# NCERT Solutions for Class 11 Physics Chapter 7 System Of Particles And Rotational Motion

NCERT Solutions for Class 11 Physics Chapter 7 System Of Particles And Rotational Motion: The chapter starts with the concepts of the rigid body. In the solutions of NCERT class 11 physics chapter 7 system of particles and rotational motion, detailed explanation of answers is given by considering the bodies to be rigid. Ideally speaking a rigid body is a body with a perfectly definite and unaltering shape. The distance between all pairs of particles of such a body does not change. According to this definition, the bodies that we see in our real life are not rigid because they will deform when a force is applied. But when this deformation is negligible we consider the bodies as rigid. The CBSE NCERT solutions for class 11 physics chapter 7 system of particles and rotational motion are important as far as board and competitive exams are considered. The two main topics that we come across in this chapter are the moment of inertia and the centre of mass. There are questions and answers based on these topics in the NCERT solutions for class 11 physics chapter 7 system of particles and rotational motion. NCERT solutions help students in self-study and also you can check the solutions of other classes and subjects by clicking on the above link. The solutions of NCERT class 12 physics are divided into two sessions which are:

Exercise

## Moment of inertia of certain bodies of mass m about a particular axis:

 Body Axis Moment of inertia A thin circular ring of radius R Perpendicular to the plane, at the centre $mR^2$ A thin circular ring of radius R Diameter $\frac{mR^2}{2}$ A thin rod of length l Perpendicular to the rod, at the midpoint $\frac{ml^2}{12}$ A circular disc of radius R Perpendicular to the disc at the centre $\frac{mR^2}{2}$ A circular disc of radius R Diameter $\frac{mR^2}{4}$ A hollow cylinder of radius R Axis of cylinder ${mR^2}$ A solid cylinder of radius R Axis of cylinder $\frac{mR^2}{2}$ A solid sphere of radius R Diameter $\frac{2mR^2}{5}$

## NCERT class 11 physics chapter 7  topics:

7.1 Introduction

7.2 Centre Of Mass

7.3 Motion Of Centre Of Mass

7.4 Linear Momentum Of A System Of Particles

7.5 Vector Product Of Two Vectors

7.6 Angular Velocity And Its Relation With Linear Velocity

7.7 Torque And Angular Momentum

7.8 Equilibrium Of A Rigid Body

7.9 Moment Of Inertia

7.10 Theorems Of Perpendicular And Parallel Axes

7.11 Kinematics Of Rotational Motion About A Fixed Axis

7.12 Dynamics Of Rotational Motion About A Fixed Axis

7.13 Angular Momentum In Case Of Rotation About A Fixed Axis

7.14 Rolling Motion

## NCERT solutions for class 11 physics chapter 7 system of particles and rotational motion exercise

For uniform mass density, the location of the centre of mass is the same as that of the geometric centre.

No, it is not necessary that the centre of mass lies inside the body. For example in the case of a ring the centre of mass outside the body.

Let us assume that the centre of mass of the molecule is x cm away from the chlorine atom.

So, we can write

$\frac{m\left ( 1.27\ -\ x \right )\ +\ 35.5mx}{m\ +\ 35.5 }\ =\ 0$

or

$x\ =\ \frac{-1.27}{\left ( 35.5-1 \right )}\ =\ -\ 0.037\AA$

Here negative sign indicates that the centre of mass lies leftward of the chlorine atom.

Since child and trolley are considered in a system thus speed of the centre of mass change only when an external force will act on the system. When the child starts to move on the trolley the force produced is the internal force of the system so this will produce no change in the velocity of the CM.

Let a and b be two vectors having $\Theta$ angle between them.

Consider $\Delta$MON,

$\sin \Theta \ =\ \frac{MN}{MO}$

or                                                      $\sin \Theta \ =\ \frac{MN}{\underset{b}{\rightarrow}}$

or                                                         $MN\ =\ b\sin \Theta$

$\left | a\times b \right |\ =\ \left | a \right |\left | b \right |\sin \Theta$

=   2 (Area of $\Delta$ MOK)

Therefore the area of $\Delta$ MOK  =   One half of      $\left | a\times b \right |$.

A parallelepiped is shown in the figure given below:-

Volume is given by :    =  abc

We can write :

$|b\times c|\ =\ |b||c|\sin \Theta\ \widehat{n}$                           (The direction of $\widehat{n}$ is in the direction of vector a.)

