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AP Physics 2 : Other Fluid Dynamics

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Example Questions

Example Question #51 : Fluid Dynamics

Viscosity is defined loosely as internal friction in a fluid. Which of the following fluids would you expect to have the highest viscosity?

Possible Answers:

Molasses

Gasoline

Air

Water

Correct answer:

Molasses

Explanation:

The correct answer is molasses. Since this liquid is notably thicker and more difficult to move, it is intuitively sensed that is has the highest viscosity of this group of fluids.

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Example Question #121 : Fluids

An object with mass $10\textup{ kg}$ $\displaystyle 10\textup{ kg}$ and volume $4\textup{ L}$ $\displaystyle 4\textup{ L}$ and reference area of $0.05\textup{ m}^2$ $\displaystyle 0.05\textup{ m}^2$ is sinking in water with a constant velocity of $2\ \frac{\textup{m}}{\textup{s}}$ $\displaystyle 2\ \frac{\textup{m}}{\textup{s}}$. What is the drag coefficient of the object?

$g=10\ \frac{\textup{m}}{\textup{s}^2}$ $\displaystyle g=10\ \frac{\textup{m}}{\textup{s}^2}$

$\rho_w = 1000\ \frac{\textup{kg}}{\textup{m}^3}$ $\displaystyle \rho_w = 1000\ \frac{\textup{kg}}{\textup{m}^3}$

Possible Answers:

0.60 $\displaystyle 0.60$

0.75 $\displaystyle 0.75$

0.45 $\displaystyle 0.45$

0.30 $\displaystyle 0.30$

0.15 $\displaystyle 0.15$

Correct answer:

0.60 $\displaystyle 0.60$

Explanation:

We will start with Newton's second law:

$F_{net}=ma$ $\displaystyle F_{net}=ma$

Since the block is sinking at a constant velocity, we know that the net force is 0:

$F_{net}=0$ $\displaystyle F_{net}=0$

There are three forces acting on the ball: gravitational, buoyant, and drag. If we designate any downward force as positive, we get:

$\displaystyle F_g -F_b - F_d=0$ (1)

Where:

$\displaystyle F_g = m_bg$ (2)

$F_b = m_{w,displaced}g$ $\displaystyle F_b = m_{w,displaced}g$

Where:

$m_w = \rho_wV_b$ $\displaystyle m_w = \rho_wV_b$

$F_b = \rho_w V_b g$ $\displaystyle F_b = \rho_w V_b g$ (3)

$F_d = \frac{1}{2}\rho_w v^2C_DA$ $\displaystyle F_d = \frac{1}{2}\rho_w v^2C_DA$ (4)

Plugging expressions 2, 3, and 4 into 1, we get:

$m_bg- \rho_w V_b g- \frac{1}{2}\rho_w v^2C_DA = 0$ $\displaystyle m_bg- \rho_w V_b g- \frac{1}{2}\rho_w v^2C_DA = 0$

Now we need to rearranging for the drag coefficient:

$m_bg- \rho_w V_b g= \frac{1}{2}\rho_w v^2C_DA$ $\displaystyle m_bg- \rho_w V_b g= \frac{1}{2}\rho_w v^2C_DA$

$C_D=\frac{2g(m_b- \rho_w V_b )}{\rho_w v^2A}$ $\displaystyle C_D=\frac{2g(m_b- \rho_w V_b )}{\rho_w v^2A}$

Plugging in our values, we get:

$C_D = \frac{2\left ( 10\frac{m}{s^2}\right )\left ( (10kg)-\left ( 1\frac{kg}{L}\right ) (4L)\right )}{\left ( 1000\frac{kg}{m^3}\right )\left ( 2\frac{m}{s}\right )^2(0.05m^2)}$ $\displaystyle C_D = \frac{2\left ( 10\frac{m}{s^2}\right )\left ( (10kg)-\left ( 1\frac{kg}{L}\right ) (4L)\right )}{\left ( 1000\frac{kg}{m^3}\right )\left ( 2\frac{m}{s}\right )^2(0.05m^2)}$

$\displaystyle C_D =0.6$

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Example Question #121 : Fluids

A spherical ball of mass 8kg $\displaystyle 8kg$ and radius 0.1m $\displaystyle 0.1m$ is sinking in water. What is the terminal velocity of the ball?

