System Of Particles And Rotational Motion Class 11 Notes Physics Chapter 7 - CBSE
Chapter : 7
What Are System Of Particles And Rotational Motion ?
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Centre Of Mass
Centre of mass is the point at which entire mass of a system may be supposed to be concentrated.
$$\text{If}\space\vec{r}_{1}\space\text{and}\space\vec{r}_{2}\space\text{are the position vectors}$$
of two particles of masses m1 and m2, then the position vector of their centre of mass is
$$\text{R}_{cm} =\frac{m_{1}r_{1} + m_{2}r_{2}}{m_{1} + m_{2}}$$
Rigid Body
A rigid body is one whose constituent particles retain their relative positions even when they more under the action of an external force.
- Equation of rotational motion.
$$\text{(i)\space}\omega = \omega_{0} + \alpha t\\\text{(ii)\space} \theta =\omega_{0}t + \frac{1}{2}\alpha t^{2}\\\text{(iii)\space}\omega^{2} -\omega_{0}^{2} = 2\alpha\theta$$
Torque
The turning effect of a force about the axis of rotation is called moment of force or torque due to the force.
$$\vec{\tau} = \vec{r}×\vec{\text{F}}$$
Couple
A pair of equal and opposite forces acting on a body along two different lines of action constitute a couple.
Work Done By A Torque
dw = τ dθ
Power of a torque is given by,
$$\text{P} =\frac{dw}{dt} =\tau\frac{d\theta}{dt} =\tau\omega$$
Angular Momentum
It is the moment of linear momentum of a particle about the axis of rotation.
Angular momentum = Linear momentum × its perpendicular distance from the axis of rotation.
$$\vec{L} = \vec{r}× \vec{p}$$
Relation Between Torque And Angular Momentum
$$\vec{\tau} =\frac{\vec{\text{dL}}}{\vec{\text{dt}}}$$
Moment Of Inertia
The moment of inertia of a rigid body about an axis of rotation is the sum of the products of masses of the various particles and squares of their perpendicular distances from the axis of rotation. Mathematically,
$$\text{I} =\displaystyle\sum^{n}_{\text{i = 1}}m_{i}r_{i}^{2}$$
SI unit of moment of inertia = Kg m2
Radius Of Gyration
It may be defined as the distance from the axis of rotation at which if whole mass of the body were supposed to be concentrated, the moment of inertia would be same as with the actual distribution of mass.
$$\text{I} = \text{Mk}^{2}, k =\sqrt{\bigg(\frac{\text{I}}{\text{M}}\bigg)}$$
Theorem Of Perpendicular Axes
It states that the moment of inertia of a plane lamina about its axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its plane and intersecting each other at the point, where the perpendicular axes pass through the lamina.
Iz = Ix + Iy
Theorem Of Parallel Axes
It states that the moment of inertia of a rigid body about any axis is equal to the sum of moment of inertia of the body about a parallel axis through its centre of mass Icmand the product of mass (m) of the body and the square of the perpendicular distance (h) between the parallel axes.
I = Icm + Mh2
Rotational K.E.
$$\text{Rotational K.E. =}\frac{1}{2}\text{I}\omega^{2}$$
Total K.E. Of A Rolling Body
Total K.E. = Translational K.E. + Rotational K.E.
$$=\frac{1}{2}\text{Mv}^{2} + \frac{1}{2}\text{I}\omega^{2}$$
Relation Between M.I And Angular Momentum
Angular momentum = M.I. × Angular velocity
L = Iω
Relation Between M.I And Torque
Torque = M.I. × Angular acceleration
i.e. τ = I α