$=\ |b||c|\sin 90^{\circ}\ \widehat{n}$

$=\ |b||c|\ \widehat{n}$

Now,

$a.(b\times c)\ =\ a.(bc)\ \widehat{n}$

$=\ abc \cos \Theta$

$=\ abc \cos 0^{\circ}$

$=\ abc$

This is equal to volume of parallelepiped.

Linear momentum of particle is given by :

$\overrightarrow{p}\ =\ p_x \widehat{i}\ +\ p_y \widehat{j}\ +\ p_z \widehat{k}$

And the angular momentum is :

$\overrightarrow{l}\ = \overrightarrow{r}\times \overrightarrow{p}$

$= \left ( x \widehat{i}\ +\ y \widehat{j}\ +\ z \widehat{k} \right ) \times \left ( p_x \widehat{i}\ +\ p_y \widehat{j}\ +\ p_z \widehat{k} \right )$

$=\begin{vmatrix} i &j &k \\ x &y &z \\ P_x &P_y &P_z \end{vmatrix}$

$=\ \widehat{i}\left ( yp_z - zp_y \right )\ -\ \widehat{j}\left ( xp_z - zp_x \right )\ +\ \widehat{k}\left ( xp_y - yp_x \right )$

When particle is confined to x-y plane then z = 0 and  pz = 0.

When we put the value of z and pz in the equation of linear momentum then we observe that only the z component is non-zero.

Assume two points (say A, B) separated by distance d.

So the angular momentum of a point about point A is given by :

$=\ mv \times d\ =\ mvd$

And about point B :                 $=\ mv \times d\ =\ mvd$

Now assume a point between A and B as C which is at y distance from point B.

Now the angular momentum becomes :          $=\ mv \times \left ( d-y \right )\ +\ mv \times y$

$=\ mv d$

Thus it can be seen that angular momentum is independent of the point about which it is measured.

The FBD of the given bar is shown below :

Since the bar is in equilibrium, we can write :

$T_1 \sin 36.9 ^{\circ}\ =\ T_2 \sin 53.1 ^{\circ}$

or                                                            $\frac{T_1}{T_2}\ =\ \frac{0.800}{0.600}\ =\ \frac{4}{3}$

or                                                            $T_1\ =\ \frac{4}{3}T_2$                                    .....................................................(i)

For the rotational equilibrium :

$T_1 \cos 36.9 ^{\circ}\times d\ =\ T_2 \cos 53.1 ^{\circ}\times \left ( 2-d \right )$                       (Use equation (i) to solve this equation)

or                                                                             $d\ =\ \frac{1.2}{1.67}\ =\ 0.72\ m$

Thus the center of gravity is at 0.72 m from the left.

The FBD of the car is shown below  :

We will use conditions of equilibrium here :

$R_f\ +\ R_b\ =\ mg$

$=\ 1800\times 9.8\ =\ 17640\ N$                ....................................(i)

For rotational equilibrium :

$R_f\left ( 1.05 \right )\ =\ R_b(1.8-1.05)$

or                                        $1.05R_f\ =\ 0.75R_b$

$R_b\ =\ 1.4R_f$                                                                         ..............................(ii)

From (i) and (ii) we get :

$R_f\ =\ \frac{17640}{2.4}\ =\ 7350\ N$

and                                    $R_b\ =\ 17640-7350\ =\ 10290\ N$

Thus force exerted by the front wheel is  =  3675 N

and force exerted by back wheel  =   5145 N.

We know that moment of inertia of a sphere about diameter is :

$=\ \frac{2}{5}MR^2$

Using parallel axes theorem we can find MI about the tangent.

Moment of inertia of a sphere about tangent   :

$=\ \frac{2}{5}MR^2\ +\ MR^2\ =\ \frac{7}{5}MR^2$

We know that moment of inertia of a disc about its diameter is :

$=\ \frac{1}{4}MR^2$

Using perpendicular axes theorem we can write :

Moment of inertia of disc about its centre:-

$=\ \frac{1}{4}MR^2\ +\ \frac{1}{4}MR^2\ =\ \frac{1}{2}MR^2$

Using parallel axes theorem we can find the required MI :

Moment of inertia about an axis normal to the disc and passing through a point on its edge is given by :

$=\ \frac{1}{2}MR^2\ +\ MR^2\ =\ \frac{3}{2}MR^2$

The moment of inertia of hollow cylinder is given by   $=\ mr^2$

And the moment of inertia of solid cylinder is given by :

$=\ \frac{2}{5}mr^2$

We know that :                $\tau \ =\ I\alpha$

Let the torque for hollow cylinder be $\tau_1$ and for solid cylinder let it be $\tau_2$.

