A body revolves with constant speed v in a circular path of radius r. the magnitude of its average acceleration during motion between two points in diametrically opposite direction is

Options

A. Zero
B. `(v^(2))/(r )`
C. `(2v^(2))/(pi r)`
D. `(v^(2))/(2r)`

Answer

To find the magnitude of the average acceleration of a body moving with constant speed \( v \) in a circular path of radius \( r \) between two points that are diametrically opposite, we can follow these steps:

### Step 1: Understand the motion
The body is in uniform circular motion, meaning it travels around a circular path with a constant speed. The points in question are diametrically opposite, which means they are on opposite ends of the diameter of the circle.

### Step 2: Define initial and final velocities
Let’s denote the initial velocity vector when the body is at point A (let's say at the top of the circle) as:
\[ \vec{v_i} = v \hat{j} \]
And the final velocity vector when the body reaches point B (at the bottom of the circle) as:
\[ \vec{v_f} = -v \hat{j} \]

### Step 3: Calculate the change in velocity
The change in velocity \( \Delta \vec{v} \) can be calculated as:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} = (-v \hat{j}) - (v \hat{j}) = -2v \hat{j} \]

### Step 4: Determine the time interval
The time taken to move from point A to point B is half the time period of the circular motion. The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi r}{v} \]
Thus, the time interval \( \Delta t \) for half a revolution is:
\[ \Delta t = \frac{T}{2} = \frac{2\pi r}{2v} = \frac{\pi r}{v} \]

### Step 5: Calculate average acceleration
Average acceleration \( \vec{a_{avg}} \) is defined as the change in velocity divided by the time interval:
\[ \vec{a_{avg}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-2v \hat{j}}{\frac{\pi r}{v}} = \frac{-2v^2 \hat{j}}{\pi r} \]

### Step 6: Find the magnitude of average acceleration
The magnitude of the average acceleration \( a_{avg} \) is:
\[ a_{avg} = \left| \frac{-2v^2}{\pi r} \right| = \frac{2v^2}{\pi r} \]

### Conclusion
Thus, the magnitude of the average acceleration during the motion between two diametrically opposite points is:
\[ \boxed{\frac{2v^2}{\pi r}} \]

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