Math Problem Statement

Problem 1: A speck of dust moves along a circular path at constant speed. The radius of the circle is 1.01m , and the distance traveled by the speck of dust as it moves from point A to point B is 3.79m .

Part (a) What is the angular displacement of the speck of dust, in radians, as it travels from A to B ? (The diagram is for visualization purposes only. The calculated angle might not appear to match the apparent angle in the diagram.) Numeric : A numeric value is expected and not an expression. θ = __________________________________________rad

Part (b) If it takes the speck of dust 0.76s to travel from A to B , what is the magnitude of its angular speed in units of radians per second? Numeric : A numeric value is expected and not an expression. ω = __________________________________________rad/s

Part (c) What is the magnitude of the average angular acceleration of the speck of dust in units of radians per second squared? Numeric : A numeric value is expected and not an expression. α = __________________________________________rad/s2

Solution

To solve this problem, we’ll tackle each part separately.

Part (a): Angular Displacement (θ\theta)

The angular displacement θ\theta in radians can be found using the formula: θ=sr\theta = \frac{s}{r} where:

  • ss is the arc length (distance traveled along the circular path) from point A to point B, which is 3.79m3.79 \, \text{m},
  • rr is the radius of the circle, given as 1.01m1.01 \, \text{m}.

Substituting the values: θ=3.791.01\theta = \frac{3.79}{1.01}

Calculating: θ3.75radians\theta \approx 3.75 \, \text{radians}

So, the answer for Part (a) is: θ=3.75rad\theta = 3.75 \, \text{rad}

Part (b): Angular Speed (ω\omega)

The angular speed ω\omega is defined as the rate of change of angular displacement over time. Since the dust particle moves at a constant speed, we can calculate ω\omega using: ω=θt\omega = \frac{\theta}{t} where:

  • θ\theta is the angular displacement we just calculated, 3.75rad3.75 \, \text{rad},
  • tt is the time taken, given as 0.76s0.76 \, \text{s}.

Substituting the values: ω=3.750.76\omega = \frac{3.75}{0.76}

Calculating: ω4.93rad/s\omega \approx 4.93 \, \text{rad/s}

So, the answer for Part (b) is: ω=4.93rad/s\omega = 4.93 \, \text{rad/s}

Part (c): Average Angular Acceleration (α\alpha)

Since the dust particle moves at a constant speed along the circular path, there is no change in angular speed, meaning there is no angular acceleration. Thus: α=0rad/s2\alpha = 0 \, \text{rad/s}^2

So, the answer for Part (c) is: α=0rad/s2\alpha = 0 \, \text{rad/s}^2


Summary of Answers:

  • (a) Angular Displacement: θ=3.75rad\theta = 3.75 \, \text{rad}
  • (b) Angular Speed: ω=4.93rad/s\omega = 4.93 \, \text{rad/s}
  • (c) Angular Acceleration: α=0rad/s2\alpha = 0 \, \text{rad/s}^2

Do you have any questions or would you like a deeper explanation for any part?


Related Questions to Expand Understanding:

  1. How does changing the radius affect the angular displacement if the distance traveled remains the same?
  2. What would be the angular displacement if the radius were halved?
  3. If the speed of the particle varied over time, how would you calculate the angular acceleration?
  4. How is linear speed related to angular speed in circular motion?
  5. How would you calculate the centripetal acceleration of the dust particle?

Tip:

In circular motion, angular displacement in radians is directly proportional to the arc length, simplifying calculations when moving between points on a circular path.

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Math Problem Analysis

Mathematical Concepts

Circular motion
Angular displacement
Angular speed
Angular acceleration

Formulas

Angular displacement formula: θ = s / r
Angular speed formula: ω = θ / t
Angular acceleration formula: α = Δω / Δt

Theorems

Relationship between linear distance and angular displacement

Suitable Grade Level

Grades 11-12