Rotation
Kinetics of Rotation 2023, 2020, 2016
Angular acceleration
𝜔𝑓 − 𝜔𝑖
𝛼=
Δt
# The angular acceleration of a body, moving along the circumference of a circle,
is: (2023)
1 Along the axis of rotation
2 Along the radius, away from centre
3 Along the radius towards the centre
4 Along the tangent to its position
Question (Angular Acceleration)
# The angular speed of a fly wheel moving with uniform angular acceleration
changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in
rad/s2 is: (2022)
1 104π
2 2π
3 4π
4 12π
Angle 2007
1
𝑄 = ω𝑡 + ∝ 𝑡2
2
# A wheel has angular acceleration of 3.0 rad/sec2 and an initial angular speed of
2.00 rad/sec. In a time of 2 sec it has rotated through an angle (in radian) of:
(2007)
1 10
2 12
3 4
4 6
Moment of Inertia 2015
IRing = MR2 2016
𝟐
Isolid sphere = 𝐌𝐑𝟐
𝟓
𝟐
IHollow Sphere = 𝐌𝐑𝟐
𝟑
# From a circular ring of mass ‘M’ and radius ‘R’ and 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 r to the plane of the ring is ‘K’ times
‘MR2’. Then the value of ‘K’ is: [MR*](2021)
7
1
8
1
2
4
1
3
8
3
4
4
Question (Sphere)
# Three identical spherical shells, each of mass m and radius r are placed as
shown in figure. Consider an axis XX’ which is touching to two shells and passing
through diameter to third shell. Moment of inertia of the system consisting of
these three spherical shells about XX’ axis is: (2015)
1 3 mr2
2 16/5 mr2
3 4 mr2
4 11/5 mr2
Question (Work energy theorem)
# A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass
has speed of 20 cm/s. How much work is needed to stop it? (2019)
1 3J
2 30 kJ
3 2J
4 1J
Radius of gyration 2023
2022
# The ratio of the radius of gyration of a thin uniform disc about an axis passing
through it centre and normal to its plane to the radius of gyration of the disc
about its diameter is: (2022)
1 1∶ 2
2 2∶1
3
2∶1
4 4∶1
M.O.I of Rod 2009
𝑴𝑳𝟐
𝑰 = 2008
𝟏𝟐
2024
# A thin rod of length L and mass M is bent at its midpoint into two halves so that
the angle between them is 90°. 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: [NR*](2008)
2𝑀𝐿2
1
24
𝑀𝐿2
2
24
𝑀𝐿2
3
12
𝑀𝐿2
4
6
M.O.I of Remaining Part 2016
Iremaining part = Icomplete part – Iremoved part
# 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?
(2016-I)
1 15 MR /32
2
2 13 MR2/32
3 11 MR2/32
4 9 MR2/32
M.O.I of Point mass 2004
𝑰 = 𝒎𝟏𝒓𝟐𝟏 + 𝒎𝟏𝒓𝟐𝟏
Question (M.O.I of Point Mass)
# Three particles, each of mass m gram, are situated at the vertices of an
equilateral triangle ABC of side l cm (as shown in the figure). The moment of the
system, about a line AX perpendicular to AB and in the plane of ABC, in gram cm2
units will be: (2004)
1 2𝑚𝑙2
5 2
2 𝑚𝑙
4
3 2
3 𝑚𝑙
2
3 2
4 𝑚𝑙
4
Perpendicular Theorem 2010
𝑰𝒐 = 𝑰𝒄𝒎 + 𝒎𝒅𝟐
# The moment of inertia of a uniform circular disc is maximum about an axis
perpendicular to the disc and passing through: (2012 Pre)
1 B
2 C
3 D
4 A
Energy 2016
𝟏 𝟐 = 𝟏𝑳𝟐 2022
𝑬 = 𝑰𝝎
𝟐 𝟐𝑰𝟐
# A solid sphere of mass m and radius R 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 (Esphere/Ecylinder) will be: (2016-II)
1 1:4
2 3:1
3 2:3
4 1:5
Work energy theorem
Work = 𝚫𝐊. 𝐄
𝟏
= 𝑰(𝝎𝒇𝟐 − 𝒘𝟐𝒊 )
𝟐
# An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to
360 rpm.
