A ring of radius R is rolling purely on the outer surface of a pipe of radius 4R At some instant the center of the ring has constant speed =v then the acceleration of the point on the ring which is in contact with the surface of the pipe is

Options

A. `4v^(2)//5R`
B. `3v^(2)//5R`
C. `v^(2)//4R`
D. zero

Explanation

To find the acceleration of the point on the ring that is in contact with the surface of the pipe, we can follow these steps:

### Step 1: Identify the parameters
- The radius of the ring is \( R \).
- The radius of the pipe is \( 4R \).
- The center of the ring is moving with a constant speed \( v \).

### Step 2: Determine the distance from the center of the ring to the center of the pipe
The distance from the center of the ring to the center of the pipe is:
\[ \text{Distance} = R + 4R = 5R \]

### Step 3: Calculate the centripetal acceleration of the center of the ring
The centripetal acceleration \( a_{cm} \) of the center of the ring is given by the formula:
\[ a_{cm} = \frac{v^2}{r} \]
where \( r \) is the radius of the circular path, which in this case is \( 5R \):
\[ a_{cm} = \frac{v^2}{5R} \]

### Step 4: Determine the angular velocity \( \omega \)
Since the ring is rolling without slipping, we can relate the linear speed \( v \) to the angular speed \( \omega \) using the formula:
\[ v = \omega R \]
Thus, we can express \( \omega \) as:
\[ \omega = \frac{v}{R} \]

### Step 5: Calculate the centripetal acceleration of the point in contact with the pipe
The point in contact with the pipe also experiences centripetal acceleration due to its circular motion around the center of the ring. This acceleration \( a_n \) is given by:
\[ a_n = \omega^2 r \]
where \( r \) is the radius of the ring \( R \):
\[ a_n = \left(\frac{v}{R}\right)^2 R = \frac{v^2}{R} \]

### Step 6: Determine the net acceleration of the point in contact
The net acceleration \( a \) of the point in contact with the pipe is the difference between the centripetal acceleration of the center of the ring and the centripetal acceleration of the point in contact:
\[ a = a_{cm} - a_n \]
Substituting the values we calculated:
\[ a = \frac{v^2}{5R} - \frac{v^2}{R} \]
To combine these fractions, we find a common denominator:
\[ a = \frac{v^2}{5R} - \frac{5v^2}{5R} = \frac{v^2 - 5v^2}{5R} = \frac{-4v^2}{5R} \]

### Final Result
The acceleration of the point on the ring which is in contact with the surface of the pipe is:
\[ a = -\frac{4v^2}{5R} \]