According to the question :                 $\tau_1\ =\ \tau_2$

So we can write the ratio of the angular acceleration of both the objects.

$\frac{\alpha _2}{\alpha _1}\ =\ \frac{I_1}{I_2}\ =\ \frac{mr^2}{\frac{2}{5}mr^2}\ =\ \frac{2}{5}$

Now for angular velocity,

$\omega \ =\ \omega _o\ +\ \alpha t$

Clearly, the angular velocity of the solid sphere is more than the angular velocity of the hollow sphere. (As the angular acceleration of solid sphere is greater).

Firstly we will calculate moment of inertia of the solid cyliner :

$I_c\ =\ \frac{1}{2}mr^2$

$=\ \frac{1}{2}(20)(0.25)^2\ =\ 0.625\ Kg\ m^2$

So the kinetic energy is given by :

$E_k\ =\ \frac{1}{2}I\omega ^2$

$=\ \frac{1}{2}\times (6.25)\times (100) ^2$

$=\ 3125\ J$

And the angular momentum is given by :     $=\ I\omega$

$=\ 0.625\times 100$

$=\ 62.5\ Js$

We are given with the initial angular speed and the relation between the moment of inertia of both the cases.

Here we can use conservation of angular momentum as no external force is acting the system.

So we can write :

$I_1w_1\ =\ I_2w_2$

$w_2\ =\ \frac{I_1w_1}{I_2}$

$=\ \frac{I(40)}{\frac{2}{5}I}$

$=\ 100\ rev/min$

The final and initial velocities are given below :

$E_f\ =\ \frac{1}{2}I_2w_2^2$                 and            $E_i\ =\ \frac{1}{2}I_1w_1^2$

Taking the ratio of both we get,

$\frac{E_f}{E_i}\ =\ \frac{\frac{1}{2}I_2w_2^2}{\frac{1}{2}I_1w_1^2}$

or                                                                  $=\ \frac{2}{5}\times \frac{100\times 100}{40\times40}$

or                                                                   $=\ 2.5$

Thus the final energy is 2.5 times the initial energy.

The increase in energy is due to the internal energy of the boy.

## Q7.14  A rope of negligible mass is wound round a hollow cylinder of mass $3kg$  and radius $40cm$ . 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.

The moment of inertia is given by :

$I\ =\ mr^2$

or                                                              $=\ 3\times (0.4)^2\ =\ 0.48\ Kg\ m^2$

And the torque is given by :

$\tau \ =\ r\times F$

$=\ 0.4\times 30\ =\ 12\ Nm$

Also,                                                $\tau\ =\ I\alpha$

So                                                   $\alpha \ =\ \frac{\tau }{I}\ =\ \frac{12}{0.48}\ =\ 25\ rad/s^{-2}$

And the linear acceleration is         $a\ =\ \alpha r\ =\ 0.4\times 25\ =\ 10\ m/s^{-2}$

The relation between power and torque is given by :

$P\ =\ \tau \omega$

or                       $=\ 180\times 200\ =\ 36000\ W$

Hence the required power is 36 KW.

## Q7.16  From a uniform disk of radius $R$ , a circular hole of radius $R/2$  is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Let the mass per unit area of the disc be $\sigma$.

So total mass is   $=\ \Pi r^2\sigma\ =\ m$

Mass of the smaller disc is given by :

$=\ \Pi \left ( \frac{r}{2} \right )^2\sigma\ =\ \frac{m}{4}$

Now since the disc is removed, we can assume that part to have negative mass with respect to the initial condition.

So, the centre of mass of the disc is given by the formula :

$x\ =\ \frac{m_1r_1\ +\ m_2r_2}{m_1\ +\ m_2}$

or                                                                    $=\ \frac{m\times 0\ -\ \frac{m}{4}\times \frac{r}{2}}{m\ -\ \frac{m}{4}}$

$=\ \frac{-r}{6}$

Hence the centre of mass is shifted   $\frac{r}{6}$   leftward from point O.

The centre of mass of meter stick is at 50 cm.

