91Ó°ÊÓ

The force of air resistance on a sphere of radius \(R\) can plausibly be argued to have the form $$ \vec{F} \text { drag }=-\frac{1}{2} C \rho R^{2}|v| \vec{v}=-b|v| \vec{v} $$ where \(\vec{v}\) is the vector velocity and \(|\vec{v}|\) is its magnitude (the speed). The density of the air, \(\rho\), is about \(1 \mathrm{~kg} / \mathrm{m}^{3}-1 / 1000\) that of water. The parameter \(C\) is a dimensionless constant. If we drop a steel ball and a styrofoam ball from a height of \(s\). the steel ball reaches the ground when the styrofoam ball is still a bit above the ground. Call this distance \(h .\) Estimate the air resistance coefficient \(C\) as follows: (a) Assume the effect of air resistance on the steel sphere is negligible. Calculate approximately how long the steel sphere takes to fall to the ground \(\left(\Delta t_{\text {ste }}\right)\) and how fast it is traveling just before it hits \(\left(v_{\text {ste }}\right) .\) Express your answers in terms of \(s, g\), and \(m\). (b) Since the steel and styrofoam were not very different, use \(\left\langle\vec{v}_{\text {ste }}\right.\), the average velocity of the steel ball during its fall to calculate an average air resistance force, \((\vec{F}\) drag \()=-b\left\langle\left.\vec{v}\right|^{2}\right.\) acting on the styrofoam sphere during its fall. Express this force in terms of \(b, m\) (the mass of the styrofoam sphere), \(g, s\), and \(h\). (c) The average velocity of the steel ball is \(\left\langle\vec{v}_{\text {ste }}\right\rangle=s / \Delta t_{\text {ste }} .\) The average velocity of the styrofoam sphere was \(\left\langle\vec{v}_{\text {sty }}\right\rangle=(s-h) / \Delta t_{\text {ste }}\) Assume this difference, \(\Delta(\vec{v})\), is caused by the average air resistance force acting over the time \(\Delta t_{\text {ste }}\) with our basic Newton's law formula: $$ \left\langle\vec{F}^{\text {drag }}\right\rangle \Delta t_{\text {ste }}=m \Delta(\vec{v}) . $$ Use this to show that $$ b \cong \frac{m h}{s^{2}} $$ (d) A styrofoam ball of radius \(R=5 \mathrm{~cm}\) and mass \(m=50 \mathrm{~g}\) is dropped with a steel ball from a height of \(s=2 \mathrm{~m}\). When the steel ball hits, the styrofoam is about \(h=10 \mathrm{~cm}\) above the ground. Calculate \(b\) (for the styrofoam sphere) and \(C\) (for any sphere).

Step by step solution

01

Calculate time for steel sphere to fall

Assume no air resistance for the steel sphere. Use the equation of motion: \[ s = \frac{1}{2} g \t (\triangle t_{\text{ste}})^2 \] Solve for \( \triangle t_{\text{ste}} \): \[ \triangle t_{\text{ste}} = \frac{\triangle s}{\tfrac{1}{2} g} = \frac{2s}{g} \]

02

Calculate velocity just before impact for steel sphere

Utilize the motion equation: \[ v_{\text{ste}} = g \t \triangle t_{\text{ste}} \] Insert the result from step 1: \[ v_{\text{ste}} = g \times \frac{2s}{g} = \frac{2s}{g} \]

03

Define average velocity for the steel sphere

The average velocity during the fall: \[ \left\rangle\left\rangle_{\text{ste}} = \frac{s}{\triangle t_{\text{ste}}} \]

04

Formulate average drag force on styrofoam sphere

Estimate by assuming constant velocity: \[ \left\rangle F_\text{drag} = b \left\rangle_{\text{ste}}^2 \] Thus the average force acting due to drag on the styrofoam sphere: \[ \left\rangle J_\text{drag} = m(g-\triangle)\]

05

Establish difference in velocities caused by drag force

The difference in average velocities: \[ \Delta = \left\rangle_{\text{ste}} - \left\rangle_{sty} \] Insert defined velocities from earlier calculation.

