Revision: Class 11 >> Rotational Motion NEET (UG) Rotational Motion

Definitions [13]

Definition: Centre of Mass

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

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.

Definition: Vector Product of two vectors

The vector product of two non-null and non-parallel vectors a and b is expressed as:

a × b = |a||b| sinθ n̂ = ab sinθ n̂

The unit vector n̂ along a × b is given by:

\[\hat{\mathbf{n}}=\frac{\mathbf{a}\times\mathbf{b}}{|\mathbf{a}\times\mathbf{b}|}\]

Definition: Angular Momentum

The product of the linear momentum of a body and the perpendicular distance of the line of action of the linear momentum vector from the axis of rotation is called Angular Momentum.

Definition: Angular Acceleration

The rate of change of angular velocity, i.e., α = dω/dt, is called Angular Acceleration. Average angular acceleration is \[\overline α\] = (ω₂ − ω₁)/(t₂ − t₁).

Definition: Angular Displacement

The change in the angle traced by the position vector of a particle about a fixed point when the particle moves along a curved path is called Angular Displacement. Its unit is radian (rad).

Definition: Torque (τ) — Moment of Force

The turning effect of a force about an axis of rotation, given as the product of the magnitude of force and the perpendicular distance of the line of action of the force from the axis of rotation, is called Torque (or Moment of Force).

\[\vec{\tau}=\vec{r}\times\vec{F},\quad\tau=rF\sin\theta\]

Definition: Angular Velocity

The angular displacement per unit time, i.e., \[\overline ω\] = Δθ/Δt, is called Angular Velocity. It is a vector quantity whose direction is normal to the rotational plane given by the right-hand screw rule.

Define the centre of gravity of a body.

The point through which the resultant of the weights of all the particles of the body acts is called its centre of gravity.

The centre of gravity is an imaginary location where the body’s whole weight is assumed to be concentrated.

Definition: Centre of Gravity

The Centre of Gravity (c.g.) of a body is the point around which the resultant torque due to the force of gravity on the body is zero.

The centre of gravity (C.G.) of a body is the point about which the algebraic sum of moments of the weights of all the particles constituting the body is zero. The entire weight of the body can be considered to act at this point, howsoever the body is placed.

The point about which the resultant torque due to force of gravity on the body is zero is called the centre of gravity.

Definition: Moment of Inertia

The new parameter characterising the mass of a rigid body and its distribution with respect to the axis of rotation, which represents the resistance of a body to change in its rotational motion (analogous to mass in linear motion), is called Moment of Inertia.

I = ∑m_ir_i²

Definition: Rolling Motion

When a body performs translational motion as well as rotational motion simultaneously, this type of motion is called rolling motion.

The combination of rotational and translational motion of a round-shaped body when the body is in contact with a surface, such that every particle of the body has two velocities (one due to rotation and one due to translation) whose vector sum gives the resultant velocity, is called Rolling Motion.

Definition: Radius of Gyration

The distance between the axis of rotation and a point at which the entire mass of the body can be supposed to be concentrated so as to possess the same moment of inertia as that of the body about that axis is called the radius of gyration of a rigid body about a given axis of rotation.

K = \[\sqrt {\frac {I}{M}\], I = MK²

Define radius of gyration.

The radius of gyration of a body is defined as the distance between the axis of rotation and a point at which the whole mass of the body is supposed to be concentrated, so as to possess the same moment of inertia as that of the body.

Formulae [8]

Formula: Centre of Mass

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

Formula: Centre of Mass for Coordinates of CM for n Particles

\[X_{cm}=\frac{m_1x_1+m_2x_2+\ldots+m_nx_n}{m_1+m_2+\ldots+m_n}\]

Formula: Centre of Mass for Continuous Mass Distribution

\[x_{cm}=\frac{\int xdm}{\int dm}\]

\[y_{cm}=\frac{\int ydm}{\int ym}\]

Formula: Centre of Mass for For Two-Particle System

\[\vec{r}_{cm}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}\]

Formula: Angle between Two Vectors(Cross)

\[\sin\theta=\frac{\left|\overline{a}\times\overline{b}\right|}{\left|\overline{a}\right|\left|\overline{b}\right|}\]

Formula: Angular Momentum

L = Iω = mvr sin θ

Formula: Velocity at Point

Velocity at point A (top): v_A = v + Rω = 2v (or \[\sqrt 2\]Rω)
Velocity at point B (side): v_B = \[\sqrt {v^2+(Rω)^2}\] = \[\sqrt 2\]Rω
Velocity at point C (bottom): v_C = v − Rω = 0
Velocity at point D (side): vD = \[\sqrt {v^2+(Rω)^2}\] = \[\sqrt 2\] Rω

Formula: Total Kinetic Energy of Rolling

Total KE = Translational KE + Rotational KE

Theorems and Laws [6]

State and prove: Law of conservation of angular momentum.

