Systems of Particles and Rotational Motion Class 11 Notes CBSE Physics Chapter 6 (Free PDF Download)

Section-A (1 Mark Questions)

1. A wheel 0.5m in radius is moving with a speed of 12m/s. find its angular speed?

Ans.

$v=r\omega$ 

$\omega =\dfrac{v}{r}=\dfrac{12}{0\cdot 5}$

$\omega =24\;rad/s$ .

2. State the condition for mechanical equilibrium of a body?

Ans. For mechanical equilibrium of a body the vector sum of all the forces and moments (torques) acting on the body must be zero.

3. How is angular momentum related to linear momentum? 

Ans. $\vec{\vec{L}}=\vec{r}\times \vec{p}$

Or L = rp sin θ

Where θ is the angle between $\vec{r}$ and $\vec{p}$ .

4. What is the position of the centre of mass of a uniform triangular lamina?

Ans. Position of the centre of mass of a uniform triangular lamina at the centroid of the triangular lamina.

5. What is the moment of inertia of a sphere of mass 20 kg and radius $\dfrac{1}{4}m$ about its diameter?

Ans. $I=\dfrac{2}{5}MR^{2}$

$I=\dfrac{2}{5}\times 20\times \left ( \dfrac{1}{4} \right )^{2}$

$I=0\cdot 5\;kgm^{2}$

6. What are the factors on which moment of inertia of a body depends?

Ans.

(i) Mass of the body

(ii) Shape and size of the body

(iii) Position of the axis of rotation

7. Two particles in an iAnsated system undergo head on collision. What is the acceleration of the centre of mass of the system?

Ans. Acceleration of center of mass is zero as all forces are internal forces.

8. Which component of a force does not contribute towards torque? 

Ans. The radial component of a force does not contribute towards torque.

9. What is the position of centre of mass of a rectangular lamina? 

Ans. The centre of mass of a rectangular lamina is the point of intersection of diagonals.

10. Does the centre of mass of a body necessarily lie inside the body?

Ans. The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.

Section-B (2 Marks Questions)

11. A planet revolves around on massive star in a highly elliptical orbit is its angular momentum constant over the entire orbit. Give reason?

Ans. A planet revolves around the star under the effect of gravitational force since the force is radial and does not contribute towards torque. Thus in the absence of an external torque angular momentum of the planet remains constant.

12. Prove the relation $vec{\tau }=\dfrac{d\vec{L}}{dt}$

Ans. We know $\vec{L}=I\vec{\omega }$

Differentiating wrt. Time

$\dfrac{d\vec{L}}{dt}=\dfrac{d}{dt}\left ( I\vec{\omega } \right )=\dfrac{Id\vec{\omega }}{dt}=1\vec{\alpha }$ …(1)

We know that

$\vec{\tau }=I\vec{\alpha }$ …(2)

From (1) and (2) $\vec{\tau }=\dfrac{d\vec{L}}{dt}$

13. What is the torque of the force $\vec{F}=\hat{2}i-\hat{3}j+\hat{4}k$ acting at the point about the $\vec{r}=\left (\hat{3}i-\hat{2}j+\hat{3}k  \right )m$ origin?

Ans. $\vec{\tau }=\vec{r}\times \vec{F}$

$\vec{\tau }=\begin{vmatrix} \hat{}i & \hat{}j & \hat{}k\\ 2 & -3 & 4\\ 3& 2 & 3 \end{vmatrix}$

$\vec{\tau }=\left ( -\hat{17}i+\hat{6}j+\hat{13}k \right )NM$

14. What is the value of linear velocity if $\vec{\omega }=\hat{3}i-\hat{4}j+\hat{}k$ and $\vec{r}=\hat{}5i-\hat{6}j+\hat{6}k$ ?

