moment of inertia for the system is the sum of the individual moments of inertia of the two masses:
Where: - , - .
Now, substituting the expressions for and into the formula for the total moment of inertia:
Substitute the values for and :
Now, simplify the expression:
Conclusion:
The total moment of inertia of the system is:
57
PYQ 2016
medium
physicsID: neet-ug-
Two rotating bodies and of masses and with momenta of inertia and have equal kinetic energy of rotation. If and be their angular momenta respectively, then -
1
2
3
4
Official Solution
Correct Option:
(3)
Rotational Motion Explanation
We are given two rotating bodies, A and B, and we are to compare their angular velocities and angular momenta. Let's go step by step through the calculations:
Step 1: Equating the Kinetic Energy
The kinetic energy of a rotating body is given by the formula:
For bodies A and B, the kinetic energy is equal, so we have:
This leads to:
Canceling out the common factors of , we get:
Rearranging for the ratio of angular velocities:
.....(\)
Step 2: Relating Angular Momentum
The angular momentum of a rotating body is given by:
Thus, for bodies A and B, the angular momenta are:
Step 3: Finding the Ratio of Angular Momentums
The ratio of angular momenta is:
Substitute equation (i) for :
Simplify the expression:
Step 4: Conclusion
Since (because ), we conclude:
Therefore, the angular momentum of body A is less than that of body B:
Conclusion:
58
PYQ 2017
medium
physicsID: neet-ug-
A rope of negligible mass is wound round a hollow cylinder of mass and radius . If the rope is pulled with a force of , then the angular acceleration produced in the cylinder is
1
2
3
4
Official Solution
Correct Option:
(3)
Here,
Force applied, Torque, Moment of inertia of hollow cylinder about its axis is
Let a is the angular acceleration produced. As
59
PYQ 2018
medium
physicsID: neet-ug-
A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy as well as rotational kinetic energy , simultaneously. The ratio for the sphere is
1
2 : 5
2
7 : 10
3
10 : 7
4
5 : 7
Official Solution
Correct Option:
(4)
So,
60
PYQ 2018
medium
physicsID: neet-ug-
A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere ?
1
Angular momentum
2
Angular velocity
3
Rotational kinetic energy
4
Moment of inertia
Official Solution
Correct Option:
(1)
So, i.e. constant So angular momentum remains constant.
61
PYQ 2019
easy
physicsID: neet-ug-
A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination 30°. The centre of mass of cylinder has speed of 4 m/s. The distance travelled by the cylinder on the incline surface will be : (Take g = 10 m/s )
1
2.2 m
2
1.6 m
3
1.2 m
4
2.4 m
Official Solution
Correct Option:
(4)
So, the correct option is (D): 2.4 m
62
PYQ 2019
easy
physicsID: neet-ug-
A hoop of radius weighs . It rolls along a horizontal floor so that its centre of mass has a speed of How much work has to be done to stop it?
Official Solution
Correct Option:
(1)
63
PYQ 2019
easy
physicsID: neet-ug-
A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2 revolutions is:
1
2
3
4
Official Solution
Correct Option:
(3)
Work energy theorem .
64
PYQ 2021
medium
physicsID: neet-ug-
From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR2 ’. Then the value of ‘K’ is
1
2
3
4
Official Solution
Correct Option:
(2)
The problem requires us to find the moment of inertia of the remaining part of a circular ring after cutting out an arc corresponding to a 90° sector.
Step 1: Understand the distribution of mass.
The full circle is initially a ring with a uniform mass distribution of 'M' and radius 'R'. The moment of inertia of a complete ring about an axis passing through its center and perpendicular to its plane is given by:
A 90° sector corresponds to of the circle. Thus, the mass of the removed sector is:
The moment of inertia of the removed part (the 90° sector) about the same axis, considering it has the same radius 'R', is:
Step 2: Calculate the moment of inertia of the remaining part of the ring.
Since the remaining mass is , and it's distributed in the remaining 270° of the ring, we need to find the adjusted moment of inertia:
Step 3: Conclusion.
The moment of inertia of the remaining part is . Comparing with the given expression , we find that the value of is:
Thus, the correct answer is .
65
PYQ 2022
easy
physicsID: neet-ug-
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is
1
2:1
2
3
4:1
4
Official Solution
Correct Option:
(2)
The problem involves understanding the concept of the radius of gyration for different axes of a uniform disc. The radius of gyration is defined as , where is the moment of inertia and is the mass of the object.
For a thin uniform disc of radius , the moment of inertia about an axis through its center and perpendicular to its plane is . Hence, the radius of gyration about this axis is:
The moment of inertia about the diameter of the disc is . Therefore, the radius of gyration about the diameter is:
Thus, the ratio of the radius of gyration about the center perpendicular axis to that about the diameter is given by:
The correct ratio is therefore .
66
PYQ 2023
hard
physicsID: neet-ug-
The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is:
1
5:2
2
3:5
3
5:3
4
2:5
Official Solution
Correct Option:
(2)
To solve this problem, we need to find the ratio of the radius of gyration of a solid sphere to that of a thin hollow sphere, both having the same mass and radius .
1. Radius of gyration: The radius of gyration of an object is defined as the distance from the axis of rotation at which the entire mass of the body can be assumed to be concentrated to provide a moment of inertia . Mathematically, .
2. Solid sphere: For a solid sphere, the moment of inertia about its axis is . Thus, the radius of gyration is:
.
3. Thin hollow sphere: For a thin hollow sphere, the moment of inertia about its axis is . Thus, the radius of gyration is:
.
4. Ratio: The ratio of the radii of gyration of the solid sphere to the hollow sphere is given by:
.
Simplifying, we multiply both numerator and denominator by to get:
.
The ratio is therefore 3:5, which matches the given correct option.
About System Of Particles Rotational Motion - NEET-UG
System Of Particles Rotational Motion is a vital chapter for NEET-UG aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
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