Let the mass of meter stick be m.

Now according to the situation given in the question, we will use the rotational equilibrium condition at the centre point of the meter stick.

$10g(45 - 12)\ -\ mg(50 - 45)\ =\ 0$

or                                                               $m\ =\ \frac{10\times 33}{5}\ =\ 66\ g$

Thus the mass of the meter stick is 66 g.

(a) Will it reach the bottom with the same speed in each case?

Let the height of the plane is h and mass of the sphere is m.

Let the velocity at the bottom point of incline be v.

So the total energy is given by :

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2$

Using the law of conservation of energy :

$\frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2\ =\ mgh$

For solid sphere moment of inertia is :

$I\ =\ \frac{2}{5}mr^2$

So,

$\frac{1}{2}mv^2\ +\ \frac{1}{2}\left ( \frac{2}{5}mr^2 \right )w^2\ =\ mgh$

Put  v  = wr   and solve the above equation.

We obtain :                              $\frac{1}{2}v^2\ +\ \frac{1}{5}v^2\ =\ gh$

or                                                   $v\ =\ \sqrt{\frac{10}{7}gh}$

Thus the sphere will reach the bottom at the same speed since it doesn't depend upon the angle of inclination.

(b) Will it take longer to roll down one plane than the other?

Let us assume the angle of inclination to be $\Theta_1$    and  $\Theta_2$.                        $\left ( \Theta_2\ > \Theta_1 \right )$

And the acceleration of the sphere will be   $g \sin \Theta_1$    and    $g \sin \Theta_2$   respectively.

Thus                             $a_2 > a_1$                                                            (since  $\sin \Theta _2 > \sin \Theta _1$)

Since their acceleration are different so the time taken to roll down the inclined plane will be different in both the cases.

(c) If so, which one and why?

The equation of motion gives :

$v\ =\ u\ +\ at$

So                                            $t \propto \frac{1}{a}$

Since we know that   $a_2 > a_1$

So        $t_2 < t_1$

Hence the inclined plane having a smaller angle of inclination will take more time.

Total energy of hoop is given by :

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}Iw^2$

Also the moment inertia of hoop is given by :    $I\ =\ mr^2$

We get,

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}(mr^2)w^2$

And      v  = rw

$T\ =\ \frac{1}{2}mv^2\ +\ \frac{1}{2}mv^2\ =\ mv^2$

So the work required is   :     $=\ mv^2\ =\ 100(0.2)^2\ =\ 4 J$

We are given the moment of inertia and the velocity of the molecule.

Let the mass of oxygen molecule be m.

So the mass of each oxygen atom is given by :      $\frac{m}{2}$

Moment of inertia is :

$I\ =\ \frac{m}{2}r^2\ +\ \frac{m}{2}r^2\ =\ mr^2$

or                                   $r\ =\ \sqrt{\frac{I}{m}}$

or                                  $r\ =\ \sqrt{\frac{1.94\times 10^{-46}}{5.36\times 10^{-26}}}\ =\ 0.60\times 10^{-10}\ m$

We are given that :

$E_{rot}\ =\ \frac{2}{3}E_{tra}$

or                                  $\frac{1}{2}I\omega ^2\ =\ \frac{2}{3}\times \frac{1}{2}mv^2$

or                                 $\omega \ =\ \sqrt{\frac{2}{3}}\times \frac{v}{r}$

or                                        $=\ 6.80\times 10^{12}\ rad/s$

(a) How far will the cylinder go up the plane?

The rotational energy is converted into the translational energy.(Law of conservation of energy)

$\frac{1}{2}I \omega ^2\ =\ \frac{1}{2}mv^2\ =\ mgh$

Since the moment of inertia for the cylinder is  :

$I\ =\ \frac{1}{2}mr^2$

Putting the value of MI and v = wr in the above equation, we get :

$\frac{1}{4}v^2\ +\ \frac{1}{2}v^2\ =\ gh$

or                                                $h\ =\ \frac{3}{4}\times \frac{v^2}{g}$

or                                                       $=\ \frac{3}{4}\times \frac{5\times 5}{g}\ =\ 1.91\ m$

Now using the geometry of the cylinder we can write :

$\sin \Theta \ =\ \frac{h}{d}$

or                                                 $\sin 30^{\circ}\ =\ \frac{h}{d}$

or                                                  $d\ =\ \frac{1.91}{0.5}$

$=\ 3.82\ m$

Thus cylinder will travel up to 3.82 m up the incline.