Ans. $\vec{v}=\vec{\omega }\times \vec{r}$

$\vec{v}=\left ( \hat{3}i-\hat{4}j+\hat{}k \right )\times \left ( \hat{}5i-\hat{6}j+\hat{6}k \right )$

$\vec{t}=\begin{vmatrix} \hat{}i & \hat{}j & \hat{}k\\ 3& -4 & 1\\ 5& -6 & 6 \end{vmatrix}$

$\vec{v}=\left ( -\hat{18}i-\hat{13}j+\hat{2}k \right )m/s$

15. Find the expression for radius of gyration of a Ansid sphere about one of its diameters?

Ans. M.I. of a Ansid sphere about its diameter $=\dfrac{2}{5}MR^{2}$

K = Radius of Gyration

$I=MK^{2}=\dfrac{2}{5}MR^{2}$

$K^{2}=\dfrac{2}{5}R^{2}$

$K=\sqrt{\dfrac{2}{5}}R$

16. Prove that the centre of mass of two particles divides the line joining the particles in the inverse ratio of their masses? 

Ans. $\vec{r}_{cm}=\dfrac{m_{1}+\vec{r}_{1}+m_{2}\vec{r}_{2}}{m_{1}+m_{2}}$

If centre of mass is at the origin

$\vec{r}_{cm}=0$

$\Rightarrow m_{1}\vec{r}_{1}+m_{2}\vec{r}_{2}=0$

$m_{1}\vec{r}_{1}=-m_{2}\vec{r}_{2}$

In terms of magnitude $m_{1}\left | \vec{r}_{1} \right |=m_{2}\left | \vec{r}_{2} \right |$

$\Rightarrow \dfrac{m_{1}}{m_{2}}=\dfrac{r_{2}}{r_{1}}$

PDF Summary - Class 11 Physics System of Particles and Rotational Motion Notes (Chapter 6) 

PDF Summary - Class 11 Physics System of Particles and Rotational Motion Notes (Chapter 6) 

System of particles and rotational motion comes under the fifth unit, Motion of system and particles. This along with Unit IV and Unit VI have a total weightage of 17 marks, which implies students will definitely find questions from this chapter. This is purely an understanding based chapter which talks about the system of particles and rotational motion. The topics extensively covered in the notes of physics class 11 chapter 6 are:

• Centre of mass and its motion.

• Centre of mass of a 2 particle system, rigid body and a uniform rod.

• Momentum of force and momentum conservation.

• Torque.

• Moment of Inertia.

• Angular momentum and laws of conservation of momentum.

• Radius of gyration.

• Parallel and perpendicular axis theorem.

Notes of System of Particles and Rotational Motion

Given below are the brief explanations of some important concepts and topics covered in the chapter. For an in-depth understanding of the same, refer to notes of system of particles and rotational motion class 11.

• Centre of Mass

The centre of mass of a body is the point at which the entire mass of the body is said to be concentrated at. It is also defined as the balancing point of the system. If any external force is to be applied at the centre of mass, the body is said to remain unaffected. It means that the body will stay in rest if at rest and there will be no change in the velocity of the body in motion.

For the centre of mass of a two particle system at motion,

\[\overline{v} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}\]

\[\overline{a} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}}\]

Where,

\[\overline{v}\] = Velocity of the centre of mass.

\[\overline{a}\] = Acceleration of the centre of mass.

• Torque

The turning effect of force about a fixed axis is defined as Torque. It can also be defined as the Moment of force. The SI unit of Torque is Nm.

\[\tau = \overline{r} \times \overline{F} = rF Sin \theta\]

Where,

τ = Moment of force or torque

r̅ = Perpendicular distance

F̅ = Force

θ = Angle between the two vectors r and F

Torque can also be calculated in  the terms of angular moment. The relationship between torque and angular momentum is defined by:

τ = dL/dt

• Moment of Inertia

Moment of Inertia is the phenomenon by which the body in motion opposes the change in its rotational motion. Mathematically, it is defined as the product of mass of particles and their distance from the axis of rotation. It is also called the rotational inertia of the body.

Importances of Systems of Particles and Rotational Motion Class 11 Notes CBSE Physics Chapter 6 (Free PDF Download)

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Conclusion 

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