In the previous section, we saw how to convert the units from one system to another using dimensional analysis. In this section, we will see another application of dimensional analysis, which is: Deducing relation among physical quantities
We know that the basic physical quantities like mass, length and time do not depend on other physical quantities.
We can say that, mass, length, time etc., are independent physical quantities
But the derived physical quantities like velocity, force etc., are dependent on other physical quantities
We will want to know the 'nature of this dependence'
A dimensional analysis can help us to achieve it
For using dimensional analysis in this regard, two conditions must be satisfied:
(i) The physical quantity must not depend on more than 3 physical quantities
That is., there must not be more than 3 quantities on the right side
(ii) The dependence must be a 'product type'
That is., the physical quantities on the right side must be multiplied together
Let us see an example:
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Solution:
1. Let the time period be denoted as T
2. Given that T depends on l, g and m
• But we do not know the powers to which 'l, g and m' have to be raised
3. We can write the dependence as: T = k × lx × gy × mz
♦ Where 'k' is a dimensionless constant
4. Considering the dimensions on both sides, we can write:
[M0L0T1] = [L1]x [LT-2]y [M]z
⇒ [M0L0T1] = [Lx] [Ly T-2y] [Mz]
⇒ [M0L0T1] = [Lx+y] [T-2y] [Mz]
5. Equating the dimensions on both sides, we get 3 linear equations:
(i) x+y = 0
(ii) -2y = 1
(iii) z = 0
• From (ii), we get: y = -1⁄2
• Substituting this in (i), we get: x - 1⁄2 = 0
⇒ x = 1⁄2
■ So we can write:
• x = 1⁄2
• y = -1⁄2
• z = 0
6. Substituting these in (3),we get:
$\mathbf\small{T=k\times l^{\frac{1}{2}}\times g^{-\frac{1}{2}}}$
$\mathbf\small{\Rightarrow T=k \sqrt{\frac{l}{g}}}$
7. Note that, the value of 'k' cannot be obtained by the method of dimensions
• Multiplying by a number on the right side does not affect the dimensions
We will now see some solved examples
Solved example 2.42
The centripetal force F acting on a particle moving uniformly in a circle depends upon mass (m), velocity (v) and radius (r) of the circle. Derive the expression for F using method of dimensions
Solution:
1. Given that F depends on m, v and r
• But we do not know the powers to which 'm, v and r' have to be raised
2. We can write the dependence as: F = k × mx × vy × rz
♦ Where 'k' is a dimensionless constant
3. Considering the dimensions on both sides, we can write:
[M1L1T-2] = [M1]x [LT-1]y [L]z
⇒ [M1L1T-2] = [Mx] [Ly T-y] [Lz]
⇒ [M1L1T-2] = [Mx] [Ly+z] [T-y]
4. Equating the dimensions on both sides, we get 3 linear equations:
(i) x = 1
(ii) y+z = 1
(iii) y = 2
Substituting 'y=2' in (ii), we get: z = -1
■ So we can write:
• x = 1
• y = 2
• z = -1
5. Substituting these in (2),we get:
$\mathbf\small{F=k\times m^1\times v^2\times r^{-1}}$
$\mathbf\small{F=k \frac{mv^2}{r}}$
• We will see the actual value of 'k' when we learn about centripetal force in a later chapter
Solved example 2.43
A gas bubble formed from an explosion under water oscillates with a period T. This period T is proportional to the pressure (p), density of water (d) and the energy of explosion (E). Derive the expression for T using method of dimensions
Solution:
1. Given that T depends on p, d and E
• But we do not know the powers to which 'p, d and E' have to be raised
2. We can write the dependence as: T = k × px × dy × Ez
♦ Where 'k' is a dimensionless constant
3. Considering the dimensions on both sides, we can write:
[M0L0T1] = [M1L-1T-2]x [M1L-3]y [M1L2T-2]z
⇒ [M0L0T1] = [MxL-xT-2x] [MyL-3y] [MzL2zT-2z]
⇒ [M0L0T1] = [Mx+y+z] [L-x-3y+2z] [T-2x-2z]
4. Equating the dimensions on both sides, we get 3 linear equations:
(i) x+y+z = 0
(ii) -x-3y+2z = 0
(iii) -2x-2z = 1
• We have 3 equations and 3 unknowns
• Solving them, we get:
x = -5⁄6
y = 1⁄2
z= 1⁄3.