The moment of inertia of the flywheel is (2022 Re)
1 0.07 kg-m2
2 0.7 kg-m2
3 3.22 kg-m2
4 30.8 kg-m2
Question (Work Energy Theorem)
# Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular
ring), each have the same mass M and radius R. They all spin with the same
angular speed ω about their own symmetry axes. The amounts of work (W)
required to bring them to rest, would satisfy the relation (2018)
1 WB > WA > WC
2 WA > WB > WC
3 WC > WB > WA
4 WA > WC > WB
Question (Work energy theorem)
# A solid sphere is in rolling motion. In rolling motion a body possesses
translational kinetic energy (Kt) as well as rotational kinetic energy (Kr)
simultaneously. The ratio Kt: (Kt + Kr) for the sphere is: (2019)
1 10 :7
2 5:7
3 7 : 10
4 2:5
Feel of E.M.I. 2022
# A circular disc is to be made by using iron and aluminum so that it acquired
maximum moment of inertia about geometrical axis. It is possible with: (2002)
1 Aluminum at interior and iron surround to it
2 Iron at interior and aluminum 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.
Conservation of Mechanical energy 1999
(K.E + P.E)i = (K.E + P.E)f
# When a stick is released (as shown in fig.) Its free end velocity when it strikes
the ground is: [MR*](1999)
1 4.2 m/s
2 1.4 m/s
3 2.8 m/s
4 6𝑚/𝑠
Torque, (Rotational equilibrium) 2021, 2020, 2019,
2018, 2017
# Find the torque about the origin when a force of 3 𝑗Ƹ N acts on a particle whose
position vector is 2 𝑘 m. (2020)
1 2 𝑗Ƹ N m
2 –6 𝑖Ƹ N m
3 6 𝑘 N m
4 6 𝑖Ƹ N m
Question (Torque)
# The moment of the force, 𝐹ത = 𝑗𝑖Ƹ + 5𝑗Ƹ − 6𝑘 at (2, 0, 03), about the point (2, –2, –2)
is given by (2018)
1 −7𝑖Ƹ − 8𝑗Ƹ − 4𝑘
2 – 4𝑖Ƹ − 𝑗Ƹ − 8𝑘
3 – 8𝑖Ƹ − 4𝑗Ƹ − 7𝑘
4 −7𝑖Ƹ − 4𝑗Ƹ − 8𝑘
Question (Rotational Equilibrium)
# A uniform rod of length 200 cm and mass 500l g is balanced on a wedge placed at
40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another
unknown mass ‘m’ is suspended from the rod at 160 cm mark as shown in the
figure. Find the value of ‘m’ such that the rod is in equilibrium. (g = 10
m/s2) [MR*](2021)
1
1 𝑘𝑔
3
1
2 𝑘𝑔
6
1
3 𝑘𝑔
12
1
4 𝑘𝑔
2
Rotational Kinematics 2023, 2019, 2017,
Z = Iα 2015, 2014
𝚫𝛚
α=
𝚫𝐭
# A constant torque of 100 N m turns N m turns a wheel of moment of inertia 300
kg m2 about an axis passing through its centre. Starting from rest, its angular
velocity after 3s is: (2023-Manipur)
1 1 rad/s
2 5 rad/s
3 10 rad/s
4 15 rad/s
Question (Combined of Spring)
# 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 2π revolution is (2019)
1 2 × 10–6 N m
2 2 × 10–3 N m
3 12 × 10–4 N m
4 2 × 10–6 N m
Angular momentum 2016
L = mvr1r 2015 Re
# Two rotating bodies A and B of masses m and 2m with moment of inertia IA and
IB (IB > IA) have equal kinetic energy of rotation. If LA and LB be their angular
momenta respectively, then: (2016 - II)
1 𝐿𝐵 > 𝐿𝐴
2 𝐿𝐴 > 𝐿𝐵
𝐿𝐵
3 𝐿𝐴 =
2
4 𝐿𝐴 = 2𝐿𝐵
Conservation of Angular Momentum 2011
Li = Lf when torque = 0
