A ball is placed on a smooth inclined plane of inclination `theta=30^(@)` to the horizontal, which is rotating at frequency `0.5` Hz about a vertical axiz passing through its lower end. At what distance from the lower end does the ball remain at rest?

Options

A. `0.87m`
B. `0.33m`
C. `0.5m`
D. `0.67m`

Explanation

To solve the problem, we need to analyze the forces acting on the ball placed on the inclined plane that is rotating about a vertical axis. Here’s a step-by-step solution:

### Step 1: Understand the System
The inclined plane is at an angle \( \theta = 30^\circ \) to the horizontal, and it is rotating with a frequency \( f = 0.5 \) Hz. We need to find the distance \( d \) from the lower end of the inclined plane where the ball remains at rest.

### Step 2: Convert Frequency to Angular Velocity
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 0.5 = \pi \, \text{rad/s} \]

### Step 3: Analyze Forces Acting on the Ball
The forces acting on the ball are:
1. The gravitational force \( mg \) acting downwards.
2. The normal force \( N \) acting perpendicular to the surface of the inclined plane.

### Step 4: Set Up the Equations
From the free body diagram, we can resolve the forces into components along the inclined plane:
- In the radial direction (perpendicular to the incline):
\[ N \cos \theta = mg \]
- In the tangential direction (along the incline):
\[ N \sin \theta = m r \omega^2 \]
where \( r \) is the horizontal distance from the axis of rotation to the ball.

### Step 5: Solve for Normal Force
From the first equation:
\[ N = \frac{mg}{\cos \theta} \]

### Step 6: Substitute Normal Force into the Second Equation
Substituting \( N \) into the second equation gives:
\[ \frac{mg}{\cos \theta} \sin \theta = m r \omega^2 \]
Cancelling \( m \) from both sides:
\[ \frac{g \sin \theta}{\cos \theta} = r \omega^2 \]
This simplifies to:
\[ g \tan \theta = r \omega^2 \]

### Step 7: Solve for \( r \)
Rearranging for \( r \):
\[ r = \frac{g \tan \theta}{\omega^2} \]

### Step 8: Substitute Known Values
Substituting \( g = 10 \, \text{m/s}^2 \), \( \theta = 30^\circ \), and \( \omega = \pi \):
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Thus,
\[ r = \frac{10 \cdot \frac{1}{\sqrt{3}}}{\pi^2} \]

### Step 9: Calculate \( r \)
Calculating \( r \):
\[ r = \frac{10}{\sqrt{3} \cdot \pi^2} \]

### Step 10: Calculate Distance \( d \)
The distance \( d \) from the lower end of the inclined plane can be found using:
\[ d = \frac{r}{\cos \theta} \]
Substituting \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\[ d = \frac{\frac{10}{\sqrt{3} \cdot \pi^2}}{\frac{\sqrt{3}}{2}} = \frac{20}{3 \pi^2} \]

### Step 11: Final Calculation
Calculating the numerical value:
\[ d \approx 0.67 \, \text{m} \]

### Final Answer
The ball remains at rest at a distance of approximately \( 0.67 \, \text{m} \) from the lower end of the inclined plane.

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