Question

In a physics class, a 2.70 -g ping-pong ball was suspended from a massless string. The string makes an angle of \(\theta=15.0^{\circ}\) with the vertical when air is blown horizontally at the ball at a speed of \(20.5 \mathrm{~m} / \mathrm{s}\). Assume that the friction force is proportional to the squared speed of the air stream. a) What is the proportionality constant in this experiment? b) What is the tension in the string?

Short answer

Question: Determine the proportionality constant (k) and the tension in the string (T) for a ping-pong ball with a mass of 2.70 g, at an angle of 15 degrees with the vertical, and an air speed of 20.5 m/s. Answer: The proportionality constant (k) is approximately \(6.07 \times 10^{-6} \mathrm{~kg/m}\) and the tension in the string (T) is approximately \(2.67 \times 10^{-2} \mathrm{~N}\).

Step-by-step solution

Draw the force diagram

Draw a diagram with the ping-pong ball at the center. Show the following forces: - Gravitational force (weight) pointing vertically downward. - Friction force (air drag), pointing horizontally and opposite to the air stream. - Tension in the string, pointing upward and at an angle of 15 degrees from vertical.

Set up the force equations

Since the ping-pong ball is in equilibrium (not accelerating), the net forces in both x (horizontal) and y (vertical) directions must be equal to zero. So, 1. For the x-direction, the friction force (air drag) must be equal and opposite to the x-component of the tension in the string: \(k v^2 = T \sin{\theta}\) 2. For the y-direction, the weight (gravitational force) must be equal and opposite to the y-component of the tension in the string: \(mg = T \cos{\theta}\)

Solve for proportionality constant (k) and tension (T)

Isolate T in the second equation: \(T = \frac{mg}{\cos{\theta}}\) Now, substitute this T value in the first equation to find k: \(k v^2 = \left(\frac{mg}{\cos{\theta}}\right) \sin{\theta}\) Now solve for k: \(k = \frac{m \sin{\theta}}{v^2 \cos{\theta}}\)

Plug in the given values and calculate k and T

We are given the mass (m), theta, and air speed (v): \(m = 2.70 \times 10^{-3} \mathrm{~kg}\) \(\theta = 15.0^{\circ} = \frac{15 \pi}{180} \mathrm{~rad}\) \(v = 20.5 \mathrm{~m/s}\) Plug these values into the equation for k: \(k = \frac{(2.70 \times 10^{-3} \mathrm{~kg}) \sin{\frac{15 \pi}{180} \mathrm{~rad}}}{(20.5 \mathrm{~m/s})^2 \cos{\frac{15 \pi}{180} \mathrm{~rad}}}\) Calculate k: \(k \approx 6.07 \times 10^{-6} \mathrm{~kg/m}\) Now, plug the given values into the equation for T: \(T = \frac{(2.70 \times 10^{-3} \mathrm{~kg}) \times 9.81 \mathrm{~m/s^2}}{\cos{\frac{15 \pi}{180} \mathrm{~rad}}}\) Calculate T: \(T \approx 2.67 \times 10^{-2} \mathrm{~N}\) So, the proportionality constant (k) is approximately \(6.07 \times 10^{-6} \mathrm{~kg/m}\) and the tension in the string (T) is approximately \(2.67 \times 10^{-2} \mathrm{~N}\).