Question

A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is its volume. The linear density of the rope µ is defined to be the mass per unit length, m/L, and can be written in the form µ = rho x A y . Using dimensional analysis, determine which expression has the correct powers of x and y.

Answer 1

The linear density of a rope is given by µ = ρ x A^y. By analyzing the units, we can determine that the correct expression is µ = ρ x A^(1/2), meaning the linear density is proportional to the square root of the cross-sectional area.

Step-by-step explanation:

The linear density of a rope is defined as the mass per unit length, m/L. In this case, the linear density is given by µ = ρ x A^y, where ρ is the density of the rope, A is the cross-sectional area, and y is a constant. To determine the values of x and y using dimensional analysis, we can look at the units of each term. The units of ρ are kg/m^3, the units of A are m^2, and the units of m/L are kg/m. Since µ = ρ x A^y, the units of µ are (kg/m^3) x m^(2y).

Comparing the units on both sides of the equation, we can see that y must be 1/2. Therefore, the correct expression is µ = ρ x A^(1/2), which means the linear density is proportional to the square root of the cross-sectional area.

Answer 2

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