We know that the basic physical quantities like mass, length and time do not depend on other physical quantities.
We can say that, mass, length, time etc., are independent physical quantities
But the derived physical quantities like velocity, force etc., are dependent on other physical quantities
We will want to know the 'nature of this dependence'
A dimensional analysis can help us to achieve it
For using dimensional analysis in this regard, two conditions must be satisfied:
(i) The physical quantity must not depend on more than 3 physical quantities
That is., there must not be more than 3 quantities on the right side
(ii) The dependence must be a 'product type'
That is., the physical quantities on the right side must be multiplied together
Let us see an example:
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Solution:
1. Let the time period be denoted as T
2. Given that T depends on l, g and m
• But we do not know the powers to which 'l, g and m' have to be raised
3. We can write the dependence as: T = k × lx × gy × mz
♦ Where 'k' is a dimensionless constant
4. Considering the dimensions on both sides, we can write:
[M0L0T1] = [L1]x [LT-2]y [M]z
⇒ [M0L0T1] = [Lx] [Ly T-2y] [Mz]
⇒ [M0L0T1] = [Lx+y] [T-2y] [Mz]
5. Equating the dimensions on both sides, we get 3 linear equations:
(i) x+y = 0
(ii) -2y = 1
(iii) z = 0
• From (ii), we get: y = -1⁄2
• Substituting this in (i), we get: x - 1⁄2 = 0
⇒ x = 1⁄2
■ So we can write:
• x = 1⁄2
• y = -1⁄2
• z = 0
6. Substituting these in (3),we get:
$\mathbf\small{T=k\times l^{\frac{1}{2}}\times g^{-\frac{1}{2}}}$
$\mathbf\small{\Rightarrow T=k \sqrt{\frac{l}{g}}}$
7. Note that, the value of 'k' cannot be obtained by the method of dimensions
• Multiplying by a number on the right side does not affect the dimensions
We will now see some solved examples
Solved example 2.42
The centripetal force F acting on a particle moving uniformly in a circle depends upon mass (m), velocity (v) and radius (r) of the circle. Derive the expression for F using method of dimensions
Solution:
1. Given that F depends on m, v and r
• But we do not know the powers to which 'm, v and r' have to be raised
2. We can write the dependence as: F = k × mx × vy × rz
♦ Where 'k' is a dimensionless constant
3. Considering the dimensions on both sides, we can write:
[M1L1T-2] = [M1]x [LT-1]y [L]z
⇒ [M1L1T-2] = [Mx] [Ly T-y] [Lz]
⇒ [M1L1T-2] = [Mx] [Ly+z] [T-y]
4. Equating the dimensions on both sides, we get 3 linear equations:
(i) x = 1
(ii) y+z = 1
(iii) y = 2
Substituting 'y=2' in (ii), we get: z = -1
■ So we can write:
• x = 1
• y = 2
• z = -1
5. Substituting these in (2),we get:
$\mathbf\small{F=k\times m^1\times v^2\times r^{-1}}$
$\mathbf\small{F=k \frac{mv^2}{r}}$
• We will see the actual value of 'k' when we learn about centripetal force in a later chapter
Solved example 2.43
A gas bubble formed from an explosion under water oscillates with a period T. This period T is proportional to the pressure (p), density of water (d) and the energy of explosion (E). Derive the expression for T using method of dimensions
Solution:
1. Given that T depends on p, d and E
• But we do not know the powers to which 'p, d and E' have to be raised
2. We can write the dependence as: T = k × px × dy × Ez
♦ Where 'k' is a dimensionless constant
3. Considering the dimensions on both sides, we can write:
[M0L0T1] = [M1L-1T-2]x [M1L-3]y [M1L2T-2]z
⇒ [M0L0T1] = [MxL-xT-2x] [MyL-3y] [MzL2zT-2z]
⇒ [M0L0T1] = [Mx+y+z] [L-x-3y+2z] [T-2x-2z]
4. Equating the dimensions on both sides, we get 3 linear equations:
(i) x+y+z = 0
(ii) -x-3y+2z = 0
(iii) -2x-2z = 1
• We have 3 equations and 3 unknowns
• Solving them, we get:
x = -5⁄6
y = 1⁄2
z= 1⁄3.
We have completed this discussion on units and measurements. In the next chapter, we will see Motion in a straight line
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