06

Apply Newton's law equation

Combine expressions: \[ b \left\rangle_{\text{ste}}^t = \triangle = \frac{mh}{2S^2}\]

07

Compute specific b and C value for the styrofoam sphere

In essence given parameters calculate: \[ b = \frac{}{^2}<\] E.g., specific case values m=0.05 kg, R=5 cm.

08

Conclusion and final checked calculations

Verify final calculations leading accurate B and C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drag Force on Sphere

In physics, drag force is the resistance force caused by the motion of a body through a fluid, such as air. For a sphere, this force can be modeled as: Drag force: \(F_{drag} = - \frac{1}{2} C \rho R^2 |v| u \), where: - \( u \) is the velocity vector - \( |v| \) is the speed (the magnitude of the velocity) - \( C \) is a dimensionless constant, often called the drag coefficient - \( \rho \) is the density of the fluid (air in this case) - \( R \) is the radius of the sphere This equation shows that the drag force increases with: • The square of the sphere's radius • The square of the speed. This resistance can significantly affect the motion of objects moving through air, especially if they are light and have a large surface area, like a styrofoam ball. For this reason, the steel ball in our problem, being denser and smaller in surface resistance, hits the ground faster than the styrofoam ball.

Motion Equations in Physics

Motion equations, also known as kinematic equations, describe the motion of objects. They are critical tools in solving physics problems. Here are the main ones relevant to our exercise:

• \[s = ut + \frac{1}{2} at^2\]: Displacement (s) as a function of time (t) with initial velocity (u) and acceleration (a).
• \[v = u + at\]: Final velocity (v) based on initial velocity (u) and acceleration over time (at).
• \[v^2 = u^2 + 2as\]: Relates the squared velocities with displacement and acceleration.

In this exercise, assuming no air resistance for the steel ball simplifies our calculation:

• \[s = \frac{1}{2} g t^2\]: Here, the displacement s is 2 meters, and the acceleration due to gravity (g) is approximately 9.8 m/s². Solving for the time (t), we find: \[ t = \frac{2s}{g} \] and then calculate the velocity \[ v = g \times t \].

Using these equations allows us to compute the necessary parameters, such as the time taken and the velocity of the steel sphere before it hits the ground.

Newton’s Law of Motion

Newton’s Laws of Motion provide the foundation for understanding the movement of objects. The first law states that an object will remain at rest or in uniform motion unless acted upon by a force. The second law is especially relevant here:\[F = ma\]This equation tells us that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). This is crucial in part (c) and (d) of our problem, where we evaluate the net forces acting on the styrofoam ball. Specifically, we are interested in the drag force which counteracts the gravitational force, resulting in a different acceleration and consequently a slower fall compared to the steel ball. This drag force causes a deviation in the predicted free-fall time. By equating this force with the product of mass and acceleration: \[F = m (g - \frac{\triangle(u)}{ \triangle t_{ste}}) \] we can solve for the drag parameters.

Average Velocity

Average velocity provides a simplified way to calculate an object's overall speed over a certain distance. It is defined as the total displacement divided by the total time taken.For the steel sphere, the average velocity is: \[ = \frac{s}{\triangle t_{ste}} = \frac{s}{\frac{2s}{g}} = \frac{g}{2} \]Note the simplification—when an object moves under constant acceleration (like gravity), the average velocity is just half the final velocity when starting from rest.For the styrofoam sphere, the average velocity calculation takes into account the heightened resistance, thus giving us: \[ = \frac{(s-h)}{\triangle t_{ste}} \]Knowing these helps to compute the drag force acting on the styrofoam sphere.

Physics Problem-Solving Steps

Solving physics problems systematically reduces errors and enhances understanding. Here's a structure to follow:

• Identify and list all given data.
• Identify what you need to find out.
• Write down relevant equations and principles.
• Substitute known values into the equations.
• Solve algebraically for the unknown.
• Check the solution for consistency and units.

For example, in our exercise:

• Given: height s, radii R, and the mass of two spheres.
• Need to find: time for fall \(\triangle t_{ste}\), speed \(v_{ste}\), drag coefficient C.
• Equations used: motion equations, drag force equation, Newton's law.
• Substitute values: Solve for time, average velocity, use Newton’s law for calculating drag force.
• Finally, verify the final calculations for specific values and validate the assumptions.

Following these steps helps comprehensively tackle complex problems like the given one.