Statement:

The angular momentum of a body remains constant if the resultant external torque acting on the body is zero.

I₁ω₁= I₂ω₂(when τ = 0)

Here I is the moment of inertia and ω is angular velocity.

Proof:

Consider a particle of mass m, rotating about an axis with torque ‘τ’.

Let `vecp` be the linear momentum of the particle and `vecr` be its position vector.

∴ Angular momentum, `vecL = vecr xx vecp` .....(1)

Differentiating equation (1) with respect to time t, we get,

`(dvecL)/(dt) = d/dt(vecr xx vecp) = vecr(dvecp)/(dt) + vecp(dvecr)/(dt)`

We know that, `(dvecp)/(dt) = vecF, (dvecr)/(dt) = vec"v", vecp = mvec"v"`

∴ `(dvecL)/(dt) = vecr xx vecF + m(vec"v" xx vec"v")`

∴ `(dvecL)/(dt) = vecr xx vecF` ....`(∵ vec"v" xx vec"v" = 0)`

∴ `(dvecL)/(dt) = vectau` ......`(∵ vecr xx vecF = vectau)`

Now, If the `vectau = 0`, then

`(dvecL)/(dt) = 0`

∴ `vecL` is constant. Hence angular momentum remains conserved.

Example:

An athlete diving off a high springboard can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall, his angular momentum remains constant.

Law: Conservation of Angular Momentum

Statement: The total angular momentum of an isolated system remains constant (conserved) if no resultant external torque acts on it.

Mathematical Form:

If \[\vec τ_{ext}\] = 0, then:

\[\frac {d\vec L}{dt}\] = 0 \[\vec L\] = constant

I₁ω₁ = I₂ω₂

Key Points:

Angular momentum \[\vec L\] = I\[\vec ω\]
When net external torque is zero, angular momentum remains constant.
Angular impulse = net torque × time = change in angular momentum: τ ⋅ Δt = ΔL

Law: Principle of Moments

Statement:

In equilibrium, the sum of anticlockwise moments equals the sum of clockwise moments about the pivot.

Explanation/Proof:

When several forces act on a pivoted body, they tend to rotate it about an axis passing through the pivot. The resultant moment is obtained by taking the algebraic sum of the moments of all the forces about the pivoted point. By convention, anticlockwise moments are taken as positive and clockwise moments as negative.

A metre rule is suspended at its centre (point O). Two weights W₁ and W₂ are hung on either side at distances l₁ and l₂ using spring balances.

W₁ creates a clockwise moment = W₁ × l₁
W₂ creates an anticlockwise moment = W₂ × l₂

By adjusting the weights or positions, the rule becomes horizontal (in equilibrium).

Conclusion:

At balance, W₁ × l₁ = W₂ × l₂, which confirms the principle of moments.

State and prove the theorem of the parallel axis about the moment of inertia.

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.

Theorem: Perpendicular Axis Theorem

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.

I_z = I_x + I_y

Theorem: Parallel Axis Theorem

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.

I_o = I_c + Mh²

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

Key Points

Key Points: Vector Product of two vectors

1. Determinant form:

If \[\overline{\mathrm{a}}=\mathrm{a}_{1}\hat{\mathrm{i}}+\mathrm{a}_{2}\hat{\mathrm{j}}+\mathrm{a}_{3}\hat{\mathrm{k}}\] and \[\overline{\mathrm{b}}=\mathrm{b}_1\hat{\mathrm{i}}+\mathrm{b}_2\hat{\mathrm{j}}+\mathrm{b}_3\hat{\mathrm{k}}\], then

\[\overline{\mathrm{a}}\times\overline{\mathrm{b}}= \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 \\ \mathbf{b}_1 & \mathbf{b}_2 & \mathbf{b}_3 \end{vmatrix}\]

2. Condition for zero cross product:

a × b = 0 ⇒ vectors are parallel (or one is zero)

Key Points: Centre of Gravity

The weight of a body acts through a single point called the centre of gravity (C.G.), where the sum of moments of all particles' weights is zero.
The position of the C.G. depends on the shape and mass distribution of the body and changes if the body is deformed.
The C.G. may lie outside the material of the body (e.g., a ring or hollow sphere).
A body balances when supported exactly at its centre of gravity, as seen in a metre rule or square lamina.
The C.G. of an irregular lamina can be found by suspending it from multiple points and tracing the intersection of plumb line paths.

Key Points: M.I. of Symmetrical Objects

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

Mechanical Energy Dynamics, Power, and Collisions Gravitational Phenomena: Laws, Effects and Applications

Revision: Class 11 >> Rotational Motion NEET (UG) Rotational Motion

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Definitions [13]

Formulae [8]

Theorems and Laws [6]

Statement:

Explanation/Proof:

Conclusion:

Key Points

Concepts [24]

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