Given that the drag force experienced by a projectile which can move in water or in air is given by F = f(p, V, D, μ) where p= density of the fluid, V= Velocity of the projectile, D= Diameter of the projectile, μ = viscosity of the fluid. Show by dimensional analysis that F / (pv²D²) = f(pVD/μ). (b) A projectile having a diameter of 1000 mm is to move in air at 50 m/s and its drag force is to be estimated from tests in water on a half-scale model. Determine: (i) the speed of the model in water and (ii) the drag force acting on the full-scale projectile given that the drag force on the half-scale model is 1153N.

Question

Asked on March 24, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-24T11:03:02.313Z

Step 1: List the primary dimensions of each variable. Drag Force (F_D): [M L T^-2] Density (): [M L^-3] Velocity (V): [L T^-1] Diameter (D): [L] Dynamic Viscosity (): [M L^-1 T^-1] Step 2: Determine the number of variables and primary dimensions. Number of variables (n) = 5 (F_D, , V, D, ) Number of primary dimensions (k) = 3 (Mass [M], Length [L], Time [T]) Number of dimensionless Pi groups () = n - k = 5 - 3 = 2 Step 3: Choose repeating variables. We choose , V, and D as repeating variables because they collectively contain all three primary dimensions (M, L, T) and are fundamental to the flow. Step 4: Form the dimensionless Pi groups. Each Pi group will be formed by combining the repeating variables with one of the remaining non-repeating variables. For _1 (using F_D): _1 = ^a V^b D^c F_D Substitute the dimensions: [M^0 L^0 T^0] = [M L^-3]^a [L T^-1]^b [L]^c [M L T^-2] Equate the exponents for each primary dimension: For M: 0 = a + 1 a = -1 For T: 0 = -b - 2 b = -2 For L: 0 = -3a + b + c + 1 Substitute a = -1 and b = -2: 0 = -3(-1) + (-2) + c + 1 0 = 3 - 2 + c + 1 0 = 2 + c c = -2 So, _1 = ^-1 V^-2 D^-2 F_D = (F_D)/( V^2 D^2). For _2 (using ): _2 = ^x V^y D^z Substitute the dimensions: [M^0 L^0 T^0] = [M L^-3]^x [L T^-1]^y [L]^z [M L^-1 T^-1] Equate the exponents for each primary dimension: For M: 0 = x + 1 x = -1 For T: 0 = -y - 1 y = -1 For L: 0 = -3x + y + z - 1 Substitute x = -1 and y = -1: 0 = -3(-1) + (-1) + z - 1 0 = 3 - 1 + z - 1 0 = 1 + z z = -1 So, _2 = ^-1 V^-1 D^-1 = ()/( V D). Step 5: Express the functional relationship. According to the Buckingham Pi theorem, the relationship between the variables can be expressed as a function of the dimensionless Pi groups: _1 = f(_2) Substituting the derived Pi groups: (F_D)/( V^2 D^2) = f(()/( V D)) Since f represents an arbitrary function, f(()/( V D)) is equivalent to f(( V D)/()) (as one is simply the inverse of the other, and the functional relationship can absorb this change). Therefore, we have shown that: (F_D)/( V^2 D^2) = f(( V D)/())

Accepted Answer · 2026-03-24T11:03:02.313Z

Step 1: List the primary dimensions of each variable. Drag Force (F_D): [M L T^-2] Density (): [M L^-3] Velocity (V): [L T^-1] Diameter (D): [L] Dynamic Viscosity (): [M L^-1 T^-1] Step 2: Determine the number of variables and primary dimensions. Number of variables (n) = 5 (F_D, , V, D, ) Number of primary dimensions (k) = 3 (Mass [M], Length [L], Time [T]) Number of dimensionless Pi groups () = n - k = 5 - 3 = 2 Step 3: Choose repeating variables. We choose , V, and D as repeating variables because they collectively contain all three primary dimensions (M, L, T) and are fundamental to the flow. Step 4: Form the dimensionless Pi groups. Each Pi group will be formed by combining the repeating variables with one of the remaining non-repeating variables. For _1 (using F_D): _1 = ^a V^b D^c F_D Substitute the dimensions: [M^0 L^0 T^0] = [M L^-3]^a [L T^-1]^b [L]^c [M L T^-2] Equate the exponents for each primary dimension: For M: 0 = a + 1 a = -1 For T: 0 = -b - 2 b = -2 For L: 0 = -3a + b + c + 1 Substitute a = -1 and b = -2: 0 = -3(-1) + (-2) + c + 1 0 = 3 - 2 + c + 1 0 = 2 + c c = -2 So, _1 = ^-1 V^-2 D^-2 F_D = (F_D)/( V^2 D^2). For _2 (using ): _2 = ^x V^y D^z Substitute the dimensions: [M^0 L^0 T^0] = [M L^-3]^x [L T^-1]^y [L]^z [M L^-1 T^-1] Equate the exponents for each primary dimension: For M: 0 = x + 1 x = -1 For T: 0 = -y - 1 y = -1 For L: 0 = -3x + y + z - 1 Substitute x = -1 and y = -1: 0 = -3(-1) + (-1) + z - 1 0 = 3 - 1 + z - 1 0 = 1 + z z = -1 So, _2 = ^-1 V^-1 D^-1 = ()/( V D). Step 5: Express the functional relationship. According to the Buckingham Pi theorem, the relationship between the variables can be expressed as a function of the dimensionless Pi groups: _1 = f(_2) Substituting the derived Pi groups: (F_D)/( V^2 D^2) = f(()/( V D)) Since f represents an arbitrary function, f(()/( V D)) is equivalent to f(( V D)/()) (as one is simply the inverse of the other, and the functional relationship can absorb this change). Therefore, we have shown that: (F_D)/( V^2 D^2) = f(( V D)/())

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