The aerodynamic drag force F acting on an object moving through a fluid of density \rho wi… — JAMB UTME

The aerodynamic drag force $F$ acting on an object moving through a fluid of density $\rho$ with cross-sectional area $A$ at speed $v$ is modeled by the equation $F = \frac{1}{2} C_d \rho^a A^b v^c$ , where $C_d$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the exponent $c$ ?

Cevap: 2

Cevap

The numerical value of the exponent

c

is 2.

By applying the principle of dimensional homogeneity, the base dimension of time on the left side is

\text{T}^{-2}

(from force

[F] = \text{M L T}^{-2}

). On the right side, the only quantity containing time is velocity

[v] = \text{L T}^{-1}

, raised to power

c

, giving

\text{T}^{-c}

. Equating the exponents gives

-2 = -c

, so

c = 2

Adım Adım Çözüm

Identify the base dimensions of each physical quantity in the given equation.

[F] = \text{M L T}^{-2}

[\rho] = \text{M L}^{-3}

[A] = \text{L}^2

, and

[v] = \text{L T}^{-1}

C_d

is dimensionless (

[C_d] = 1

Dimensional homogeneity requires both sides of a physical equation to have identical base dimensions.

Substitute the base dimensions into the formula

F = \frac{1}{2} C_d \rho^a A^b v^c

and simplify.

\text{M L T}^{-2} = (\text{M L}^{-3})^a (\text{L}^2)^b (\text{L T}^{-1})^c = \text{M}^a \text{L}^{-3a + 2b + c} \text{T}^{-c}.

Combining powers of base dimensions allows direct comparison of corresponding exponents.

Equate the exponent of time (T) on both sides of the dimensional equation.

-2 = -c \implies c = 2.

The exponent of T on the left side is -2, which must equal the exponent of T on the right side (-c).

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:1m 15s

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