The velocity of cylinder is given by :

$v\ =\ \left ( \frac{2gh}{1\ +\ \frac{k^2}{r^2}} \right )^ \frac{1}{2}$

or                                                          $v\ =\ \left ( \frac{2gd\sin \Theta }{1\ +\ \frac{k^2}{r^2}} \right )^ \frac{1}{2}$

We know that for cylinder :

$K^2\ =\ \frac{R^2}{2}$

Thus                                                    $v\ =\ \left ( \frac{4}{3} gd \sin \Theta \right )^\frac{1}{2}$

Required time is :

$t\ =\ \frac{d}{v}$

or                                                              $=\ \left ( \frac{11.46}{19.6} \right )^\frac{1}{2}\ =\ 0.764\ s$

Hence required time is  0.764(2)  =  1.53  s.

NCERT solutions for class 11 physics chapter 7 system of particles and rotational motion additional exercise

The FBD of the figure is shown below :

$AH\ =\ \sqrt{\left ( AD^2\ -\ DH^2 \right )}$                                            (AD and DH can be found using geometrical analysis.)

$=\ \sqrt{\left ( 0.8^2\ -\ 0.25^2 \right )}\ =\ 0.76\ m$

Now we will use the equilibrium conditions :

(i) For translational equilibrium :

$N_c\ +\ N_b\ =\ mg\ =\ 40\times 9.8\ =\ 392\ N$                    ......................................(i)

(ii)  For rotational equilibrium :

$-N_b\times BI\ +\ mg\times FG\ +\ N_c \times CI \ +\ T\times AG\ -\ T\times AG\ =\ 0$

or                                                            $(N_c\ -\ N_b)\times 0.5\ =\ 49$

or                                                             $N_c\ -\ N_b\ =\ 98$                       ............................................................(ii)

Using (i) and (ii) we get :

$N_c\ =\ 245\ N$      and         $N_b\ =\ 147\ N$

Now calculate moment about point A :

$-N_b \times BI \ +\ mg\times FG\ +\ T\times AG\ =\ 0$

Solve the equation :                                   $T\ =\ 96.7\ N$

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

Moment of inertia when hands are stretched :

$=\ 2\times mr^2\ =\ 2\times (0.5)9^2$

$=\ 8.1\ Kg\ m^2$

So the moment of inertia of system (initial )  =   7.6  +  8.1  =  15.7 Kg m2.

Now, the moment of inertia when hands are folded :

$=\ 2\times mr^2\ =\ 2\times 5(0.2)^2\ =\ 0.4\ Kg\ m^2$

Thus net final moment of inertia is :     =   7.6  +  0.4  =  8 Kg m2.

Using conversation of angular momentum we can write :

$I_1\omega_1\ =\ I_2 \omega_2$

or                                                                       $\omega_2 \ =\ \frac{15.7\times 30}{8}\ =\ 58.88\ rev/min$

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

No, the kinetic is not constant. The kinetic energy increases with decrease in moment of inertia. The work done by man in folding and stretching hands is responsible for this result.

The imparted angular momentum is given by :

$\alpha \ =\ mvr$

Putting all the given values in the above equation we get :

$=\ (10\times 10^{-3})\times500\times \frac{1}{2}$

$=\ 2.5\ Kg\ m^2\ s^{-1}$

Now, the moment of inertia of door is :

$I\ =\ \frac{1}{3}ML^2$

or                                                       $=\ \frac{1}{3}(12)1^2\ =\ 4\ Kg\ m^2$

Also,                                    $\alpha\ =\ I \omega$

or                                         $\omega\ =\ \frac{\alpha}{I}\ =\ \frac{2.5}{4}\ =\ 0.625\ rad\ s^{-1}$

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

Let the moment of inertia of disc I and disc II be I1 and I2 respectively.

Similarly, the angular speed of disc I and disc II be w1 and w2 respectively.

So the angular momentum can be written as :

$L_1\ =\ I_1 \omega_1$             and             $L_2\ =\ I_2 \omega_2$

Thus the total initial angular momentum is :      $=\ I_1 \omega_1\ +\ I_2 \omega_2$

Now when the two discs are combined the angular momentum is :

$L_f\ =\ (I_1\ +\ I_2)w$

Using conservation of angular momentum :

$I_1\omega_1\ +\ I_2\omega_2 =\ (I_1\ +\ I_2)w$

Thus angular velocity is :

$\omega \ =\ \frac{I_1\omega_1\ +\ I_2\omega_2}{I_1\ +\ I_2}$

(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 $\omega _{1}\neq \omega _{2}$ .

