**NCERT Solutions for Class 11 Physics Physics Chapter 7 System of particles and Rotational Motion** includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for **Class 11 Physics Physics Chapter 7 System of particles and Rotational Motion**. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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## NCERT Solutions for Class 11 Physics Physics Chapter 7 System of particles and Rotational Motion

NCERT Solutions for Class 11 Physics Physics Chapter 7 System of particles and Rotational Motion are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

**NCERT Exercises**

**Question 1.**

Give the location of the center of mass of a

(1) sphere,

(2) cylinder,

(3) ring, and

(4) cube, each of uniform mass density. Does the center of mass of a body necessarily lie inside the body?

**Answer:**

In all the four cases, center of mass is located at geometrical center of each. No, the center of mass may lie outside the body, as in case of ring, a hollow sphere, a hollow cylinder, a hollow cube etc.

**Question 2.**

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 A (1 Å = 10^{-10} m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

**Answer:
**Let C.M. be at a distance x A from H-atom

Distance of C.M. from Cl atom = (1.27 -x) Å

Let the mass of H-atom = m units

The mass of the Cl-atom = 35.5 m units

If C.M. is taken at the origin, then

mx + (1.27 – x) 35.5 m

= 0 mx = – (1.27 – x) 35.5 m

Negative sign indicates that if Cl atom is on the right side of C.M. (+), the hydrogen atom is on the left side of C.M. So, avoiding, if we get

x + 35.5x = 1.27 x 35.5

36.5x = 45.085

\(x=\cfrac { 45.005 }{ 36.5 } =1.235\)Å

Therefore, the center of mass located on the line joining H and Cl nuclei at a distance of 1.24 A from the H atom.

**Question 3.**

A child sits stationary at one end of a long trolley moving uniformly with a speed v on 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 speed of the CM of the (trolley + child) system remains unchanged (equal to v) because no external force acts on the system. The forces involved in running on the trolley are internal to this system.

**Question 4.**

Show that the area of the triangle contained between the vectors \(\vec { L } \) and \( \vec {b}\) and is one half of the magnitude of \(\vec { a } \) x \( \vec {b}\)

**Answer:
**

**Question 5.**

Show that \(\vec { a } .(\vec { b } X\vec { C } )\) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, \(\vec {a},\vec {b}\) and \( \vec {c}\).

**Answer:
**

**Question 6.**

Find the components along the x, y, z axes of the angular momentum \(\vec {L}\) of a particle, whose position vector is \(\vec {r}\) with components x, y, z and momentum is

\(\vec {p} \) with components p_{x′} , p_{y} and p_{z}. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

**Answer:
**

Question 7.

Question 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 vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

**Answer:**

As shown in figure given below, suppose the two particles move parallel to the y-axis

Clearly, \(\vec {i} \) does not depend on x and hence on the origin O. Thus the angular momentum of the two particle system is same whatever be point about which the angular momentum is taken.

**Question 8.**

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from its left end.

**Answer:**

Let AB be the uniform bar of weight W suspended at rest by the two strings OA and O’B which make angles 36.9° and 53.1° respectively with the vertical.

∴ ∠OAA’ = 90° -36.9° =53.1°

Similarly ∠ O’BB’ = 36.9°

AB = 2 m, AC = d m.

Let T_{1 }and T_{2} be the tensions in the strings OA and O’B respectively and their rectangular components are shown in the figure.

As the rod is at rest, so the vector sum of the forces acting along A’B’ axis and ⊥ to it are zero i.e

**Question 9.**

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

**Answer:
**

**Question 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 2MR^{2/5}, 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 MR^{2}/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 sphere about any diameter = \( \cfrac { 2 }{ 5 }\) MR^{2
}Using parallel axes theorem,

Moment of inertia of sphere about a tangent to the sphere

= \( \cfrac {2}{ 5 }\) MR^{2} + MR^{2} =\( \cfrac { 7 }{ 5 }\) MR^{2
}**(b)** We are given, moment of inertia of the disc about any of its diameter = \( \cfrac { 1 }{ 4 }\) MR^{2}

- Applying perpendicular axes theorem, moment of inertia of the disc about an axis passing through its center and normal to the disc =2x \( \cfrac { 1 }{ 4 }\) MR
^{2}= \( \cfrac { 1 }{ 2 }\) MR^{2 }Applying parallel axes theorem,moment of inertia of the disc passing through a point on its edge and normal to the disc

= \( \cfrac { 1 }{ 2 }\)MR^{2}+MR^{2}= \( \cfrac { 3 }{ 2 }\)MR^{2 }

**Question 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 center. Which of the two will acquire a greater angular speed after a given time.

