NCERT Solutions for Class 11 Physics Chapter 6 System Of Particles And Rotational Motion

Q6.1 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:

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.

Q6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about $1.27 AA (1 AA =10^{-10}m)$ . 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 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.

Q6.3 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:

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.

Q6.4 Show that the area of the triangle contained between the vectors $a\: and\: b$ is one half of the magnitude of $a\times b$ .

Answer:

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 $\left | a\times b \right |$ .

Q6.5 Show that $a.(b\times c)$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors , $a$ , $b$ and $c$ .

Answer:

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.

Q6.6 Find the components along the $x,y,z$ axes of the angular momentum $1$ of a particle, whose position vector is $r$ with components $x,y,z$ and momentum is $p$ 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:

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 p _z = 0.

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

Q6.7 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:

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.

Q6.8 A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are $36.9^{0}$ and $53.1^{0}$ respectively. The bar is $2m$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

Answer:

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.

Q6.9 A car weighs $1800 kg$ . The distance between its front and back axles is $1.8 m$ . Its centre 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:

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.

Q6.10 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:

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).

Q6.11 A solid cylinder of mass $20 kg$ rotates about its axis with angular speed $100 \: rad\: s^{-1}$ . The radius of the cylinder is $0.25m$ . 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:

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 : $=\ I\omega$

$=\ 0.625\times 100$

$=\ 62.5\ Js$

Q6.12 (a) A child stands at the centre 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.

Answer:

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 :

$I_1w_1\ =\ I_2w_2$

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

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

$=\ 100\ rev/min$

Q6.12(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:

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.

Q6.13 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.

Answer:

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 $a\ =\ \alpha r\ =\ 0.4\times 25\ =\ 10\ m/s^{-2}$

Q6.14 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 $100^{0}/_{0}$ efficient.

Answer:

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.

Q6.15 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.

Answer:

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 :

$x\ =7 \frac{m_1r_1\ +\ m_2r_2}{m_1\ +\ m_2}$

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.

Q6.16 A metre stick is balanced on a knife edge at its centre. 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 metre stick?

Answer:

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.

Q6.17 The oxygen molecule has a mass of $5.30\times 10^{-26}kg$ and a moment of inertia of $1.94\times 10^{-46}kg\: m^{2}$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500m/s$ 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:

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

Class 11 physics NCERT Chapter 6: Higher Order Thinking Skills (HOTS) Questions

Question 1) A solid sphere of mass $M$ and radius $R$ is rolling without slipping with a velocity $v_0$ on a horizontal surface. It encounters an inclined plane making an angle $\theta$ with the horizontal. Assuming no energy loss due to friction, find the maximum height $h$ the sphere will reach before coming to rest.
1) $
\frac{5 v_0^2}{6 q_2}
$

2) $
\frac{3 v_0^2}{4 g}
$

3) $
\frac{7 v_0^2}{10 g}
$

4) $
\frac{v_0^2}{2 g}
$

Answer:

For a rolling solid sphere, the total kinetic energy is the sum of translational and rotational kinetic energy:

$
\begin{gathered}
K E_{\text {total }}=K E_{\text {translational }}+K E_{\text {rotational }} \\
K E_{\text {total }}=\frac{1}{2} M v_0^2+\frac{1}{2} I \omega^2
\end{gathered}
$

For a solid sphere, the moment of inertia about its center is:

$
I=\frac{2}{5} M R^2
$

Using the rolling without slipping condition, $v_0=R \omega$, so:

$
\omega=\frac{v_0}{R}
$

Substituting $I$ and $\omega$ :

$
\begin{aligned}
K E_{\text {rotational }} & =\frac{1}{2} \times \frac{2}{5} M R^2 \times\left(\frac{v_0}{R}\right)^2 \\
& =\frac{1}{5} M v_0^2
\end{aligned}
$

So, the total kinetic energy:

$
\begin{gathered}
K E_{\text {total }}=\frac{1}{2} M v_0^2+\frac{1}{5} M v_0^2 \\
K E_{\text {total }}=\frac{5}{10} M v_0^2+\frac{2}{10} M v_0^2=\frac{7}{10} M v_0^2
\end{gathered}
$

Step 2: Converting Kinetic Energy to Potential Energy
At maximum height, all kinetic energy converts into gravitational potential energy:

$
P E=M g h
$

Equating energy:

$
\begin{gathered}
\frac{7}{10} M v_0^2=M g h \\
h=\frac{7 v_0^2}{10 g}
\end{gathered}
$

Hence, the answer is the option (3).

Question 2) A rod starts rotating with angular velocity $\omega_o$ but due to viscous force, its angular speed starts decreasing with variable retardation: $\alpha=-c \omega$
Find out the angle traversed by the rod as a function of time.
1) $
\theta=2 \frac{\omega_0}{c}\left[1-e^{-c t}\right]
$

2) $
\theta=\frac{\omega_0}{c}\left[1-e^{-c t}\right]
$

3) $
\theta=\frac{\omega_0}{2 c}\left[1-e^{-c t}\right]
$

4) $
\theta=\frac{\omega_0}{3 c}\left[1-e^{-c t}\right]
$

Answer:

We know

$
\begin{aligned}
& \alpha=-c \omega \\
& \frac{d \omega}{d t}=-c \omega \\
& \Rightarrow \int_{a_0}^\omega \frac{d \omega}{\omega}=-\int_0^t c d t \\
& \ln \frac{\omega}{\omega_0}=-c t \\
& \Rightarrow \omega=\omega_0 e^{-c t} \\
& \Rightarrow \quad \frac{d \theta}{d t}=\omega_0 e^{-c t} \\
& \Rightarrow \int_0^\theta d \theta=\omega_0 \int_0^t e^{-c t} d t \\
& \Rightarrow \quad \theta=\frac{\omega_0}{c}\left[1-e^{-c t}\right]
\end{aligned}
$
Hence, the answer is the option (2).