The initial kinetic energy is written as :

$K_i\ =\ \frac{1}{2}I_1\omega_1^2\ +\ \frac{1}{2}I_2\omega_2^2$

or                                             $K_i\ =\ \frac{1}{2}\left ( I_1\omega_1^2\ +\ I_2\omega_2^2 \right )$

Now the final kinetic energy is :

$K_f\ =\ \frac{1}{2}\left ( I_1\ +\ I_2 \right )\omega^2$

Put the value of final angular velocity from part (a).

We need to find :

$K_i\ -\ K_f\ =\ \frac{1}{2}\left ( I_1\omega_1^2\ +\ I_2\omega_2^2 \right )\ -\ \frac{1}{2}\left ( I_1\ +\ I_2 \right )\omega^2$    

Solve the above equation, we get :

$=\ \frac{I_1 I_2}{2(I_1\ +\ I_2)}(\omega_1^2\ +\ \omega_2^2\ -\ 2\omega_1\omega_2)$

or                                           $=\ \frac{I_1 I_2}{2(I_1\ +\ I_2)}(\omega_1\ -\ \omega_2)^2$

$>\ 0$                                                                             (Since none of the quantity can be negative.)

Thus initial energy is greater than the final energy.  (due to frictional force).

Consider the figure given below:

The moment of inertia about x axis is given by :

$I_x\ =\ mx^2$

And the moment of inertia about y-axis is :

$I_y\ =\ my^2$

$I_z\ =\ m\left ( \sqrt{\left ( x^2\ +\ y^2 \right )} \right )^2$

or                                    $I_z\ =\ I_x\ +\ I_y$

Consider the figure given below :

The moment of inertia about RS axis  :-

$I_{RS}\ =\ \sum m_ir_i^2$

Now the moment of inertia about QP axis :-

$I_{QP}\ =\ \sum m_i(a+r_i)^2$

or                                        $=\ \sum m_i(a^2 +r_i^2 + 2ar_i)$

or                                        $=\ I_{RS} + \sum m_i a^2 + 2 \sum m_iar_i^2$

Thus                         $I_{QP} =\ I_{RS} + M a^2$                                                           $\left ( \because 2\sum m_iar_i^2 \right )\ =\ 0$

Hence proved.

$v^{2}=\frac{2gh}{(1+k^{2}/R^{2)}}$
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.

Consider the given situation :

The total energy when the object is at the top (potential energy)  =  mgh.

Energy when the object is at the bottom of the plane :

$E_b\ =\ \frac{1}{2}I\omega^2\ +\ \frac{1}{2}mv^2$

Put    $I\ =\ mk^2$       and      $v\ =\ \omega r$, we get :

$E_b\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )$

By the law of conservation of energy we can write :

$mgh\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )$

or

$v\ =\ \frac{2gh}{\left ( 1\ +\ \frac{k^2}{r^2} \right )}$                                        .

Let the angular speed of the disc is  $\omega$.

So the linear velocity can be written as    $v\ =\ \omega r$

(a)  Point A:-

The magnitude of linear velocity is $\omega r$ and it is tangentially rightward.

(b)  Point B:-

The magnitude linear velocity is $\omega r$ and its direction is tangentially leftward.

(c) Point C:-

The magnitude linear velocity is   $\frac{ \omega r}{2}$  and its direction is rightward.

The disc cannot roll as the table is frictionless.

(a) Give the direction of frictional force at $B$ , and the sense of frictional torque, before perfect rolling begins.

Since the velocity at point B is tangentially leftward so the frictional force will act in the rightward direction.

The sense of frictional torque is perpendicular (outward) to the plane of the disc.

(b) What is the force of friction after perfect rolling begins?

Perfect rolling will occur when the velocity of the bottom point (B) will be zero. Thus the frictional force acting will be zero.

Friction is the cause for motion here.

