A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. If the centripetal force F is kept constant but the angular velocity is doubled , the new radius of the path (original radius R) will be

Options

A. `R//4`
B. `R//2`
C. 2R
D. 4R

Answer

To solve the problem step by step, we will use the concept of centripetal force in uniform circular motion.

### Step 1: Understand the formula for centripetal force
The centripetal force \( F \) acting on a particle moving in a circle of radius \( R \) with angular velocity \( \omega \) is given by the formula:
\[ F = m R \omega^2 \]
where \( m \) is the mass of the particle.

### Step 2: Set up the initial conditions
Let:
- The initial radius be \( R = 20 \, \text{cm} \).
- The initial angular velocity be \( \omega_1 \).
- The centripetal force be \( F \) (which is constant).

From the formula, we can write the initial centripetal force as:
\[ F = m R \omega_1^2 \]

### Step 3: Set up the conditions after doubling the angular velocity
If the angular velocity is doubled, then:
\[ \omega_2 = 2 \omega_1 \]
Let the new radius be \( R_2 \). The centripetal force in this case can be expressed as:
\[ F = m R_2 \omega_2^2 \]
Substituting \( \omega_2 \):
\[ F = m R_2 (2 \omega_1)^2 = m R_2 \cdot 4 \omega_1^2 \]

### Step 4: Set the two expressions for centripetal force equal to each other
Since the centripetal force is constant, we can set the two expressions for \( F \) equal to each other:
\[ m R \omega_1^2 = m R_2 \cdot 4 \omega_1^2 \]

### Step 5: Cancel out common terms
Since \( m \) and \( \omega_1^2 \) are common in both sides, we can cancel them:
\[ R = 4 R_2 \]

### Step 6: Solve for the new radius \( R_2 \)
Rearranging the equation gives:
\[ R_2 = \frac{R}{4} \]
Substituting \( R = 20 \, \text{cm} \):
\[ R_2 = \frac{20 \, \text{cm}}{4} = 5 \, \text{cm} \]

### Final Answer
The new radius of the path \( R_2 \) will be \( 5 \, \text{cm} \).

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