# A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a
radius R0 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
𝑅0
moves in a circle of radius The final value of the kinetic energy is: (2015)
2
1 1
𝑚𝑣02
4
2 𝑢ሷ 02
1
3 𝑚𝑣02
2
4 𝑚𝑣02
Question (Conservation of Angular momentum)
# The instantaneous angular position of a point on a rotating wheel is given by the
equation θ(t) = 2t3 – 6t2. The torque on the wheel becomes zero at: (2011 Pre)
1 t=1s
2 t = 0.5 s
3 t = 0.25 s
4 t = 2s
Question (Conservation of Angular momentum)
# A circular disk of moment of inertia I prime is rotating in a horizontal plane,
about its symmetry axis, with a constant angular speed ωi. Another disk of moment
of inertia Ib is dropped coaxially into the rotating disk. Initially the second disk has
zero angular speed. Eventually both the disks rotate with a constant angular speed
ωf The energy lost by the initially rotating disc due to friction is: [MR*](2010)
1 𝐼𝑏2
1 2 𝐼𝑡+𝐼𝑏
𝜔𝑖2
1 𝐼𝑡2
2 𝜔𝑖2
2 𝐼𝑡+𝐼𝑏
𝐼𝑏 −𝐼𝑡
3 𝐼𝑡+𝐼𝑏
𝜔𝑖2
1 𝐼𝑏𝐼𝑡
4 𝜔𝑖2
2 𝐼𝑡+𝐼𝑏
Question (Conservation of Angular Momentum)
# Two discs of same moment of inertia rotating about their regular axis passing
through centre and perpendicular to the plane of disc with angular velocities ω1
and ω2. They are brought into contact face to face coinciding the axis of rotation.
The expression for loss of energy during this process is: [MR*](2017-Delhi)
1 1
𝐼 𝜔1 − 𝜔2 2
4
2 𝐼(𝜔1 – 𝜔2)2
1 2
3 𝐼 𝜔1 − 𝜔2
8
1 2
4 2
𝐼 𝜔1 − 𝜔2
Ring/Disc → Conservation of Angular Momentum 2010, 2011
𝑰𝟏𝒘𝟏 + 𝑰𝟐𝒘𝟐
𝑾𝒇 = 2009, 2004
𝒘𝟏 + 𝒘𝟐
𝟏 𝑰 𝟏𝑰 𝟐 2003, 1998
𝚫𝑲. 𝑬 = 𝒘𝟏 − 𝒘𝟐
𝟐 𝑰 𝟏 + 𝑰𝟐
Question (Ring Disc Conserved Angular Momentum)
# A thin circular ring of mass M and radius r is rotating about its axis with constant
angular velocity w. The objects each of mass m are attached gently to the opposite
ends of a diameter - of the ring. The ring now rotates with angular velocity given
by: (2010 mains)
𝑀+2𝑚 𝜔
1
2𝑚
2𝑀𝜔
2
𝑀 + 2𝑚
𝑀 + 2𝑚 𝜔
3
𝑀
𝑀𝜔
4
𝑀 + 2𝑚
Rolling Motion
2024, 2001
V𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭 𝐏𝐨𝐢𝐧𝐭
2 Vcm
Vcm
0
Question (Rolling Motion velocity)
# A disc is rolling the velocity of its centre of mass is vcm then which one will be
correct: (2001)
1 The velocity of highest point is 2 vcm 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 2vcm and point of contact is 2vcm
Work energy theorem in Rolling 2019
𝟏 𝑲 𝟐
𝑾 = 𝚫𝑲. 𝑬 = 𝒎𝒖𝟐𝒂 𝟏 + 𝟐 2012
𝟐 𝑹
Acceleration of Rolling 2014
𝒈𝒔𝒊𝒏𝜽 2018
𝒂𝒄𝒎 =
𝑲𝟐
𝟏+ 𝟐
𝒓
Question
# The ratio of the accelerations for a solid sphere (mass m and radius R) rolling
down an incline of angle ‘’ without slipping and slipping down the incline without
rolling is: (2014)
1 5:7
2 2:3
3 2:5
4 7:5
Question
# Small object of uniform density rolls up a curved surface with an initial velocity v.
3𝑣 2
It reaches up to a maximum height of with respect to the initial position. The
4𝑔
object is: (2013)
1 Disc
2 Ring
3 Solid sphere
4 Hollow sphere
