A stone of mass M tied at the end of a string, is moving in a circular of radius R, with a constant angular velocity `omega`. The work done on the stone, in any half circle I s

Options

A. `piMR^(2)omega^(2)`
B. `2MR^(2)omega^(2)`
C. `MR^(2)omega^(2)`
D. zero

Answer

To solve the problem of the work done on a stone of mass \( M \) tied at the end of a string moving in a circular path of radius \( R \) with a constant angular velocity \( \omega \), we follow these steps:

### Step 1: Understand the Concept of Work Done
Work done \( W \) is defined as the dot product of force \( F \) and displacement \( d \). Mathematically, it can be expressed as:
\[ W = F \cdot d = F \cdot d \cos(\theta) \]
where \( \theta \) is the angle between the force and the displacement vectors.

### Step 2: Analyze the Forces in Circular Motion
In circular motion, the stone experiences a centripetal force directed towards the center of the circular path. This force is responsible for keeping the stone in circular motion. The tension in the string provides this centripetal force.

### Step 3: Determine the Direction of Displacement
When the stone moves in a circular path, its displacement during any half-circle is tangential to the circle. The direction of the centripetal force (tension in the string) is always directed towards the center of the circle.

### Step 4: Calculate the Angle Between Force and Displacement
In this scenario, the angle \( \theta \) between the centripetal force (tension) and the displacement of the stone is \( 90^\circ \) because the force is directed radially inward while the displacement is tangential.

### Step 5: Substitute Values into the Work Done Formula
Since \( \theta = 90^\circ \):
\[ \cos(90^\circ) = 0 \]
Thus, the work done can be calculated as:
\[ W = F \cdot d \cdot \cos(90^\circ) = F \cdot d \cdot 0 = 0 \]

### Conclusion
The work done on the stone in any half-circle is zero.

### Final Answer
The work done on the stone in any half-circle is \( 0 \).

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