So using Newton's law of motion we can write :

$f\ =\ ma$

$\mu _k mg\ =\ ma$

or                                           $a\ =\ \mu _k g$

Now by the equation of motion, we can write :

$v\ =\ u\ +\ at$

or                                            $v\ =\ \mu _k gt$

The torque is given by :

$\tau\ =\ I \alpha$

or                                  $r\times f \ =\ -I \alpha$

or                                   $\mu _k mgr \ =\ -I \alpha$

or                                             $\alpha\ =\ -\frac{\mu _k mgr}{I}$

Now using the equation of rotational motion we can write :

$\omega\ =\ \omega _o\ +\ \alpha t$

or                                      $\omega\ =\ \omega _o \ +\ -\frac{\mu _k mgr}{I}t$

Condition for rolling is         $v\ =\ \omega r$

So we can write :

$v =\ r\left ( \omega _o \ +\ -\frac{\mu _k mgr}{I}t \right )$

For ring the moment of inertia is :   $mr^2$

So we have :

$t\ =\ \frac{r \omega _o}{2 \mu _k g}$

or                                            $=\ \frac{0.1\times 10\times 3.14}{2\times 0.2 \times 10}\ =\ 0.80\ s$

Now in case of the disc, the moment of inertia is :

$I\ =\ \frac{1}{2}mr^2$

Thus                                $t\ =\ \frac{r \omega _o}{3 \mu _k g}$

or                                          $=\ \frac{0.1\times 10\times 3.14}{3 \times 0.2 \times 9.8}\ =\ 0.53\ s$

Hence disc will start rolling first.

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

Consider the following figure :

Moment of inertia of cylinder is   :

$I\ =\ \frac{1}{2}mr^2$

Thus acceleration is given by :

$a\ =\ \frac{mg \sin \Theta}{m\ +\ \frac{1}{2}\times \frac{ mr^2}{r^2}}$

or                                                                 $=\ \frac{2}{3}\ g \sin 30^{\circ}\ =\ 3.27\ m/s^2$

Now using Newton's law of motion :

$F\ =\ ma$

or                                                 $mg \sin 30^{\circ}\ -\ f\ =\ ma$

or                                                        $f\ =\ mg \sin 30^{\circ}\ -\ ma$

or                                                              $=\ 49\ -\ 32.7\ =\ 16.3\ N$

Hence frictional force is 16.3 N.

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

We know that the bottommost point of body (which is in contact with surface) is at rest during rolling. Thus work done against the frictional force is zero.

## Q7.31 (c)  A cylinder of mass $10kg$  and radius $15cm$  is rolling perfectly on a plane of inclination $30^{0}$. The coefficient of static friction $\mu _{s}=0.25$.

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

In case of rolling without any skidding is given by :

$\mu\ =\ \frac{1}{3}\tan \Theta$

Thus                                  $\tan \Theta\ =\ 3 \mu$

or                                                      $=\ 3 \times (0.25)\ =\ 0.75$

or                                              $\Theta \ =\ 37.87 ^{\circ}$

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the $CM$ of the body.

False. Friction also opposes the relative motion between the contacted surfaces. In the case of rolling, the cm is moving in backward direction thus the frictional force is directed in the forward direction.

(b) The instantaneous speed of the point of contact during rolling is zero.

True. This is because the translational speed is balanced by rotational speed.

(c) The instantaneous acceleration of the point of contact during rolling is zero.

False. The value of acceleration at contact has some value, it is not zero as the frictional force is zero but the force applied will give some acceleration.

## Q7.32 (d)  Read each statement below carefully, and state, with reasons, if it is true or false;

(d) For perfect rolling motion, work done against friction is zero.

True. As the frictional force at the bottommost point is zero, so the work done against it is also zero.

(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

True. Since it is a frictionless plane so frictional force is zero thus torque is not generated. This results in slipping not rolling.

(a) Show $P=P_{i}^{^{'}}+m_{i}V$
where $P_{i}$ is the momentum of the ith particle (of mass $m_{i}$)and  $P_{t}^{'}=m_{t}V_{t}$ Note$V_{t}^{'}$  is the velocity
of the ith particle relative to the centre of mass.Also, prove using the definition of the centre of mass$\sum P_{t}'=0$

The momentum of ith particle is given by :             $p_i\ =\ m_iv_i$

The velocity of the centre of mass is V.