**Answer:**

Let M and R be mass and radius of the hollow cylinder and the solid sphere, then, Moment of inertia of the hollow cylinder about its axis of symmetry,

l_{1}, = MR^{2 }Moment of inertia of the solid sphere about

from ω ω_{0} + at, we find that for given ω_{0} and t, ω_{2} > ω_{1}, angular speed of solid sphere will be greater than the angular speed of hollow cylinder.

**Question 12.**

A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^{-1 } The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

**Answer:**

Mass of solid cylinder = 20 kg

Radius of solid cylinder = 0.25 m

Angular velocity co = 100 rad s^{-1
}Therefore, MI of the cylinder about its axis

**Question 13.**

**(a)** A child stands at the center of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

**(b)** Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

**Answer:**

**(a)** Given that, initial angular speed,

ω_{1} = 40 rev/min, ω_{2} = ?

Suppose that initial moment of inertia of the child is l_{1} Then, final moment of inertia of the child

The new kinetic energy is 2.5 times initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.

**Question 14.**

A rope of negligible mass is wound round 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 of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

**Answer:**

Mass of hollow cylinder, M = 3 kg

Radius of hollow cylinder, R = 40 cm = 0.4 m

M.I. of the hollow cylinder about its axis

I = MR^{2} = 3 kg x (0.4 m)^{2} = 0.48 kg m^{2
}Force F = 30 N

.’. Torque, τ=FxR = 30N x 0.4 m = 12 Nm

**Question 15.**

To maintain a rotor at a uniform angular speed of 200 rad s^{-1}, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

**Answer:**

Here, angular velocity, ω = 200 rad s^{-1
}Torque, τ= 180 Nm

Power, P = ?

Power, P = τω

P = 180 x 200 = 36000 watt = 36 kW

**Question 16.**

From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The center of the hole is at R/2 from the center of the original disc. Locate the center of gravity of the resulting flat body.

**Answer:**

Suppose mass per unit area of the disc = m

Mass of original disc M = πR^{2} x m

Mass of hole removed from the disc,

**Question 17.**

A meter sticks is balanced on a knife edge at its center. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the meter stick?

**Answer:**

Let m be the mass of meter stick concentrated at C, the 50 cm mark as shown in figure

**Question 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:**

Let θ_{1}, l_{1 }be the angle of inclination and distance travelled from top to bottom respectively on plane (1) and θ_{2}, l_{2} be the angle of inclination and distance travelled from top to bottom respectively on plane (2)

where I = MI of the sphere, ω = its angular speed, k is the radius of gyration

From (ii) and (iii) it is clear that the sphere reaches the bottom with same speed in each case.

**(b)** **To find the time of rolling motion :** Yes, it will take longer time down one plane than the other. It will be longer for the plane having smaller angle of inclination.

**(c) Explanation :** Let t_{1} and t_{2} be the time taken by the sphere in rolling on plane (1) and (2) respectively. Acceleration of solid sphere on an inclined plane is given by,

Question 19.

A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its center of mass has a speed of 20 cm s^{-1}. How much work has to be done to stop it?

**Answer:**

Radius of hoop, R = 2 m Mass of hoop, M = 100 kg Velocity of center of mass = 20 cm s^{-1 }= 0.2 m s^{_1
}The total kinetic energy of the hoop

**Question 20.**

The oxygen molecule has a mass of 5.30 x 10^{-26} kg and a moment of inertia of 1.94 x 10^{-26} kg m^{2} about an axis through its center perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m s^{-1} and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

**Answer:**

Here, m = 5.30 x 10^{-26} kg,

I = 1.94 x 10^{-46} kg m^{2}, v = 500 m s^{-1 }m

If \( \frac { m }{ 2 } \) is mass of each atom of oxygen and 2r is distance between the two atoms as shown in figure, then

**Question 21.**

A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the center of mass of the cylinder has a speed of 5 m s^{-1
}(a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?

**Answer:**

Given that, 0 = 30°

Speed of C.M. of cylinder at the bottom, υ = 5 m s^{-1
}**(a)** As cylinder goes up, it attains potential energy at the expense of its kinetic energy of transnational and rotational motion. Suppose that the cylinder goes up to the height h on the inclined plane.