$g=10\frac{m}{s^2}$ $\displaystyle g=10\frac{m}{s^2}$

$\rho_w = 1000\frac{kg}{m^3}$ $\displaystyle \rho_w = 1000\frac{kg}{m^3}$

$C_{D_{sphere}} = 0.47$ $\displaystyle C_{D_{sphere}} = 0.47$

Possible Answers:

$0.79\frac{m}{s}$ $\displaystyle 0.79\frac{m}{s}$

$1.14\ \frac{\textup{m}}{\textup{s}}$ $\displaystyle 1.14\ \frac{\textup{m}}{\textup{s}}$

$2.81\frac{m}{s}$ $\displaystyle 2.81\frac{m}{s}$

$2.64\frac{m}{s}$ $\displaystyle 2.64\frac{m}{s}$

$1.30\frac{m}{s}$ $\displaystyle 1.30\frac{m}{s}$

Correct answer:

$1.14\ \frac{\textup{m}}{\textup{s}}$ $\displaystyle 1.14\ \frac{\textup{m}}{\textup{s}}$

Explanation:

We will start with Newton's second law for this problem:

$F_{net}=ma$ $\displaystyle F_{net}=ma$

When the ball reaches terminal velocity, we know that the net force on the ball is equal to 0:

$F_{net}=0$ $\displaystyle F_{net}=0$

There are three forces acting on the ball: gravitational, buoyant, and drag. If we designate any downward force as positive, we get:

$\displaystyle F_g -F_b - F_d=0$ (1)

Where:

$\displaystyle F_g = m_bg$ (2)

$F_b = m_{w,displaced}g$ $\displaystyle F_b = m_{w,displaced}g$

Where:

$m_w = \rho_wV_b$ $\displaystyle m_w = \rho_wV_b$

$m_w = \rho_w\left ( \frac{4}{3}\pi r^3\right )$ $\displaystyle m_w = \rho_w\left ( \frac{4}{3}\pi r^3\right )$

$F_b = \frac{4}{3}\pi r^3\rho_wg$ $\displaystyle F_b = \frac{4}{3}\pi r^3\rho_wg$ (3)

$F_d = \frac{1}{2}\rho_w v^2C_DA$ $\displaystyle F_d = \frac{1}{2}\rho_w v^2C_DA$

$F_d = 2\pi r^2 \rho_wC_D v^2$ $\displaystyle F_d = 2\pi r^2 \rho_wC_D v^2$ (4)

Substituting expressions (2), (3), and (4) into (1), we get:

$m_bg- \frac{4}{3}\pi r^3\rho_wg - 2\pi r^2 \rho_wC_D v^2=0$ $\displaystyle m_bg- \frac{4}{3}\pi r^3\rho_wg - 2\pi r^2 \rho_wC_D v^2=0$

Now we just need to rearranging for velocity.

$3m_bg- 4\pi r^3\rho_wg = 6\pi r^2 \rho_wC_D v^2$ $\displaystyle 3m_bg- 4\pi r^3\rho_wg = 6\pi r^2 \rho_wC_D v^2$

$v^2 = \frac{3m_bg- 4\pi r^3\rho_wg}{6\pi r^2 \rho_wC_D}$ $\displaystyle v^2 = \frac{3m_bg- 4\pi r^3\rho_wg}{6\pi r^2 \rho_wC_D}$

We have all of these values, so time to plug and chug:

$v^2 = \frac{3(8kg)\left ( 10\frac{m}{s^2}\right )-4\pi(0.1m)^3\left ( 1000\frac{kg}{m^3}\right )\left ( 10\frac{m}{s^2}\right )}{6\pi (0.1m)^2\left ( 1000\frac{kg}{m^3}\right )(0.47)}$ $\displaystyle v^2 = \frac{3(8kg)\left ( 10\frac{m}{s^2}\right )-4\pi(0.1m)^3\left ( 1000\frac{kg}{m^3}\right )\left ( 10\frac{m}{s^2}\right )}{6\pi (0.1m)^2\left ( 1000\frac{kg}{m^3}\right )(0.47)}$

$v = 1.14\frac{m}{s}$ $\displaystyle v = 1.14\frac{m}{s}$

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AP Physics 2 : Other Fluid Dynamics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #51 : Fluid Dynamics

Example Question #121 : Fluids

Example Question #121 : Fluids

All AP Physics 2 Resources

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