Revision: Motion of System of Particles and Rigid Body Physics (Theory) ISC (Science) ISC Class 11 CISCE

"Centre of mass is a point about which the summation of moments of masses in the system is zero."

OR

A point at which the total mass (MM) of a finite body is supposed to be concentrated, and about which the summation of moments of masses in the system is zero, is called the centre of mass.

Torque is defined as the moment of the externally applied force about a point or axis of rotation. The expression for torque is, `vectau = vecr xx vecF`.

i. For n particle system,

Position vector \[\vec r_{C.M.}\] = \[\frac{\sum_{i}^{n}m_{i}\overset{\to}{\operatorname*{r_{i}}}}{\sum_{i}^{n}m_{i}}=\frac{\sum_{i}^{n}m_{i}\overset{\to}{\operatorname*{r_{i}}}}{M}\]

ii. For continuous distribution,

Positive Vector \[\vec r_{C.M}\] = \[\frac {\int{\vec r}\mathrm dm}{M}\]

A body's moment of inertia along an axis is equal to the product of two things: Its moment of inertia about a parallel axis through its centre of mass and the product of the body's mass and the square of the distance between the two axes. This is known as the parallel axis theorem.

Proof: Let I_CM represent a body of mass M moment of inertia (MI) about an axis passing through its centre of mass C and let I stand for that body's MI about a parallel axis passing through any point O. Let h represent the separation of the two axis.

Think about the body's minuscule mass element dm at point P. It is perpendicular to the rotation axis through point C and to the parallel axis through point O, with a corresponding perpendicular distance of OP. CP² dm is the MI of the element about the axis through C. As a result, `I_(CM) = int CP^2  dm` is the body's MI about the axis through the CM. In a similar vein, `I = int OP^2  dm` is the body's MI about the parallel axis through O.

Draw PQ perpendicular to OC produced, as shown in the figure. Then, from the figure,

`I = int OP^2  dm`

= `int (OQ^2 + PQ^2) dm`

= `int [(OC + CQ)^2 + PQ^2] dm`

= `int (OC^2 + 2OC.CQ + CQ^2 + PQ^2) dm`

= `int (OC^2 + 2OC.CQ + CP^2)dm`   ...(∵ CQ² + PQ² = CP²)

= `int OC^2  dm + int 2OC.CQ  dm + int CP^2  dm`

= `OC^2 int dm + 2OC int CQ  dm + int CP^2  dm`

Since OC = h is constant and `int dm = M` is the mass of the body,

`I = Mh^2 + 2h int CQ  dm + I_(CM)`

The integral `int CQ  dm` now yields mass M times a coordinate of the CM with respect to the origin C, based on the concept of the centre of mass. This position and the integral are both zero because C is the CM in and of itself.

∴ I = I_CM + Mh²

This proves the theorem of the parallel axis.

Statement: The moment of inertia (I_o​) of an object about any axis is equal to the sum of its moment of inertia (I_c) about an axis parallel to the given axis and passing through the centre of mass, and the product of the mass of the object and the square of the distance between the two axes.

where M = mass of the object, h = distance between the two parallel axes.

Statement: The moment of inertia (I_z​) of a laminar object about an axis (z) perpendicular to its plane is equal to the sum of its moment of inertias about two mutually perpendicular axes (x and y) in its plane, all three axes being concurrent.

• Thin ring / Hollow cylinder (Central axis) → I = MR²
• Thin ring (Diameter) → I = \[\frac {1}{2}\]M R²
• Annular ring / Thick hollow cylinder (Central) → I = \[\frac{1}{2}M (R_1^2+R_2^2)\]
• Uniform disc / Solid cylinder (Central) → I = \[\frac {1}{2}\]M R²
• Uniform disc (Diameter) → I = \[\frac {1}{4}\]M R²
• Solid sphere (Central) → I = \[\frac {2}{5}\]M R²
• Thin walled hollow sphere (Central) → I = \[\frac {2}{3}\]M R²
• Thin rod (Centre, ⊥ to length) → I = \[\frac {1}{12}\]M L²
• Thin rod (One end, ⊥ to length) → I = \[\frac {1}{3}\]M L²
• Solid cone (Central) → I = \[\frac {3}{10}\]M R², Hollow cone (Central) → I = \[\frac {1}{2}\]M R²