According to the principle of conservation of energy

**Question 22.**

As shown in the figure, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F. 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 ms^{-2})

(**Hint:** Consider the equilibrium of each side of the ladder separately.)

**Answer:
**Here, BA = CA = 1.6 m; DE = 0.5 m;

M = 40 kg; BF = 1.2 m Let T = tension in the rope,

N

_{1},N

_{2}=normal reaction at B and C respectively, i.e., forces exerted by the floor on the ladder. In figure, we find

**Question 23.**

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg in each hand. The angular speed of the platform is 30 revolutions per minute.The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m^{2}.

(a) What is his new angular speed? (Neglect friction).

(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

**Answer:**

Here, Mass in each hand = 5 kg

Moment of inertia of the man together with the platform, I = 7.6 kg m^{2
}Distance of the weight from the axis, r_{1} = 90 cm = 0.9 m

Distance of the weight from the axis, r_{2} = 20 cm = 0.2 m

Initial moment of inertia of man, platform and weights

l_{1 }= I + Mr_{1}^{2} = 7.6 + 2 x 5 x (0.9)^{2
}= 7.6 + 8.1 = 15.7 kg m^{2
}Final moment of inertia of man, platform and weights

I_{2} = 7.6 + 2 x Mr^{2} = 7.6 + 2 x 5 x (0.2)^{2} = 8.0 kg m^{2
}According to Principle of conservation of angular momentum,

No, kinetic energy is not conserved in the process. Infact, as moment of inertia decreases kinetic energy of rotation increases. This change in K.E. is due to the work done by the man in decreasing the MI of the body.

**Question 24.**

A bullet of mass 10 g and speed 500 m s^{-1} is fired into a door and gets embedded exactly at the center of the door. The door is 0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

(**Hint:** The moment of inertia of the door about the vertical axis at one end is ML^{2/3})

**Answer:**

Here, Mass of the bullet,

m = 10 g = 0.01 kg

Velocity of bullet, υ = 500 m s^{-1
}Width of door = 1 m

The distance from the axis, where the bullet gets embedded in the door,

r = \( \cfrac { 1}{ 2 } \) = 0.5 m

**Question 25.**

Two discs of moments of inertia l_{1} and l_{2} about their respective axes (normal to the disc and passing through the center), and rotating with angular speeds ω_{1} and ω_{2} are brought into contact face to face with their axes of rotation coincident. Calculate

(a) What is the angular speed of the two-disc system?

(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take to, ω_{1} ≠ ω_{2
}**Answer:**

Let l_{1} and I_{2} be the moments of inertia of two discs have angular speed ω_{1} and ω_{2}. When they are brought in contact, the M.I. of the two discs system will be l_{1} + l_{2} .

Let ω= angular speed of the combined system,

Hence, K_{1} – K_{2} > 0 or K1> K_{2
}or K_{2} < K_{1} i.e. rotational K.E. of the combined system is less than the sum of the initial energies of the two discs.

Hence there occurs a loss of K.E. on combining the two discs and is the dissipation of energy because of the frictional forces between the faces of the two discs. These forces bring about a common angular speed of the two discs on combining. This however is an internal loss and angular momentum remains conserved.

**Question 26.**

**(a)** Prove the theorem of perpendicular axes,

**(b)** Prove the theorem of parallel axes.

**Answer:**

**(a) Theorem of perpendicular axes :** It states that the moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its plane and intersecting each other at the point, where the perpendicular axis passes through it.

Let OZ be the axis perpendicular to the plane lamina and passing through the point O. Let OX and OY be two mutually perpendicular axes in the plane of the lamina and intersecting at the point O.

If I_{x}, I_{y} and I_{z} are the moments of inertia of the plane lamina about the axes OX, OY and OZ respectively, then according to theorem of perpendicular axes,

I_{z}=I_{x }+I_{y}…………(i)

**Proof :** Suppose that the rigid body is made of n particles of masses m_{1} ,m_{2} … m_{n} lying at distance

r_{1}, r_{2}……. r_{n} from the axis of rotation OZ Further suppose that the i^{th} particle of mass m, lies at point

P(x_{i} y_{i}), such that OP = r_{i}. Then,

**(b) Theorem of parallel axes :** It states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis through its center of mass plus the product of the mass of the body and the square of the perpendicular distance between the two axes.

