Question

A point marked on a ring of radius \(2\,\text{cm}\) is in contact with a horizontal plane. Now the ring is rolled forward half a revolution along the positive X direction. Then the angle made by the displacement vector of the point with the X-axis is:

• A. \(\theta = \tan^{-1}\left(\frac{2}{3\pi}\right)\)
• B. \(\theta = \tan^{-1}\left(\frac{2}{\pi}\right)\)
• C. \(\theta = \tan^{-1}\left(\frac{2\pi}{3}\right)\)
• D. \(\theta = \cot^{-1}\left(\frac{2}{\pi}\right)\)

Hint:

For rolling without slipping, distance travelled by centre equals arc length \(R\theta\). For half revolution, horizontal displacement is \(\pi R\), while the marked bottom point rises by \(2R\).

Answer 1

The Correct Option is B

Step 1: Understand the motion of the marked point.
Initially, the marked point is at the point of contact with the horizontal plane. When the ring rolls without slipping, the marked point follows a cycloidal path.

Step 2: Find horizontal displacement after half revolution.
For half revolution, the centre of the ring moves forward by half the circumference:
\[ x = \pi R \]
Given,
\[ R = 2\,\text{cm} \]
So,
\[ x = \pi \times 2 = 2\pi\,\text{cm} \]

Step 3: Find vertical displacement after half revolution.
After half revolution, the point which was initially at the bottom reaches the top of the ring. Hence its vertical displacement is equal to diameter:
\[ y = 2R \]
\[ y = 2 \times 2 = 4\,\text{cm} \]

Step 4: Use angle formula.
The angle made by displacement vector with the X-axis is:
\[ \tan\theta = \frac{y}{x} \]

Step 5: Substitute values.
\[ \tan\theta = \frac{4}{2\pi} \]
\[ \tan\theta = \frac{2}{\pi} \]

Step 6: Find angle.
\[ \theta = \tan^{-1}\left(\frac{2}{\pi}\right) \]

Step 7: Final answer.
\[ \boxed{\theta = \tan^{-1}\left(\frac{2}{\pi}\right)} \]