The dimensionless parameters for a ball released and falling from rest in a fluid are $$C_{D}, \quad \frac{g t^{2}}{D}, \quad \frac{\rho}{\rho_{b}}, \quad \text { and } \quad \frac{V_{t}}{D}$$ where \(\left.C_{\mathbf{D}} \text { is a drag coefficient (assumed to be constant at } 0,4\right)\) \(g\) is the acceleration of gravity, \(D\) is the ball diameter, \(t\) is the time after it is released, \(\rho\) is the density of the fluid in which it is dropped, and \(\rho_{b}\) is the density of the ball. Ball 1 , an aluminum ball \(\left(\rho_{b_{1}}=2710 \mathrm{kg} / \mathrm{m}^{3}\right)\) having a diameter \(D_{1}=1.0 \mathrm{cm},\) is dropped in water \(\left(\rho=1000 \mathrm{kg} / \mathrm{m}^{3}\right) .\) The ball velocity \(V_{\mathrm{t}}\) is \(0.733 \mathrm{m} / \mathrm{s}\) at \(t_{1}=\) 0.10 s. Find the corresponding velocity \(V_{2}\) and time \(t_{2}\) for ball 2 having \(D_{2}=2.0 \mathrm{cm}\) and \(\rho_{b_{2}}=\rho_{b_{1}} .\) Next, use the computed value of \(V_{2}\) and the equation of motion,$$\rho_{b} V_{g}-\frac{\mathrm{C}_{\mathrm{D}}}{2} \rho A V^{2}=\rho_{b} V \frac{d V}{d t}$$ where \(Y\) is the volume of the ball and \(A\) is its cross-sectional area. to verify the value of \(t_{2}\). Should the two values of \(t_{2}\) agree?

Step by step solution

01

Establish the equation

Start by expressing the dimensionless parameter using given parameters for the first ball. The four dimensionless parameters, using the known parameters for ball 1, will be expressed as \(x=C_D\), \(y=\frac{g t_1^{2}}{D_1}\), \(z=\frac{\rho}{\rho_{b_1}}\), \(w=\frac{V_{t_1}}{D_1}\).

02

Solve for the velocity of the second ball

Since the dimensionless parameter \(w=\frac{V_t}{D}\) retains its value for identical balls in the same liquid, it can be concluded that \(w_1 = w_2\), hence \(\frac{V_{t_1}}{D_1}=\frac{V_{t_2}}{D_2}\). Solving this for \(V_{t_2}\) yields \(V_{t_2}=\frac{V_{t_1}D_2}{D_1}\). Substitute the given values to get the numerical value of \(V_{t_2}\).

03

Solve for the time for the second ball

Similarly, use the first ball's values to form the dimensionless parameter \(\frac{g t^{2}}{D} = \frac{g t_{1}^{2}}{D_{1}} = \frac{g t_{2}^{2}}{D_{2}}\). Solving this for \(t_{2}\) yields \(t_{2}=\sqrt{t_{1}^{2} \frac{D_{2}}{D_{1}}}\). Substitute the given values to get the numerical value of \(t_{2}\).

04

Verification of the value of \(t_2\)

Use the equation given by the problem to substitute the known values and solve for \(t_2\). The equation is \(\rho_{b} V_{g}-\frac{\mathrm{C}_{\mathrm{D}}}{2} \rho A V^{2}=\rho_{b} V \frac{d V}{d t}\), solve it for \(t\) when \(V = V_{t_2}\).The values of \(t_2\) obtained from both methods should agree.

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