I can clearly see the questions from the images. Let's break them down one by one and provide solutions.

Question 1:

A rider notices that the wheels of his bicycle make 12 revolutions in 15 seconds.

• a. What is the average angular speed of the wheel in radians?
• b. What distance in meters does the wheel travel if its radius is 33 cm?

Solution:

• Part a: Angular speed
  Angular speed $\omega$ is given by the formula: $\omega = \frac{\theta}{t}$ Where:
• $\theta$ is the angular displacement in radians (1 revolution = $2\pi$ radians).
• $t$ is the time in seconds.

Given:

• The wheel makes 12 revolutions.
• Time $t = 15 \, \text{seconds}$.

So, $\theta = 12 \times 2\pi \, \text{radians} = 24\pi \, \text{radians}$.

Now, $\omega = \frac{24\pi}{15} \, \text{radians/second} = \frac{8\pi}{5} \, \text{radians/second} \approx 5.03 \, \text{radians/second}$

• Part b: Distance traveled
  The distance traveled by the wheel is the circumference multiplied by the number of revolutions: $\text{Distance} = \text{Circumference} \times \text{Number of revolutions}$ Circumference $C$ of the wheel: $C = 2\pi r$ Given radius $r = 33 \, \text{cm} = 0.33 \, \text{m}$, $C = 2\pi \times 0.33 \approx 2.073 \, \text{m}$ Now, the total distance for 12 revolutions is: $\text{Distance} = 2.073 \times 12 = 24.876 \, \text{m}$

Question 2:

What is the angular velocity of an object traveling in a circle of radius 0.75 m with a linear speed of 3.5 m/s?

Solution:

The relation between linear speed $v$ and angular velocity $\omega$ is: $v = r \omega$ Given:

• $v = 3.5 \, \text{m/s}$
• $r = 0.75 \, \text{m}$

We can solve for $\omega$: $\omega = \frac{v}{r} = \frac{3.5}{0.75} = 4.67 \, \text{radians/second}$

Question 3:

Two students are balancing on a 10m seesaw.

• The student on the left weighs 60 kg and stands 3 meters away from the center.
• The student on the right weighs 45 kg.
• What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?

Solution:

For the seesaw to balance, the torques on both sides should be equal. Torque $\tau$ is given by: $\tau = r \times F$ Where:

• $r$ is the distance from the pivot (center of the seesaw),
• $F$ is the force (which is the weight, $F = mg$, where $m$ is the mass and $g$ is acceleration due to gravity).

Given:

• $m_1 = 60 \, \text{kg}$, $r_1 = 3 \, \text{m}$
• $m_2 = 45 \, \text{kg}$, $r_2 = ?$

For balance: $m_1 r_1 = m_2 r_2$ Substitute the values: $60 \times 3 = 45 \times r_2$ $180 = 45 \times r_2$ $r_2 = \frac{180}{45} = 4 \, \text{meters}$

So, the student on the right should stand 4 meters away from the center.

Would you like further details on any of these solutions?

Here are 5 related questions to expand your understanding:

1. How would the angular speed change if the number of revolutions per second increased?
2. What is the relation between linear and angular velocity in terms of rotational motion?
3. What would happen if the student on the seesaw moved closer to the center?
4. How do mass and distance influence the balance of a lever in physics?
5. What is the importance of radians in measuring angular displacement?

Tip: In rotational dynamics, always remember that linear and angular quantities are connected through the radius of rotation.