The torque \tau required to rotate a thin flat disk of radius r at a constant angular velo… — JAMB UTME

The torque $\tau$ required to rotate a thin flat disk of radius $r$ at a constant angular velocity $\omega$ in a fluid of dynamic viscosity $\eta$ is expressed by the dimensional formula $\tau = k \eta^x \omega^y r^z$ , where $k$ is a dimensionless constant. What is the value of the sum of the exponents $x + y + z$ ?

Answer: 5

Answer

The sum of the exponents

x + y + z

is 5.

By substituting the base dimensions into

\tau = k \eta^x \omega^y r^z

, we get

M L^2 T^{-2} = (M L^{-1} T^{-1})^x (T^{-1})^y L^z = M^x L^{-x+z} T^{-x-y}

. Equating exponents of

M

gives

x = 1

. Equating exponents of

T

gives

-1 - y = -2 \implies y = 1

. Equating exponents of

L

gives

-1 + z = 2 \implies z = 3

. Thus,

x + y + z = 1 + 1 + 3 = 5

Step-by-Step Solution

Determine the dimensions of torque, dynamic viscosity, angular velocity, and radius in base mechanical dimensions (M, L, T).

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

[\eta] = M L^{-1} T^{-1}

[\omega] = T^{-1}

, and

[r] = L

Dimensional analysis requires converting all parameters into base dimensions.

Apply the principle of dimensional homogeneity to set up exponential equations for each base dimension.

M^1 L^2 T^{-2} = M^x L^{-x+z} T^{-x-y}

Both sides of a physically valid equation must share identical net dimensions.

Solve for each exponent individually by comparing indices.

x = 1

y = 1

z = 3

Matching powers of M yields

x=1

, matching powers of T yields

y=1

, and matching powers of L yields

z=3

Sum the three calculated exponent values.

1 + 1 + 3 = 5

The question asks specifically for the value of

x + y + z

Key Concept

Dimensional analysis and dimensional homogeneity

Estimated Time:1m 30s

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