NCERT Solution (Class 11) > Physics Chapter 7 Systems Of Particles And Rotational Motion
NCERT Solution Class 11 Physics Chapter 7 Systems Of Particles And Rotational Motion: National Council of Educational Research and Training Class 11 Physics Chapter 7 Solution – Systems Of Particles And Rotational Motion. Free PDF Download facility available at our website.
Board |
NCERT |
Class |
11 |
Subject |
Physics |
Chapter |
7 |
Chapter Name |
Systems Of Particles And Rotational Motion |
Topic |
Exercise Solution |
Systems Of Particles And Rotational Motion Chapter all Questions and Numericals Solution
Chapter7-system of particles and rotational motion
7.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv)cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
ANSWER-
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For given shapes all are symmetric in nature and having uniform mass density hence their centre of mass at their respective geometric centres. The centre of mass of a body need not necessarily lie within it. For example, the C.M. of a ring lies outside the body.
7.2)
7.3) A child sits stationary at one end of a long trolley moving uniformly with a speed Von a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
ANSWER-
The child started running on a trolley moving with speed v. The force due to the boy’s motion is internal. which produces no effect on the motion of the trolley.as no external force is involved in the system, the there is no change in the velocity of the centre of mass of the trolley.
7.4) Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
ANSWER-
Consider two vectors having angle q between them as shown below,
Answer –
7.5) Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.
ANSWER-
7.6) Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
ANSWER-
Linear momentum and position vector of the particle in vector form is given as,
P = PXi + PY j+ PZ k and r = x i + y j+ zk
Angular momentum is given by,
Therefore momentum is having only z component hence having direction along z direction only.
7.7) Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
ANSWER-
Let one particle is placed at point A and other at point B separated by distance d. both particle are having mass m and velocity v hence
Angular momentum of the system about point A is given by,
∴ LA= mv x 0 + mvd
∴ LA= mvd
Angular momentum of the system about point B is given by,
∴ LB= mv x 0 + mvd
∴ LB= mvd
∴ LA = LB
Angular momentum remains the same independent of the point about which it is taken.
7.8)
7.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
ANSWER-
a)
Moment of inertia of the sphere about any of its diameters to be 2MR2/5. To find moment of inertia about axis tangential to the sphere and parallel to the diameter using parallel axis theorem
the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
ITANGENT = 2MR2/5 + MR2 = (7/5)MR2 .
b)
Moment of inertia of the DISC about any of its diameters to be MR2/4.According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
The M.I. of the disc about its centre = MR2/4 +MR2/4 = MR2/2.
Using parallel axis theorem,
ITANGENT = MR2/2 + MR2 = (3/2)MR2.
7.11) Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Answer –
Given torque for hollow cylinder and solid sphere is same hence,
7.12)
7.13)
7.18) A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
ANSWER-
Suppose a solid sphere on inclined surface having radius R from height h. according to conservation of energy
Total energy at top = total energy at bottom
∴ Kinetic energy at top + potential energy at top = Kinetic energy at bottom + potential energy bottom
At top only potential energy is present which gets converted into rolling kinetic energy at bottom
From above formula velocity of the sphere is constant as height (h) and acceleration due to gravity (g) are constants. Therefore, the velocity at the bottom remains the same independent of the inclination of plane.
For finding out the time taken using equation
As inclination changes it has effect on sinθ hence time will be different for different angle of inclination. as θ increases sinθ increases and it in turn decreases the time taken hence time will be longer for small angle of inclination.
7.19)
7.20)
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7.21)
Additional Exercises –
7.22)
7.23)
7.25)
7.26 (a) Prove the theorem of perpendicular axes.(Hint : Square of the distance of a point (x, y) in the x–y plane from an axis through the origin and perpendicular to the plane is x2+y2).
ANSWER-
(a) Theorem of perpendicular axes.
The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes passing through same point and lying in the plane of the body.
Consider a planar body with IX IY and IZ moment of inertia about x axis y axis and z axis respectively. Consider a point p at a distance x from y axis and at a distance y from x axis and at the distance Z from z axis as shown in figure.
Hence theorem is proved.
(a) Theorem of parallel axes.
The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between those two parallel axes.
Consider a axis RS passing through centre of mass o . let the axis PQ parallel to axis RS at a distance a.
Suppose a rigid body is made up of n particles, having masses m1, m2, m3, … ,mn, at perpendicular distances r1, r2, r3, … , rn respectively from the centre of mass O of the rigid body
∴ Moment of inertia about axis passing through point o is given by ;
Hence theorem is proved.
7.27) Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by
V2 = 2gh/1+K2/R2
Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
ANSWER-
Suppose a body rolling on inclined surface having radius R from height h. according to conservation of energy
Total energy at top = total energy at bottom
∴ Kinetic energy at top + potential energy at top = Kinetic energy at bottom + potential energy bottom
At top only potential energy is present which gets converted into rolling kinetic energy at bottom
7.28) A disc rotating about its axis with angular speed wo is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?
7.29) Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?
ANSWER-
For rotating the disc about centre of mass necessary torque is required. For providing this torque force should be present at the contact surface. This force is provided by frictional force hence friction is necessary to roll the given disc in the direction indicated.
(a)Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward and the sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction according to right hand screw rule.
(b)After perfect rolling begins the velocity at point B becomes equal to zero. This will make the frictional force acting on the disc zero.
7.30)
7.31)
7.32) Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
ANSWER-
(a)False. Frictional force acts opposite to the direction of motion of the centre of mass of a body.
(b) True. The instantaneous speed of the point of contact during rolling is zero as Rolling is the rotation of a body about an axis passing through the point of contact of the body with the ground.
(c) False. When a body is rolling, its instantaneous acceleration is not equal to zero.
(d) True. For perfect rolling the frictional force at the lowermost point is zero and the work done against friction is also zero.
(e) True. Body rolls when a frictional force acts between the body and the inclined surface. This frictional force provides the torque necessary for rolling. For perfectly frictionless inclined plane, the body undergo slipping not rolling.
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