A body completes one revolution along a horizontal circle in 10 s when tied to a cord. If the radius of the circle is decreased to one-third of its original value, then what will be the time period of revolution of the body?

Answer

To solve the problem, we need to understand the relationship between the time period of revolution, the radius of the circle, and angular momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
The body completes one revolution in 10 seconds with an initial radius \( r \). The time period \( T_1 \) is given as:
\[ T_1 = 10 \, \text{s} \]

### Step 2: Relate Time Period to Angular Velocity
The angular velocity \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Thus, the initial angular velocity \( \omega_1 \) can be calculated as:
\[ \omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{10} = \frac{\pi}{5} \, \text{rad/s} \]

### Step 3: Understand the Change in Radius
The radius of the circle is decreased to one-third of its original value:
\[ r' = \frac{r}{3} \]

### Step 4: Apply Conservation of Angular Momentum
Since no external torque is acting on the system, angular momentum is conserved. The initial angular momentum \( L_i \) can be expressed as:
\[ L_i = I_1 \omega_1 \]
Where \( I_1 \) is the moment of inertia at radius \( r \).

The final angular momentum \( L_f \) when the radius is \( r' \) is:
\[ L_f = I_f \omega_f \]
Where \( \omega_f \) is the final angular velocity and \( I_f \) is the moment of inertia at radius \( r' \).

### Step 5: Calculate Moment of Inertia
The moment of inertia \( I \) for a point mass at a distance \( r \) is given by:
\[ I = m r^2 \]
Thus,
- Initial moment of inertia:
\[ I_1 = m r^2 \]
- Final moment of inertia:
\[ I_f = m (r')^2 = m \left(\frac{r}{3}\right)^2 = m \frac{r^2}{9} \]

### Step 6: Set Up the Conservation Equation
Using the conservation of angular momentum:
\[ I_1 \omega_1 = I_f \omega_f \]
Substituting the values:
\[ m r^2 \cdot \frac{\pi}{5} = m \frac{r^2}{9} \cdot \omega_f \]

### Step 7: Simplify and Solve for Final Angular Velocity
Canceling \( m \) and \( r^2 \) from both sides:
\[ \frac{\pi}{5} = \frac{\omega_f}{9} \]
Rearranging gives:
\[ \omega_f = \frac{9\pi}{5} \, \text{rad/s} \]

### Step 8: Calculate the Final Time Period
Now, we can find the new time period \( T_2 \) using the relationship between angular velocity and time period:
\[ T_2 = \frac{2\pi}{\omega_f} = \frac{2\pi}{\frac{9\pi}{5}} = \frac{2 \cdot 5}{9} = \frac{10}{9} \, \text{s} \]

### Final Answer
The time period of revolution of the body when the radius is decreased to one-third of its original value is:
\[ T_2 = \frac{10}{9} \, \text{s} \]

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