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AP Physics 2 : Other Fluid Dynamics

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Example Question #1 : Other 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:

Water

Air

Molasses

Gasoline

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 #53 : Fluid Dynamics

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

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

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

Possible Answers:

0.45

0.60

0.15

0.30

0.75

Correct answer:

0.60

Explanation:

We will start with Newton's second law:

$F_{net}=ma$

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

$F_{net}=0$

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

F_g -F_b - F_d=0 (1)

Where:

F_g = m_bg (2)

$F_b = m_{w,displaced}g$

Where:

$m_w = \rho_wV_b$

$F_b = \rho_w V_b g$ (3)

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

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

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

Now we need to rearranging for the drag coefficient:

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

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

Plugging in our values, we get:

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

C_D =0.6

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Example Question #52 : Fluid Dynamics

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

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

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

$C_{D_{sphere}} = 0.47$

Possible Answers:

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

$0.79\frac{m}{s}$

$2.81\frac{m}{s}$

$2.64\frac{m}{s}$

$1.30\frac{m}{s}$

Correct answer:

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

Explanation:

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

$F_{net}=ma$

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

$F_{net}=0$

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

F_g -F_b - F_d=0 (1)

Where:

F_g = m_bg (2)

$F_b = m_{w,displaced}g$

Where:

$m_w = \rho_wV_b$

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

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

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

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

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

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

Now we just need to rearranging for velocity.

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

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

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

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

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

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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 #1 : Other Fluid Dynamics

Example Question #53 : Fluid Dynamics

Example Question #52 : Fluid Dynamics

All AP Physics 2 Resources

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