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

System of Particles and Rotational Motion NCERT Solutions: 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 are developed in such a way that they enhance theoretical knowledge, as well as assess the application of learning. These questions can be much more than simple problem-solving ones, and invite the students to analyse, evaluate and apply concepts of rotational motion and centre 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) $\frac{5v_0^2}{6q_2}$

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

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

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

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 centre 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 $\omega$ :

$\begin{array}{rl}K{E}_{\text{rotational}}& =\frac{1}{2}\times \frac{2}{5}M{R}^2\times {\left(\frac{v_0}{R}\right)}^2\\ =\frac{1}{5}M{v}_0^2\end{array}$

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:

$PE=Mgh$

Equating energy:

$\begin{array}{c}\frac{7}{10}M{v}_0^2=Mgh\\ h=\frac{7v_0^2}{10g}\end{array}$

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) $\theta =2\frac{{\omega }_0}{c}[1-e^{-ct}]$

2) $\theta =\frac{{\omega }_0}{c}[1-e^{-ct}]$

3) $\theta =\frac{{\omega }_0}{2c}[1-e^{-ct}]$

4) $\theta =\frac{{\omega }_0}{3c}[1-e^{-ct}]$

Answer:

We know

$\begin{array}{r}\alpha =-c\omega \\ \frac{d\omega }{dt}=-c\omega \\ \Rightarrow {\int }_{a_0}^{\omega }\frac{d\omega }{\omega }=-{\int }_0^{t}cdt\\ \ln \frac{\omega }{{\omega }_0}=-ct\\ \Rightarrow \omega ={\omega }_0{e}^{-ct}\\ \Rightarrow \frac{d\theta }{dt}={\omega }_0{e}^{-ct}\\ \Rightarrow {\int }_0^{\theta }d\theta ={\omega }_0{\int }_0^t{e}^{-ct}dt\\ \Rightarrow \theta =\frac{{\omega }_0}{c}[1-e^{-ct}]\end{array}$
Hence, the answer is the option (2).

Question 3) A wedge of mass M=2kg carries two blocks A and B, as shown in the figure. Block A lies on a smooth-incline surface of the wedge inclined at an 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 the 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 the horizontal direction.

$\begin{array}{r}\therefore \mathrm{\Delta }{x}_{\mathrm{COM}}=0\\ \Rightarrow m_1\mathrm{\Delta }{x}_1+m_2\mathrm{\Delta }{x}_2+m_3\mathrm{\Delta }{x}_3=0\\ \Rightarrow M\cdot x+m(x-l)+m(l\cos \theta +x)=0\\ \Rightarrow x=\frac{ml(1-\cos \theta )}{M+2m}\end{array}$
Now, putting $\mathrm{m}=1\mathrm{kg},\mathrm{M}=2\mathrm{kgI}=2\mathrm{m}$ and $\theta ={60}^{\circ }$ in (i), we get

$\begin{array}{r}x=\frac{1\times 2(1-\cos {60}^{\circ })}{2+2\times 1}\\ x=0.25\mathrm{m}\end{array}$
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 the 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 the moment of inertia will be,$I=km{l}^2$

$\tau$= 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 a plate of double-sided length

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

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

Hence, the answer is 2.

Class 11 Physics Chapter 6 - System of Particles and Rotational Motion: Topics

In Class 11 Physics Chapter 6 - System of Particles and Rotational Motion, 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

System of Particles and Rotational Motion NCERT Solutions: Key Formulas

System of Particles and Rotational Motion class 11 question answers: 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: ${\overrightarrow{R}}_{\mathrm{COM}}=\frac{\sum m_i{\overrightarrow{r}}_i}{\sum m_i}$

2. Motion of COM

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

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

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

3. Angular Velocity & Linear Velocity

Average angular velocity =∆θ/∆t

Instantaneous angular velocity, ω=dθ/dt

4. Moment of Inertia (MI)

$\begin{array}{r}I=\sum m_i{r}_i^2\text{(Discrete system)}\\ I=\int r^{2}dm\text{(Continuous body)}\end{array}$

5. Moment of Inertia of a few useful configurations

6. Torque (Moment of Force)

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=rF\sin \theta$

7. Angular Momentum

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}=m\overrightarrow{r}\times \overrightarrow{v}$
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)

$\begin{array}{r}\omega ={\omega }_0+\alpha t\\ \theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2\\ {\omega }^2={\omega }_0^2+2\alpha \theta \end{array}$

Approach to Solve Questions of Class 11 Physics Chapter 6 - System of Particles and Rotational Motion

In Physics, Chapter 11, System of Particles and Rotational Motion, Chapter 6 is concerned with the motion of rigid bodies, torque, angular momentum and moment of inertia. This chapter requires conceptual clarity and practice in problem-solving methods to address the questions effectively in this chapter. The systematic approach can be used to make sure that there are no inaccuracies in derivations, numericals, and applications.

• Knowledge of Rigid Body Assumption: It is always good to start by noting that the chapter assumes that the bodies are rigid (no deformation). This makes the problems of torque and rotational motion easier.
• Divide Problems into Translational and Rotational: When solving, separate linear motion (using Newton's laws) and rotational motion (using torque = Iα). They should only be combined when necessary.
• Use Vector Representation: Appropriate rules of vectors should be applied to answer questions on torque, angular momentum and cross products. Practice right-hand rule applications for direction.
• Learn and Use Theorems of Inertia: The theorems which are commonly applied in numerical questions are the parallel axis and the perpendicular axis theorems. Use them directly in practice in order to save time during exams.
• Study Motion About the Center of Mass (COM): In problem of particle systems find the COM first. Simplification can then be made on motion of the entire system by taking translation + rotation of a system about the COM.
• Focus on Derivations and Proofs: Derivations such as the correlation between torque and angular momentum, kinetic energy in rotation and rolling motion are common exam problems to be prepared.
• Connect with Real-Life Applications: Comparing rotational motion to real-life examples such as wheels, flywheels, ceiling fans or gyroscopes can be beneficial when it comes to writing a better response to an explanation.