Then the velocity of ith particle with respect to the center of mass will be :     $v_i'\ =\ v_i\ -\ v$

Now multiply the mass of the particle to both the sides, we get :

$mv_i'\ =\ mv_i\ -\ mv$

or                                                             $p_i'\ =\ p_i\ -\ p$                                          (Here p'i is the momentum of ith particle with respect to center of mass.)

or                                                             $p_i\ =\ p_i'\ +\ p$

Now consider   $p_i'$  :

$\sum p_i'\ =\ \sum m_iv_i'\ =\ \sum m_i\frac{dr_i}{dt}$

But as per the definition of centre of mass, we know that :

$\sum m_ir_i'\ =\ 0$

Thus                                                             $\sum p_i\ =\ 0$

(b) show$K=K^{'}+1/2MV^{2}$ 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 centre of mass and $MV^{2}/2$  is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.

From the first part we can write :

$\sum mv_i'\ =\ \sum mv_i\ -\ \sum mv$

or                                                     $\sum mv_i\ =\ \sum mv_i'\ +\ \sum mv$

Squaring both sides (in vector form taking dot products with itself), we get :

$\sum mv_i. \sum mv_i\ =\ \sum m(v_i'\ +\ v)\ .\ \sum m(v_i'\ +\ v)$

or                                                      $M^2 \sum v_i^2\ =\ M^2 \sum v_i'^2\ +\ M^2v^2$

or                                                      $\frac{1}{2}M \sum v_i^2\ =\ \frac{1}{2}M \sum v_i'^2\ +\ \frac{1}{2}Mv^2$

Hence                                                                 $K\ =\ K'\ +\ \frac{1}{2}Mv^2$

(c) Show $L=L^{'}+R\times MV$
where $L=\sum r_{i}^{'}\times P_{i}^{'}$
is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $r_{i}^{'}=r_{i}-R$ ; rest of the notation is the standard notation used in the chapter. Note$L^{'}$  and $MR\times V$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

The position vector of the ith particle (with respect to the center of mass) is given by :

$r_i'\ =\ r_i\ -\ R$

or                                                                  $r_i\ =\ r_i'\ +\ R$

From the first case we can write :

$p_i\ =\ p_i'\ +\ p$

Taking cross product with position vector we get  ;

$\sum r_i\times p_i\ =\ \sum r_i\times p_i'\ +\ \sum r_i\times p$

or                                                                $L\ =\ L'\ +\ \sum R\times p_i'\ +\ \sum r_i\times m_iv\ +\ \sum R\times m_iv$

or                                                                  $L=L^{'}+R\times MV$

(d) Show $\frac{dL^{'}}{dt}=\sum r_{i}^{'}\times \frac{dp^{'}}{dt}$
Further, show that

$\frac{dL^{'}}{dt}=\tau ^{'}_{ext}$

where $\tau ^{'}_{ext}$is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the definition of centre of mass and third law of motion. Assume the
internal forces between any two particles act along the line joining the particles.)

Since we know that :

$L'\ =\ \sum r_i' \times p_i'$

Differentiating the equation with respect to time, we obtain :

$\frac{dL'}{dt}\ =\ d\frac {\left ( \sum r_i' \times p_i' \right )}{dt}$

or                                                  $=\ \frac{d\left ( \sum m_ir_i' \right )}{dt}\times v_i'\ +\ \sum r_i'\ \times \frac{d}{dt}p_i'$

or                                          $\frac {dL'}{dt} =\ \sum r_i'\ \times m_i \frac{d}{dt}v_i'$                                                                      $\left ( \because \sum m_ir_i\ =\ 0 \right )$

Now using Newton's law of motion we can write :

$\sum r_i'\ \times m_i \frac{d}{dt}v_i'\ =\ \tau _{ext}'$

Thus                                         $\frac{dL^{'}}{dt}=\tau ^{'}_{ext}$

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## NCERT Solutions for Class 11 Subject wise

 CBSE NCERT solutions for class 11 biology NCERT solutions for class 11 maths CBSE NCERT solutions for class 11 chemistry NCERT solutions for class 11 physics

## Significance of NCERT solutions for class 11 physics chapter 7 system of particles and rotational motion:

• To understand the NCERT solutions for class 11 physics chapter 7 system of particles and rotational motion knowledge of previous chapters of class 11 physics is a prerequisite.
• The solutions of NCERT for class 11 physics chapter 7 system of particles and rotational motion help to understand how to apply concepts studied in numerical problems.
• For exams like NEET and JEE Mains, two or more questions are expected from the chapter systems of particles and rotational motion.