A rolling object rolls without slipping down an inclined plane (angle of inclination `theta`), then the minimum acceleration it can have is ?

Options

A. `gsintheta`
B. `(2gsintheta)/3`
C. `(gsintheta)/2`
D. zero

Explanation

To find the minimum acceleration of a rolling object down an inclined plane without slipping, we can follow these steps:

### Step 1: Understand the dynamics of rolling motion
When an object rolls down an incline, both translational and rotational motions are involved. The acceleration of the center of mass of the object can be expressed in terms of gravitational acceleration and the geometry of the incline.

### Step 2: Write down the equation for acceleration
The acceleration \( a \) of a rolling object down an incline can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
where:
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of inclination,
- \( I \) is the moment of inertia of the object,
- \( m \) is the mass of the object,
- \( r \) is the radius of the object.

### Step 3: Identify the condition for minimum acceleration
To find the minimum acceleration, we need to maximize the term \( \frac{I}{m r^2} \). The maximum moment of inertia \( I \) for a ring (which is a common rolling object) is given by:
\[ I_{\text{ring}} = m r^2 \]

### Step 4: Substitute the moment of inertia into the acceleration formula
Substituting \( I = m r^2 \) into the acceleration formula gives:
\[ a = \frac{g \sin \theta}{1 + \frac{m r^2}{m r^2}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]

### Step 5: Conclusion
Thus, the minimum acceleration \( a_{\text{min}} \) of the rolling object down the incline is:
\[ a_{\text{min}} = \frac{g \sin \theta}{2} \]

### Final Answer
The minimum acceleration the rolling object can have is \( \frac{g \sin \theta}{2} \).

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