Question 3) A wedge of mass M=2kg carries two blocks A and B, as shown in figure. Block A lies on a smooth-incline surface of the wedge inclined at angle $\theta$ to the horizontal. B lies inside a horizontal groove made in the wedge. The two blocks have m=1kg each and are connected by a light string. Length of the groove is l=2m as shown. The entire system is released from rest. What is the distance travelled by the wedge (in m) on the smooth horizontal surface by the time block B comes out of the groove?

Answer:

Let the displacement of the wedge be $x$ towards the right. Displacement of $B=(x-1)$ towards right.
Displacement of A w.r.t wedge (on the incline) is I.
$\therefore$ Displacement of A in horizontal direction $=l \cos \theta+x$
The COM of the entire system suffers no displacement in horizontal direction.

$
\begin{aligned}
& \therefore \Delta x_{\mathrm{COM}}=0 \\
& \Rightarrow m_1 \Delta x_1+m_2 \Delta x_2+m_3 \Delta x_3=0 \\
& \Rightarrow M \cdot x+m(x-l)+m(l \cos \theta+x)=0 \\
& \Rightarrow x=\frac{m l(1-\cos \theta)}{M+2 m}
\end{aligned}
$
Now, putting $\mathrm{m}=1 \mathrm{~kg}, \mathrm{M}=2 \mathrm{kgI}=2 \mathrm{~m}$ and $\theta=60^{\circ}$ in (i), we get

$
\begin{aligned}
& x=\frac{1 \times 2\left(1-\cos 60^{\circ}\right)}{2+2 \times 1} \\
& x=0.25 \mathrm{~m}
\end{aligned}
$
Hence, the answer is 0.25 .

Question 4)

A sphere is released on a smooth inclined plane from the top. When it moves down its angular momentum is:
1) Conserved about every point
2) Conserved about the point of contact only
3) Conserved about the centre of the sphere only
4) Conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball.

Answer:

As the inclined plane is smooth, the sphere can never roll, rather it will just slip down. Hence, the angular momentum remains conserved at any point on a line parallel to the inclined plane and passing through the centre of the ball.

Hence, the answer is the option (4).

Question 5) A square plate of mass $m$ and length $I$ is rotated with angular velocity ' $w$ ' with a vertical axis passing through its centre, and it is kept on a rough surface. It takes ' $t_0$ ' time to stop because of uniform friction. Now if we hinge it at the corner \& give same angular velocity, In $K t_0$ time it will take to stop, Find $k$.

Answer:

The torque by friction will be τ = Cμmgl and moment of inertia will be,$I=k m l^2$

τ = Iα

Hence, angular acceleration will be

$\alpha=\frac{C \mu g}{k}\left(\frac{1}{l}\right)$

$\Rightarrow \alpha$ is inversely proportional to l

Now as it is hinged at one corner, we can assume plate of double side length

Now this is again the central axis, the value of angular acceleration will be half.

Hence, it will take $2t_0$ time to stop.

Hence, the answer is 2.

Approach to Solve Questions of NCERT Solutions for Class 11 Physics Chapter 6 System Of Particles And Rotational Motion

Conceptual Understanding

• Revise key concepts: centre of mass, torque, angular momentum, moment of inertia, rolling motion.
• Understand the vector representation of torque and angular momentum.
• Practice the use of conservation laws, specifically the one that preserves angular momentum.

Approach to Solving Numerical Questions

• Carefully examine the question to identify which information the problem gives you and which information you need to determine.
• Create a straightforward diagram with proper labels that especially includes torque and rotational motion when possible.
• The solution requires the conversion of every quantity into the SI measurement system.

For Centre of Mass Problems

Use the formula:

$
\vec{R}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+\cdots+m_n \vec{r}_n}{m_1+m_2+\cdots+m_n}
$

Apply separately for $\mathbf{x}$ and $\mathbf{y}$ components if needed.

For Torque and Rotational Equilibrium

Use:

$
\tau=r \times F=r F \sin \theta
$
Check for rotational equilibrium:

$
\sum \tau=0
$
For Moment of Inertia (MI) Problems

Memorise standard MI formulas for rod, disc, ring, etc.
Use the parallel axis or perpendicular axis theorem when needed:

For Conservation of Angular Momentum

Use:

$
I_1 \omega_1=I_2 \omega_2
$

Applicable when no external torque is acting on the system.

• Revise the theory before attempting questions.
• Solve all examples and exercises given in the NCERT – they often cover exam-oriented concepts.
• Practice time-bound problem solving to improve speed and accuracy.