Math Problem Statement

A4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius2cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The wheel rotates counterclockwise at2.5revolutions per second. Thus, when t=1/15 sec, the rod is situated as in the diagram at the right below. Note that the wheel has its center at the origin.

Solution

To analyze the motion of the rod and its attachment to the rotating wheel, we can break down the problem into a few key components:

  1. Wheel Rotation: The wheel rotates at 2.5 revolutions per second. This corresponds to an angular velocity ω\omega given by: ω=2.5×2π rad/s=5π rad/s\omega = 2.5 \times 2\pi \text{ rad/s} = 5\pi \text{ rad/s}

  2. Position of Point A: At any time tt, the position of point AA on the circumference of the wheel can be described using the parametric equations: A(t)=(2cos(ωt),2sin(ωt))A(t) = (2 \cos(\omega t), 2 \sin(\omega t))

  3. Rod Movement: The rod length is 4 cm. The position of point BB can vary based on the angle that the rod makes with the horizontal bar.

At t=0t = 0 seconds, we can denote the initial position of AA as: A(0)=(2,0)A(0) = (2, 0) and point BB is at (xB,0)(x_B, 0) on the horizontal bar.

At t=115t = \frac{1}{15} seconds, we calculate the new position of AA: A(115)=(2cos(5π115),2sin(5π115))A\left(\frac{1}{15}\right) = (2 \cos\left(5\pi \cdot \frac{1}{15}\right), 2 \sin\left(5\pi \cdot \frac{1}{15}\right))

Calculating these: ωt=5π115=π3\omega t = 5\pi \cdot \frac{1}{15} = \frac{\pi}{3} A(115)=(2cos(π3),2sin(π3))=(1,3)A\left(\frac{1}{15}\right) = (2 \cos(\frac{\pi}{3}), 2 \sin(\frac{\pi}{3})) = (1, \sqrt{3})

  1. Position of B: Given that the length of the rod is 4 cm, the position of BB will depend on the angle of the rod with respect to the horizontal. As BB moves along the horizontal bar, the constraint is that the distance ABAB equals 4 cm. The coordinates of BB will need to satisfy: (xBxA)2+(yByA)2=42(x_B - x_A)^2 + (y_B - y_A)^2 = 4^2

If you want to delve into specifics like the position of BB at t=115t = \frac{1}{15} seconds or the subsequent motion, please clarify your focus.

Do you have any specific questions or need details on a particular part of this analysis? Here are some related questions you might consider:

  1. What is the position of point BB at t=0t = 0?
  2. How does the angle of the rod affect the position of BB?
  3. What are the coordinates of BB when t=115t = \frac{1}{15} seconds?
  4. How can we describe the motion of BB over time?
  5. What other factors could influence the movement of the rod?

Tip: Visualize the problem by sketching the positions of AA and BB at different times to better understand the relationships.

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Math Problem Analysis

Mathematical Concepts

Rotational Motion
Parametric Equations
Geometry

Formulas

\(\omega = 2.5 \times 2\pi\)
\(A(t) = (2 \cos(\omega t), 2 \sin(\omega t))\)
\((x_B - x_A)^2 + (y_B - y_A)^2 = 4^2\)

Theorems

Motion of Rigid Bodies
Geometry of Circles

Suitable Grade Level

Grades 11-12