Question

Difficulty: MediumPhysical Quantities, Units and Dimensions

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

Answer: -2

Answer

The numerical value of the product x⋅yx \cdot y is −2-2.
By dimensional analysis, centripetal acceleration has dimensions [a]=L T−2[a] = \text{L T}^{-2}, velocity [v]=L T−1[v] = \text{L T}^{-1}, and radius [r]=L[r] = \text{L}. Substituting into a=kvxrya = k v^x r^y gives L T−2=Lx+yT−x\text{L T}^{-2} = \text{L}^{x+y} \text{T}^{-x}. Equating exponents of time yields −x=−2  ⟹  x=2-x = -2 \implies x = 2. Equating exponents of length gives x+y=1  ⟹  2+y=1  ⟹  y=−1x + y = 1 \implies 2 + y = 1 \implies y = -1. Consequently, x⋅y=2⋅(−1)=−2x \cdot y = 2 \cdot (-1) = -2.

Step-by-Step Solution

1
Identify the base dimensions for each physical quantity.
[a]=M0L1T−2[a] = \text{M}^0 \text{L}^1 \text{T}^{-2}, [v]=M0L1T−1[v] = \text{M}^0 \text{L}^1 \text{T}^{-1}, and [r]=M0L1T0[r] = \text{M}^0 \text{L}^1 \text{T}^0.
Dimensional analysis requires substituting fundamental dimensions of mass, length, and time.
2
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 kk is dimensionless, the net dimensions on both sides of the equation must be identical.
3
Solve for exponents xx and yy by equating powers of corresponding base units.
For \text{T}: −x=−2  ⟹  x=2-x = -2 \implies x = 2. For \text{L}: x+y=1  ⟹  2+y=1  ⟹  y=−1x + y = 1 \implies 2 + y = 1 \implies y = -1.
Equating coefficients of identical base dimensions gives a system of linear equations.
4
Multiply the derived values of xx and yy.
x⋅y=2⋅(−1)=−2x \cdot y = 2 \cdot (-1) = -2.
The question asks specifically for the product of exponents xx and yy.

Key Concept

Dimensional Homogeneity and Exponent Analysis
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