Relation between Angular Momentum and Moment of inertia

When a body moves along a straight line, then the product of the, mass m and the linear velocity v of the body is called the “linear momentum” p of the body (p = m * v), if a body is rotating about an axis, then the sum of the moments of the linear momenta of all the particles about the given axis is called the “angular momentum” of the body about that axis. It is represented by ‘J’.

Let a body be rotating about an axis with an angular velocity ω. All the particles of the body will have the same angular velocity, but their linear velocities will be different. Let a particle be at a distance r₁ from the axis of rotation. The linear velocity of this particle is given by

v₁ = r₁ ω

If the mass of the particle be m₁, then its linear momentum = m₁ v₁.

The moment of this momentum about the axis of rotational

= momentum * distance

= m₁ v₁ * r₁

= m₁ (r₁ ω) * r₁

= m₁ r₁² ω

Similarly, if the masses of other particles be m₂, m₃, … and their respective distances from the axis of rotation be r₂, r₃, … the moments of their linear momenta about the axis of rotation will be m₂ r₂² ω, m₃ r₃² ω, … respectively. The sum of the moments of linear momenta of all particles, that is, the angular momentum of the body is given by

L = m₁ r₁² ω + m₂ r₂² ω + m₃ r₃² ω + …

= (m₁ r₁² ω + m₂ r₂² ω + m₃ r₃² ω + …) ω

= (∑mr²) ω.

But ∑mr² is the moment of interia I of the body about the axis of rotation. Hence the angular momentum of the body about the axis of rotation is

L = I * ω

The S.I. unit of angular momentum is kgm²s^-1 of Js. Its dimensional formula is [ML²T^-1].

It is clear from the above formula that just as the linear momentum (m v) of a body is equal to the product of the mass of the body and its linear velocity; in the same way the angular momentum (I ω) of a body about an axis is equal to the product of the moment of interia of the body and its angular velocity about that axis.

Product of determinants

Let  and  be two determinants. Then, the product Δ₁Δ₂ is defined as 

This is row-by-column multiplication value for finding the product of two determinants and it is same as the rule of multiplication of two matrices. Since the value of a determinant does not alter by interchanging the rows and columns. So, we can also follow the row-by-row or the column –by-row or column-by-column multiplication rule.

Ex:

If . Show that 

Sol:

We have,

 

Laws of chemical combinations

1. Law of conservation of mass:
   Given by Lavoisier: “In all chemical changes, the total mass of the system remains constant.”i.e., Matter can neither be created nor destroyed. Ex: 2H2 + O2 = 2H2O
   Reactants mass = 4+32 = 36g
   Products mass = 2 X (2+16) = 36g
   Reactants mass = Products mass
   
2. Law of definite or constant proportions:
   Given by Proust: “A given compound always contains its constituent elements in the same ratio by weight.”i.e., the method of preparation of a compound does not affect the combining ratio of the elements (weight).
   
3. Law of multiple proportions:
   Given by Dalton: “When two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers”.
   For example, there are five distinct oxides of nitrogen, and the weights of oxygen in combination with 14 grams of nitrogen are, in increasing order, 8, 16, 24, 32 and 40 grams, or in a ratio of 1, 2, 3, 4 and 5.
   
4. Law of gaseous volumes:
   Given by Gay-Lussac: “The volume of gases taking part in a chemical reaction show simple whole number ratios to one another when those volumes are measured at the same temperature and pressure”.
   1 liter of nitrogen gas reacts with 3 litres of hydrogen gas to produce 2 litres of ammonia gas
   N2 (g) + 3H2 (g) → 2NH3 (g)
   Since all the reactants and products are gases, the mole ratio of
   N2 (g): H2 (g): NH3 (g) of 1:3:2 is also the ratio of the volumes of gases.
   So, 10 mL of nitrogen gas would react with 10 x 3 = 30 mL of hydrogen gas to produce 10 x 2 = 20 mL ammonia gas
   
5. Avogadro’s law:
   Given by Avogadro: Under the same conditions of temperature and pressure, equal volumes of different gases contain an equal number of molecules.
   The volume occupied by one gram-mole of gas is about 22.4 L (0.791 cubic feet) at standard temperature and pressure (0 °C, 1 atmosphere) and is the same for all gases, according to Avogadro’s law and they all contain 6.023 x 1023 molecules.

Projectile motion

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown near the earth's surface, and it moves along a curved path under the action of gravity only.

Motion along Horizontal direction:

A body projected from the ground with a velocity u at an angel  with the horizontal.

Horizontal components of the velocity at time t is v_x = u cos θ

The horizontal distances covered in time t is x = (u cos θ) t

Motion along Vertical direction:

Vertical components of the velocity at time t is v_y = u sinθ - gt

The vertical distance covered in time t is  

y = (u sin θ) t – ½ gt²

Equation of trajectory:  

y = (tan θ) x – gx² / 2u² cos²θ

It is in the form of y = Ax + Bx², which represents a parabola (where A = tan , B = - g / 2u² cos²θ)

Resultant velocity at time t:

Magnitude of resultant velocity at time t is v = (v²_x + v²_y) ^½

The angle  subtended by the resultant velocity vector with the horizontal is given by tan α = v_y / v_x

Time of flight (T):

We know the general equation in vertical direction is y = (u sin θ) t – ½ gt²

For time of flight y=0
0= u sin θT – ½ gT²

U sinθ = gT/2
T = 2usinθ /g

          
Maximum height attained:

We know the general equation in vertical direction is V²_y – U²_y = 2a_ys

For maximum height 0 – u² sin²θ = 2 (-g) h_max

h_max = u² sin²θ/2g

Horizontal range:

We know the general equation in vertical direction is x = u cos θt

For range t=T

R = ucosθ * 2usinθ/g

R = u² sin (2θ)/g.

