Motion of System of Particles and Rigid Bodies (Physics)
Choose the correct answers
1. The center of mass of a system of particles does not depend upon,
(a) position of particles
(b) relative distance between particles
(c) masses of particles
(d) force acting on particle
Answer : d) force acting on
particle
2. A couple produces,
(a) pure rotation
(b) pure translation
(c) rotation and translation
(d) no motion
Answer : a) pure rotation
3. A particle is moving with a constant velocity along a line parallel to positive X-axis. The magnitude of its angular momentum with respect to the origin is,
(a) zero
(b) increasing with x
(c) decreasing with x
(d) remaining constant
Answer : d) remaining
constant
4. A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s–2
(b) 25 rad s–2
(c) 5 m s–2
(d) 25 m s–2.
Answer : b) 25 rad s-2
Solution:
τ = Iα ;
F × R = MR2α
30 × 0.4 = 3(0.4)2α
12 = 3 × 0.16α
400 =16 α
α = 25rad s-2
5. A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia,
(a) increases
(b) decreases
(c) remains constant
(d) depends on direction of rotation.
Answer: a) increases
6. A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes,
(a) L
(b) L/2
(c) 2L
(d) L/ 2
Answer: d) L/√2
Solution:
L2 = 2I(KE) ;
KE = Half
L2 = 2I (KE / 2)
L2 = L2 / 2
L2 = 2I × [ L2 / 2I ] × [ 1/2 ]
L = √[L2 / 2]
L = L / √2
7. A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about,
(a) the center point of the circle.
(b) the point on the circumference of the circle.
(c) any point inside the circle.
(d) any point outside the circle.
Answer : a) the center
point of the circle
8. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along,
(a) a line perpendicular to the plane of rotation
(b) the line making an angle of 45° to the plane of rotation
(c) the radius
(d) tangent to the path
Answer : a) a line
perpendicular to the plane of rotation
9. Two discs of same moment of inertia rotating about their regular axis passing through center and perpendicular to the plane of disc with angular velocities ω1 and ω2. They are brought in to contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is,
(a) 1/4 I (ω1- ω2)2
(b) I (ω1- ω2)2
(c) 1/8 I (ω1- ω2)2
(d) 1/2 I (ω1- ω2)2
Answer: a) 1/4 I (ω1 – ω2)2
Solution:
Δ KE = 1/2 ([I1 I2]
/ [I1 + I2]) (ω1 – ω2)2
= 1/2 (I 2 / (2I))
(ω1 – ω2)2
= 1/4 I (ω1 – ω2)2
10. A disc of moment of inertia Ia is rotating in a horizontal plane about its symmetry axis with a constant angular speed ω. Another discinitially at rest of moment of inertia Ib is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed. The loss of kinetic energy due to friction in this process is,
Answer: d) 1/2 { [IbIb] / (Ia +
Ib) } ω2
Solution:
Li = Lf
Iaωa = Ifωf
Iaωa = (Ia+Ib)ωf
11. The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle θ without slipping and slipping down the incline without rolling is,
(a) 5:7
(b)2:3
(c) 2:5
(d) 7:5
Answer: a) 5:7
Solution:
a1 = [ g sin θ ] / [ 1 + K2/r2
]
a2 = g sin θ
a1 / a2 = 1 / { 1 + K2/r2
} = 1 / { 1 + 2/5} = 5 / 7 = 5 : 7
12. From a disc of radius R a mass M, a circular hole of diameter R, whose rim passes through the center is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through it
(a) 15MR2/32
(b) 13MR2/32
(c) 11MR2/32
(d) 9MR2/32
Answer: b) 13 MR2 / 32
Solution:
M.I of total disc about perpendicular axis
Ir = 1/2 MR2
Mass of the removed portion
M1 = [ M / 4πR2 ] × 4π(R/2)2 =
M / 4
Moment of inertia of a removed disc
IR = [ 1/2 M1(R/2)2
] + [ M1(R/2)2 ]
= [ 1/2 M/4 R2/4 ] + [ MR2 / 16 ] = 3/32
MR2
M.I of the remaining portion
IRemain = Ir − IR = 1/2 MR2
– 3/32 MR2 = 13/32 MR2
13. The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is,
Ans: a
Answer: a / b) √[ (4 / 3) gh ] / √[ (10 / 7) gh ]
14. The speed of the center of a wheel rolling on a horizontal surface is vo. A point on the rim in level with the center will be moving at a speed of speed of,
(a) zero
(b) vo
(c) √2vo
(d) 2vo
Answer : c) √(2 vo)
Solution: v = √[ V02
+ V2 ]
v = √[2V02]
v = √2 V0
15. A round object of mass M and radius R rolls down without slipping along an inclined plane. The fractional force,
(a) dissipates kinetic energy as heat.
(b) decreases the rotational motion.
(c) decreases the rotational and transnational motion
(d) converts transnational energy into rotational energy
Answer : d) converts
translational energy into rotational energy
Answers:
1) d 2) a 3) d 4) b
5) a 6) d 7) a 8) a
9) a 10) d 11) a 12) b
13) a 14) c 15) d
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.