Math Problem Statement

A 10-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 5 cm. The other end B is free to move along a horizontal line. The wheel rotates at 2.0 revolutions per second. Determine the coordinates, angular velocity, and velocity of points A and B as functions of time t.

Solution

Let's analyze the answers and corrections based on the given diagram and provided details.

Problem Summary:

A rod of 10 cm length is attached at one end (Point A) to a point on a wheel of radius 5 cm rotating counterclockwise. Point A moves in a circular path, while Point B (the other end of the rod) moves horizontally along a line passing through the wheel's center. The wheel completes 2.0 revolutions per second.

Solutions Review:

Here are the provided answers along with my analysis:


(a) Coordinates of Point A and B at t=112t = \frac{1}{12} sec:

  1. Correct answer for Point A: (52,532)\left( \frac{5}{2}, \frac{5 \sqrt{3}}{2} \right).
  2. Incorrect answer for Point B:
    • The coordinates should reflect Point B's horizontal displacement relative to Point A's location.
    • Correct position of Point B should be: (52+5cos(60),532+5sin(60))\left( \frac{5}{2} + 5\cos(60^\circ), \frac{5\sqrt{3}}{2} + 5\sin(60^\circ) \right).

(b) Angular Velocity:

  • Answer Provided: ω=4π\omega = 4\pi radians per second.
  • Analysis: This is correct. Since the wheel rotates at 2 revolutions per second, the angular velocity is ω=2×2π=4π\omega = 2 \times 2\pi = 4\pi radians per second.

(c) Coordinates of Point A as a function of tt:

  • Correct Answer:
    • x=5cos(4πt)x = 5 \cos(4\pi t)
    • y=5sin(4πt)y = 5 \sin(4\pi t)
  • Explanation: These are accurate expressions for the coordinates of Point A, considering the angular velocity ω=4π\omega = 4\pi.

(d) x-coordinate of Point B:

  • Incorrect Answer: 5cos(4πt)+105 \cos(4\pi t) + 10.
  • Correct Form: To express the x-coordinate of Point B, consider the full rod length and rotation.

(e) Velocity of Point B:

  • Correct Answer: 20πsin(4πt)-20\pi \sin(4\pi t) cm/sec.
  • Explanation: This represents the horizontal velocity of Point B correctly.

Let me know if you'd like further breakdowns of any part of the solution or related explanations.


Follow-up Questions:

  1. What is the full derivation of the x-coordinate of Point B as a function of tt?
  2. How is the y-coordinate of Point B calculated for any given time tt?
  3. Can we verify Point A’s position using trigonometric identities?
  4. How does the length of the rod influence Point B's movement?
  5. Why is the angular velocity given as 4π4\pi, and how does this relate to the rotation period?

Quick Tip:

When dealing with rotational motion, remember that angular velocity (ω\omega) relates directly to the number of revolutions per second and is given by ω=2π×frequency\omega = 2\pi \times \text{frequency}.

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Math Problem Analysis

Mathematical Concepts

Circular Motion
Angular Velocity
Trigonometry
Coordinate Geometry

Formulas

x = r * cos(ωt) and y = r * sin(ωt) for circular motion
Angular velocity ω = 2 * π * frequency
Velocity formula for linear motion derived from trigonometric functions

Theorems

Trigonometric identities for coordinate transformation
Relationship between angular velocity and linear velocity

Suitable Grade Level

Grades 11-12 or early college