System of Particles and Rotational Motion
CBSE Class 11 Physics • Chapter 06
Chapter Overview
This chapter covers the fundamental concepts of System of Particles and Rotational
Motion . Ensure you understand the definitions and derivations thoroughly.
06.2 Center of Mass (CM)
Definition: Point where entire mass of the system is supposed to be concentrated.
$$ \vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i} $$
Two particles: $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.
Continuous distribution: $\vec{R}_{CM} = \frac{1}{M} \int \vec{r} dm$.
Motion of CM:
$M \vec{V}_{CM} = \sum m_i \vec{v}_i = \vec{P}_{total}$.
$M \vec{A}_{CM} = \vec{F}_{ext}$. (Internal forces cancel out).
06.6 Angular Velocity & Acceleration
Angular Velocity ($\omega$): Rate of change of angular displacement. $\omega = d\theta/dt$. (Rad/s).Relation with Linear Velocity: $\vec{v} = \vec{\omega} \times \vec{r}$. ($v = r\omega$).Angular Acceleration ($\alpha$): $\alpha = d\omega/dt$. ($a_t = r\alpha$).
Q: Two particles of mass 1kg and 2kg are at (1,0) and (2,2) respectively. Find CM.
Ans: $x_{cm} = \frac{1(1) + 2(2)}{1+2} = \frac{5}{3}$.
06.7 Torque and Angular Momentum
Torque ($\vec{\tau}$): Turning effect of force. Moment of Force.$\vec{\tau} = \vec{r} \times \vec{F} = rF \sin \theta \hat{n}$. SI Unit: Nm.
Angular Momentum ($\vec{L}$): Moment of Linear Momentum.$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$. SI Unit: $kg m^2/s$.
O
r
F
$\theta$
Figure 6.1: Torque $\vec{\tau} = \vec{r} \times \vec{F}$
Relation: $\vec{\tau} = \frac{d\vec{L}}{dt}$. (Analogous to $F = dp/dt$).
Conservation of Angular Momentum: If $\tau_{ext} = 0$, then $\frac{dL}{dt} = 0 \Rightarrow L = \text{constant}$.
$I_1 \omega_1 = I_2 \omega_2$. (e.g., Ice skater spinning).
06.9 Moment of Inertia
Moment of Inertia ($I$): Analogue of mass in rotational motion. Measure of resistance to change in rotational motion.$I = \sum m_i r_i^2$. SI Unit: $kg m^2$.
Radius of Gyration ($k$): $I = Mk^2$.
Theorems:
Perpendicular Axes Theorem (Planar Body): $I_z = I_x + I_y$.Parallel Axes Theorem (Any Body): $I_{z'} = I_{cm} + Md^2$.
06.2 Center of Mass (CM)
Definition: Point where entire mass of the system is supposed to be concentrated.
$$ \vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i} $$
Two particles: $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.
Continuous distribution: $\vec{R}_{CM} = \frac{1}{M} \int \vec{r} dm$.
Motion of CM:
$M \vec{V}_{CM} = \sum m_i \vec{v}_i = \vec{P}_{total}$.
$M \vec{A}_{CM} = \vec{F}_{ext}$. (Internal forces cancel out).
06.6 Angular Velocity & Acceleration
Angular Velocity ($\omega$): Rate of change of angular displacement. $\omega = d\theta/dt$. (Rad/s).Relation with Linear Velocity: $\vec{v} = \vec{\omega} \times \vec{r}$. ($v = r\omega$).Angular Acceleration ($\alpha$): $\alpha = d\omega/dt$. ($a_t = r\alpha$).
Q: Two particles of mass 1kg and 2kg are at (1,0) and (2,2) respectively. Find CM.
Ans: $x_{cm} = \frac{1(1) + 2(2)}{1+2} = \frac{5}{3}$.
06.7 Torque and Angular Momentum
Torque ($\vec{\tau}$): Turning effect of force. Moment of Force.$\vec{\tau} = \vec{r} \times \vec{F} = rF \sin \theta \hat{n}$. SI Unit: Nm.
Angular Momentum ($\vec{L}$): Moment of Linear Momentum.$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$. SI Unit: $kg m^2/s$.
O
r
F
$\theta$
Figure 6.1: Torque $\vec{\tau} = \vec{r} \times \vec{F}$
Relation: $\vec{\tau} = \frac{d\vec{L}}{dt}$. (Analogous to $F = dp/dt$).
Conservation of Angular Momentum: If $\tau_{ext} = 0$, then $\frac{dL}{dt} = 0 \Rightarrow L = \text{constant}$.
$I_1 \omega_1 = I_2 \omega_2$. (e.g., Ice skater spinning).
06.9 Moment of Inertia
Moment of Inertia ($I$): Analogue of mass in rotational motion. Measure of resistance to change in rotational motion.$I = \sum m_i r_i^2$. SI Unit: $kg m^2$.
Radius of Gyration ($k$): $I = Mk^2$.
Theorems:
Perpendicular Axes Theorem (Planar Body): $I_z = I_x + I_y$.Parallel Axes Theorem (Any Body): $I_{z'} = I_{cm} + Md^2$.
06.11 Dynamics & Rolling Motion
Rotational Analogues:
Force $F \rightarrow$ Torque $\tau$.
Mass $m \rightarrow$ Inertia $I$.
Newton's 2nd Law: $\tau = I \alpha$.
Kinetic Energy: $K_{rot} = \frac{1}{2}I\omega^2$.
Rolling without Slipping:
Condition: $v_{cm} = R\omega$.
Total KE = Translational + Rotational.
$$ K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv_{cm}^2 (1 + \frac{k^2}{R^2}) $$
$v_{cm}$
$\omega$
Rolling Motion ($v = \omega R$)
Figure 6.2: Rolling Motion without Slipping
Q: A solid cylinder of mass 2kg and radius 0.1m rolls without slipping at 3 m/s. Find total KE.
Ans: For solid cylinder, $I = \frac{1}{2}MR^2 \Rightarrow k^2/R^2 = 1/2$.
Study Tip: Master the derivations in this chapter as they are frequently asked in exams.
Practice numericals based on the formulas.
