A wheel rotates counterclockwise about its center O. A point P on the rim starts at the rightmost point of...

1. TRANSLATE the problem information

• Given information:
  • Wheel rotates counterclockwise about center O
  • Point P starts at rightmost position
  • Point P ends at diametrically opposite position (leftmost)
  • Need positive angle measure in radians

2. INFER the relationship between positions

• Key insight: Diametrically opposite points are \(\pi\) radians apart
• However, the wheel can complete multiple full rotations (\(2\pi\) radians each) before reaching the final position
• Therefore: \(\mathrm{\theta} = \pi + 2\pi\mathrm{k}\) where \(\mathrm{k} = 0, 1, 2, 3, ...\)
• This simplifies to: \(\mathrm{\theta} = (2\mathrm{k} + 1)\pi\) (odd multiples of \(\pi\))

3. SIMPLIFY each answer choice

Check which option gives an integer value for k in \(\mathrm{\theta} = (2\mathrm{k} + 1)\pi\):

• (A) \(\frac{23\pi}{2} = 11.5\pi\) → \(11.5 = 2\mathrm{k} + 1\) → \(\mathrm{k} = 5.25\) ✗
• (B) \(12\pi\) → \(12 = 2\mathrm{k} + 1\) → \(\mathrm{k} = 5.5\) ✗
• (C) \(\frac{25\pi}{2} = 12.5\pi\) → \(12.5 = 2\mathrm{k} + 1\) → \(\mathrm{k} = 5.75\) ✗
• (D) \(13\pi\) → \(13 = 2\mathrm{k} + 1\) → \(\mathrm{k} = 6\) ✓

4. APPLY CONSTRAINTS to select the valid answer

• Only choice D yields an integer value for k
• This confirms that \(13\pi\) represents exactly 6 full rotations plus one half-rotation

Answer: D (\(13\pi\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students recognize that \(\pi\) radians moves the point to the diametrically opposite position, but fail to consider that multiple full rotations are possible before reaching that final position.

They think only \(\pi\) radians works and become confused when \(\pi\) isn't among the answer choices. This leads to confusion and guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students understand the concept that \(\mathrm{\theta} = (2\mathrm{k} + 1)\pi\) but make calculation errors when checking which answer choices work, particularly with the fractional coefficients in choices A and C.

For example, they might incorrectly conclude that \(\frac{23\pi}{2}\) works because they miscalculate \(\frac{23}{2} = 2\mathrm{k} + 1\). This may lead them to select Choice A (\(\frac{23\pi}{2}\)).

The Bottom Line:

This problem tests whether students understand that rotational motion is periodic - the same final position can be reached through different total rotation amounts. The key insight is recognizing that "diametrically opposite" allows for multiple valid rotation angles, not just the minimum angle.