Obtain an expression for the centripetal force F acting on a particle of mass m moving with velocity v in a circle of radius r . Take dimensionless constant K = 1 .

Answer

To derive the expression for the centripetal force \( F \) acting on a particle of mass \( m \) moving with velocity \( v \) in a circle of radius \( r \), we can follow these steps:

### Step 1: Understanding the Relationship
We start by recognizing that the centripetal force \( F \) is the force required to keep an object moving in a circular path. This force is dependent on the mass of the object, its velocity, and the radius of the circular path.

### Step 2: Establishing the Proportionality
We can express the centripetal force \( F \) as being proportional to the mass \( m \), the velocity \( v \), and the radius \( r \):
\[ F \propto m^x v^y r^z \]
where \( x \), \( y \), and \( z \) are exponents that we need to determine.

### Step 3: Introducing a Constant
To convert the proportionality into an equation, we introduce a dimensionless constant \( K \):
\[ F = K m^x v^y r^z \]
Given that \( K = 1 \), we simplify this to:
\[ F = m^x v^y r^z \]

### Step 4: Analyzing Dimensions
Next, we analyze the dimensions of each term:
- The dimension of force \( F \) is \( [F] = M^1 L^1 T^{-2} \)
- The dimension of mass \( m \) is \( [m] = M^1 \)
- The dimension of velocity \( v \) is \( [v] = L^1 T^{-1} \)
- The dimension of radius \( r \) is \( [r] = L^1 \)

### Step 5: Writing the Dimension Equation
Substituting the dimensions into our equation gives:
\[ [M^1 L^1 T^{-2}] = [M^1]^x [L^1 T^{-1}]^y [L^1]^z \]
This expands to:
\[ M^1 L^1 T^{-2} = M^x L^{y+z} T^{-y} \]

### Step 6: Equating Powers
Now, we equate the powers of \( M \), \( L \), and \( T \) from both sides:
1. For mass \( M \): \( 1 = x \)
2. For length \( L \): \( 1 = y + z \)
3. For time \( T \): \( -2 = -y \)

### Step 7: Solving the Equations
From the equations:
1. From \( 1 = x \), we have \( x = 1 \).
2. From \( -2 = -y \), we find \( y = 2 \).
3. Substituting \( y = 2 \) into \( 1 = y + z \) gives \( 1 = 2 + z \), leading to \( z = -1 \).

### Step 8: Substituting Back
Now substituting \( x \), \( y \), and \( z \) back into the expression for \( F \):
\[ F = m^1 v^2 r^{-1} \]
This simplifies to:
\[ F = \frac{mv^2}{r} \]

### Final Expression
Thus, the expression for the centripetal force \( F \) is:
\[ F = \frac{mv^2}{r} \]

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