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

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

The Higher Order Thinking Skills (HOTS) questions of Class XII Physics Chapter 6: System of Particles and Rotational Motion is developed in such a way that it enhances theoretical knowledge, as well as assesses the application learning. These questions can be much more than simple problem solving ones, and invite the students to analyze, evaluate and apply concepts of rotational motion and center of mass to new and complicated scenarios, which are perfect to practice a competitive exam such as JEE.

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:

$

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
$

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:

$

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

$

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 $l$.
$\therefore$ 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 $l$ 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 $\tau$ = Cμmgl and moment of inertia will be,$I=k m l^2$

$\tau$= 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.

NCERT Chapter 6 System of Particles and Rotational Motion Topics

In Systems of particles and rotational motion, NCERT Class 11 Physics Chapter 6, the topics that are covered deal with centre of mass, motion of system of particles, torque, angular velocity, angular momentum and rotational dynamics. These topics constitute the basis of learning complex physical systems and are of great value in board exams and competitive exams such as JEE and NEET.

6.1 Introduction
6.1.1 What Kind Of Motion Can A Rigid Body Have?
6.2 Centre Of Mass
6.3 Motion Of the Centre Of Mass
6.4 Linear Momentum Of A System Of Particles
6.5 Vector Product Of Two Vectors
6.6 Angular Velocity And Its Relation With Linear Velocity
6.6.1 Angular Acceleration
6.7 Torque And Angular Momentum
6.7.1 Moment Of Force (Torque)
6.7.2 Angular Momentum Of A Particle
6.8 Equilibrium Of A Rigid Body
6.8.2 Centre Of Gravity
6.9 Moment Of Inertia
6.10 Kinematics Of Rotational Motion About A Fixed Axis
6.11 Dynamics Of Rotational Motion About A Fixed Axis
6.12 Angular Momentum In Case Of Rotation About A Fixed Axis
6.12.1 Conservation Of Angular Momentum

NCERT Solutions for Class 11 Physics Chapter 6: Key Formulas

NCERT Solutions for Class 11 Physics Chapter 6: Key Formulas include important equations related to centre of mass, torque, moment of inertia, angular momentum, and rotational kinetic energy. These formulas act as quick revision tools and help students solve numerical problems efficiently in exams like CBSE, JEE, and NEET.

1. Centre of mass

For a system of particles: $\vec{R}_{\mathrm{COM}}=\frac{\sum m_i \vec{r}_i}{\sum m_i}$

2. Motion of COM

$M=\sum m_i \\$
$\vec{v}_{\mathrm{cm}}=\frac{\sum m_i \vec{v}_i}{M}$

$\vec{p}_{\mathrm{cm}}=M \vec{v}_{\mathrm{cm}}$

$\vec{a}_{\mathrm{cm}}=\frac{\vec{F}_{\mathrm{ext}}}{M}$

3. Angular Velocity & Linear Velocity

Average angular velocity =∆θ/∆t

Instantaneous angular velocity, ω=dθ/dt

4. Moment of Inertia (MI)

$\begin{aligned} & I=\sum m_i r_i^2 \text { (Discrete system) } \\ & I=\int r^2 d m \text { (Continuous body) }\end{aligned}$

5. Moment of Inertia of few useful configurations

6. Torque (Moment of Force)

$\vec{\tau}=\vec{r} \times \vec{F}=r F \sin \theta$

7. Angular Momentum

$
\vec{L}=\vec{r} \times \vec{p}=m \vec{r} \times \vec{v}
$
For a rotating body:

$
L=I \omega
$

8. Torque-Angular Acceleration Relation

$\tau=I \alpha$

9. Rotational Kinetic Energy

$K \cdot E_{\cdot \mathrm{rot}}=\frac{1}{2} I \omega^2$

10. Angular Equations of Motion (Uniform Angular Acceleration)

$\begin{aligned} & \omega=\omega_0+\alpha t \\ & \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\ & \omega^2=\omega_0^2+2 \alpha \theta\end{aligned}$

Approach to Solve Questions of NCERT Solutions for Class 11 Physics Chapter 6

To solve questions based on NCERT Class 11 Physics Chapter 6: the first step to solving these questions is to gain an understanding of some of the simpler concepts such as the center of mass, the torque, angular velocity, moment of inertia, and the rotational equivalents of the motion. Look keenly to determine whether the problem is a linear motion, a rotational motion or both. Employ the free-body diagrams to depict forces and torque. Use Newton laws, conservation of angular momentum and rotational laws of motion where applicable. In numerical problems make sure to convert the units properly and use of parallel or perpendicular axis theorems to compute moment of inertia. Organization and practice of various problems are important to succeed in this chapter.