Math Problem Statement

A rod of 8 cm is attached at one end to a point A rotating counterclockwise on a wheel of radius 4 cm. The other end, B, is free to move along a horizontal bar through the center of the wheel. When t = 1/12 sec, find the coordinates of points A and B, the angular velocity of point A, and express the coordinates of point A as a function of time t.

Solution

Let's solve each part of the problem.

Given:

  • Rod length = 8 cm
  • Wheel radius = 4 cm
  • Wheel rotation = 2.0 revolutions per second, counterclockwise
  • Initial time t=0t = 0 is the left position shown in the diagram.

Since the wheel rotates at 2.02.0 revolutions per second, we can convert this into radians per second, as each revolution is 2π2\pi radians.

Solution

Part (a): Coordinates of Points A and B at t=112t = \frac{1}{12} seconds

  1. Angular Velocity Calculation: ω=2.0×2π=4π radians per second\omega = 2.0 \times 2\pi = 4\pi \text{ radians per second}

  2. Angular Position of Point A: The angle θ\theta of point A at t=112t = \frac{1}{12} seconds can be calculated using: θ=ωt=4π112=π3 radians\theta = \omega \cdot t = 4\pi \cdot \frac{1}{12} = \frac{\pi}{3} \text{ radians}

  3. Coordinates of Point A: Since the center of the wheel is at the origin and the radius is 44 cm, the coordinates of point AA (located on the circle's circumference) are: xA=4cos(π3)=4×12=2x_A = 4 \cos\left(\frac{\pi}{3}\right) = 4 \times \frac{1}{2} = 2 yA=4sin(π3)=4×32=23y_A = 4 \sin\left(\frac{\pi}{3}\right) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} So, AA is at (2,23)(2, 2\sqrt{3}).

  4. Coordinates of Point B: Since BB is 88 cm away from AA along the rod in a horizontal direction, the x-coordinate of BB would be: xB=xA+8=2+8=10x_B = x_A + 8 = 2 + 8 = 10 yB=0y_B = 0 So, BB is at (10,0)(10, 0).

Part (b): Angular Velocity ω\omega

As calculated above, ω=4π radians per second\omega = 4\pi \text{ radians per second}

Part (c): Parametric Equations for Point A

To express the x and y coordinates of point AA as a function of time tt, we use the general parametric form for circular motion: x=Rcos(ωt)x = R \cos(\omega t) y=Rsin(ωt)y = R \sin(\omega t) Substituting R=4R = 4 cm and ω=4π\omega = 4\pi: x=4cos(4πt)x = 4 \cos(4\pi t) y=4sin(4πt)y = 4 \sin(4\pi t)


Final Answers

  • (a) Coordinates at t=112t = \frac{1}{12} sec:
    • A=(2,23)A = (2, 2\sqrt{3})
    • B=(10,0)B = (10, 0)
  • (b) Angular velocity ω=4π\omega = 4\pi radians per second
  • (c) Parametric equations for AA:
    • x=4cos(4πt)x = 4 \cos(4\pi t)
    • y=4sin(4πt)y = 4 \sin(4\pi t)

Would you like further clarification or details on any part of this solution?


Related Questions

  1. How would the coordinates of BB change if the rod length was different?
  2. What would be the coordinates of AA and BB after one full rotation?
  3. How does changing the wheel's rotational speed affect the angular velocity?
  4. How can you calculate the instantaneous velocity of point AA at any given time?
  5. What would happen to the motion if the rod were rigidly fixed rather than sliding?

Tip

Always remember that in circular motion, angular velocity ω\omega is directly proportional to the rotational speed. More rotations per second mean a higher angular velocity.

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Math Problem Analysis

Mathematical Concepts

Circular Motion
Angular Velocity
Trigonometry
Parametric Equations

Formulas

Angular velocity formula: ω = 2π × (rotations per second)
Position formula for circular motion: x = R cos(ωt), y = R sin(ωt)
Trigonometric identities for coordinates

Theorems

Properties of Circular Motion
Trigonometric Formulas for Coordinates

Suitable Grade Level

Grades 10-12