Let I_{c} be the moment of inertia of a body of mass M about an axis LM passing through its center of mass C. Let I be the moment of inertia of the body about an axis ZZ‘ parallel to the axis LM and at a distance h from it as shown in the figure.

Then, according to the theorem of parallel axes,

I = I_{c} + Mh^{2} …(iii)

**Proof :** Consider that i^{th} particle located at the point P in the body is of mass m, and lies at a distance r_{i }from the axis LM. Then, the distance of i^{th} particle from axis ZZ’ is (r, + h). The moment of inertia of the i^{th} particle about the axis LM is m_{i}-r^{2}_{i}. Therefore, moment of inertia of the body about the axis LM is given by

Question 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

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:**

Let M, R, k be the mass, radius and radius of gyration of a body placed at the top A of the inclined plane of height h and angle of inclination θ.

**Question 28**

A disc rotating about its axis with angular speed ω_{0} is placed (without any translational push) on a perfectly frictionless table. The radius of the disc is What are the linear velocities of the points A, B and C on the disc shown in the figure.

**Answer:**

Using the relation υ= rω, we get

For point A,υ_{A} = R ω_{0}, along AX

For point B,υ_{B} = Rω_{0}, along BX’

For point C, υ_{c}=(\( \cfrac {R}{2} \)) ω_{0} parallel to AX,

The disc will not rotate, because it is placed on a perfectly frictionless table. Without friction, rolling is not possible.

**Question 30.**

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10π rad s^{-1}. Which of the two will start to roll earlier? The co-efficient of kinetic friction is = 0.2. μ_{k}= 0.2

**Answer:**

Here, radius of both the ring and solid disc

R = 10 cm = 0.1 m

μ_{k} = 0.2

Moment of inertia of the solid disc = \( \cfrac { 1}{ 2 } \) MR

Initial angular velocity ω_{0} = 10π rad s^{_1 }Initial velocity of center of mass is zero. Frictional force causes the C.M. to accelerate

**Question 31.**

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction = 0.25.

(a) How much is the force of friction acting on the cylinder?

(b) What is the work done against friction during rolling?

(c) If the inclination 6 of the plane is increased, at what value of 6 does the cylinder begin to skid, and not roll perfectly?

**Answer:**

Mass of cylinder M = 10 kg

Radius of cylinder, R = 0.15 m

Angle of inclination, 0 = 30°

Coefficient of static friction, μ_{g} = 0.25

**(a)** Force of friction on the cylinder is given by

**Question 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) It is true.** During rolling, the force of friction acts in the same direction as the direction of motion of the C.M. of the body.

**(b) It is true.** A rolling body can be imagined to be rotating about an axis passing through the point of contact of the body with the ground and hence its instantaneous speed is zero.

**(c) It is false.** Since the body is rotating, its instantaneous acceleration is not zero.

**(d) It is true.** In case of perfect rolling, work done against friction is zero as point of contact does not move.

**(e) It is true.** A body rolls due to the force of friction acting on it. If the wheel is moving down a perfectly frictionless inclined plane, it will be under the effect of its weight only. Since the weight of the wheel acts along the vertical through its center of mass, the wheel will not rotate. It will keep on slipping.

**Question 33.**

Separation of motion of a system of particles into motion of the center of mass and motion about the center of mass:

**(a)
**

where p

_{i}is the momentum of the i

^{th}particle (of mass m

_{i}) and p′

_{i}= m

_{i}v′

_{i}. Note v′

_{i}is the velocity of i

^{th }the particle relative to the center of mass. Also, prove using the definition of the center of mass

Σ p′

_{i}=0.

**(b)**

where K is the total kinetic energy of the system of particles, K’ is the total kinetic energy of the system when the particle velocities are taken with respect to the center of mass and MV

^{2}/2 is the kinetic energy of the translation of the system as a whole (i.e. of the center of mass motion of the system).

**(c)**

where .is the angular momentum of the system about the center of mass with velocities taken relative to the center of mass. Remember rest of the notation is the standard notation used in the chapter. Note L’ and MR x V can be said to be angular momenta, respectively, about and of the center of mass of the system of particles

the sum of all external torques acting on the system about the center of mass.

(**Hint :** Use the definition of center of mass and Newton’s Third law. Assume the internal forces between any two particles act along the line joining the particles.)

**Answer:
**

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