Application:

•  The horizontal range is the same for angles θ and (90⁰ - θ).
• The horizontal range is maximum for θ = 45⁰. R_max = u²/g.
• When horizontal range is maximum, h_max = R_max /4.
• At the point of projection, KE = ½ mu², PE = 0. Total energy E = ½ mu².
• At the highest point, KE = ½ mu² cos²θ and PE = total energy – KE = ½ mu² - ½ mu² cos²θ  = ½ mu² sin²θ
• To find R and  from the equation of trajectory y = ax – bx² Where a and b are constants, refer to this figure.
  
  
• At O and B, y = 0. Putting y = 0 in the above equation, we have 0 = ax - bx² => x = 0, x = a/b. Therefore R = a/b
• At A y = h_max and x = R/2 = a/2b. Using these values in y = ax - bx², we get h_max = a²/4b
• If A and B are two points at the same horizontal level on a trajectory at a height h from the ground,
  
  
• t_f  = 2usinθ/g = t₁ + t₂
• h = ½ g t₁t₂
• Average velocity during time interval (t₂ – t₁) is v_av = ucosθ (·.· during this interval, the vertical displacement is zero).

Minor and Co-factors of determinants

Minor:

Let A = [a_ij] be a square matrix of order n. Then the minor M_ij of a_ij in A is the determinant of the square sub-matrix of order (n - 1) obtained by leaving i^th row and j^th column of A.

Let

M₁₁ = Minor of

M₁₂ = Minor of  

M₁₃ = Minor of  

M₂₁ = Minor of  

Similarly for the elements a₂₂, a₂₃, a₃₁, a₃₂ and a₃₃.

Cofactor:

Let A = [a_ij] be a square matrix of order n. Then the cofactor of C_ij of a_ij in A is equal to  (- 1)^i+j times the determinant of the sub-matrix of order (n - 1) obtained by leaving i^th row and j^th column of A.

C_ij = cofactor of a_ij in A.

= (- 1) ^i+j M^ij, where M_ij is minor of a_ij in A.

Balancing Redox Reactions

A)     Oxidation number method

1.   Write correct formula for each reactant & product.

Ex: K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2

2.   Write oxidation numbers for all elements

K+12Cr+62O-27 + H+1Cl-1 → K+1Cl-1 + Cr+3Cl-13 + H+12O-2 + Cl02

3.   Calculate the increase or decrease in oxidation number per atom

• Oxidation number of Cr decreased from +6 to +3
  K2Cr+62O7 → 2Cr+32Cl3

• Oxidation number of Cl increased from -1 to 0
  HCl-1 → Cl02

4.   The decrease in oxidation number should be equal to increase in oxidation number

• Decrease in Oxid. num of Cr = 6 units per molecule of K2Cr2O7. Increase in oxid. num of Cl = 1 unit per molecule of HCl
  We need 6 HCl for 6 units increase in oxid num. so the reaction becomes

K2Cr2O7 + 6 HCl → 2CrCl3 + 3Cl2

5.   Balance the remaining atoms. Potassium, Hydrogen & oxygen in this case.

K2Cr2O7 + 14 HCl → +2KCl + 2CrCl3 + 7 H2O + 3Cl2

The reaction is balanced.

6.   If the reacting is taking place in water, add H+ or OH- ions on the appropriate side so that the total ionic charges of reactants & products are equal.

7.   If reaction is carried out in acidic solution, use H+ ions in the equation.

If in basic solution, use OH- ions.

B)     Half reaction method

The 2 half reactions are balanced separately and then added together
Let’s balance the following reaction
H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)

Step 1

• Write half-cell reactions
  

1. Oxidation reaction Fe2+ (aq) → Fe3+ (aq) → eq (i)
2. Reduction reaction H2O2 (aq) → H2O (l) → eq (ii)

Step 2

• Balance the atoms other than O & H in each half reaction.
  Here Fe atoms are already balanced.

Step 3

• For acidic medium, add H2O to balance O atoms and H+ to balance H atoms
• For basic medium, add OH- to balance H atoms and H2O to balance O atoms.
  The given reaction is taking place in acidic medium so to balance O atoms we add H2O & to balance H atoms we add H+

.·. Eq (ii) becomes

H2O2 + 2H+ → H2O + H2O

=> H2O2 + 2H+ → 2H2O

Step 4

• To balance the charges, add electrons to the side deficient of electrons

Fe2+ → Fe3+ + e-             → (iii)

H2O2 + 2H+ + 2e- → 2H2O         → (iv)

Step 5

• Make the number of electrons in the two half reactions equal by multiplying the reactions by appropriate coefficients. Here, multiply equation (iii) by 2

2Fe2+ → 2Fe3+ + e-

Step 6

• Add the two half-cell reactions

2Fe2+ + H2O2 + 2H+ → 2Fe3+ + 2H2O

Step 7

• Verify that the equation contains same type and number of atoms and same charges on both equations. If not, the equation is not balanced so check the steps again.
  Thus, the balanced equation is

2Fe2+ (aq) + H2O2 (aq) + 2H+ → 2Fe3+ (aq) + 2H2O (l)