The centripetal acceleration a of a particle moving in a circular path depends on its line… — JAMB UTME

The centripetal acceleration $a$ of a particle moving in a circular path depends on its linear speed $v$ and the radius $r$ of the path according to the formula $a = k v^x r^y$ , where $k$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the product $x \cdot y$ ?

Answer: -2

Answer

The numerical value of the product

x \cdot y

-2

By dimensional analysis, centripetal acceleration has dimensions

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

, velocity

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

, and radius

[r] = \text{L}

. Substituting into

a = k v^x r^y

gives

\text{L T}^{-2} = \text{L}^{x+y} \text{T}^{-x}

. Equating exponents of time yields

-x = -2 \implies x = 2

. Equating exponents of length gives

x + y = 1 \implies 2 + y = 1 \implies y = -1

. Consequently,

x \cdot y = 2 \cdot (-1) = -2

Step-by-Step Solution

Identify the base dimensions for each physical quantity.

[a] = \text{M}^0 \text{L}^1 \text{T}^{-2}

[v] = \text{M}^0 \text{L}^1 \text{T}^{-1}

, and

[r] = \text{M}^0 \text{L}^1 \text{T}^0

Dimensional analysis requires substituting fundamental dimensions of mass, length, and time.

Set up the dimensional homogeneity equation.

\text{L}^1 \text{T}^{-2} = (\text{L T}^{-1})^x \cdot (\text{L})^y = \text{L}^{x+y} \text{T}^{-x}.

Since

k

is dimensionless, the net dimensions on both sides of the equation must be identical.

Solve for exponents

x

and

y

by equating powers of corresponding base units.

For \text{T}:

-x = -2 \implies x = 2

. For \text{L}:

x + y = 1 \implies 2 + y = 1 \implies y = -1

Equating coefficients of identical base dimensions gives a system of linear equations.

Multiply the derived values of

x

and

y

x \cdot y = 2 \cdot (-1) = -2

The question asks specifically for the product of exponents

x

and

y

Key Concept

Dimensional Homogeneity and Exponent Analysis

Question

Answer

Step-by-Step Solution

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