A ball tied to a string takes us to complete revolution along a horizontal circle. If, by pulling the cord, the radius of the circle is reduceed to half of the previous value, then how much time the ball will take in one revolution?

Answer

To solve the problem, we need to analyze the situation using the principles of rotational motion and angular momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the initial conditions
The ball is tied to a string and is moving in a horizontal circle. Initially, it takes 4 seconds to complete one revolution.

### Step 2: Calculate the initial angular velocity
The angular velocity (ω_initial) can be calculated from the time taken for one revolution. Since one complete revolution corresponds to an angle of \(2\pi\) radians:
\[ \omega_{\text{initial}} = \frac{2\pi \text{ radians}}{4 \text{ seconds}} = \frac{\pi}{2} \text{ radians/second} \]

### Step 3: Analyze the change in radius
When the radius of the circle is reduced to half, the new radius (R_final) is:
\[ R_{\text{final}} = \frac{R_{\text{initial}}}{2} \]

### Step 4: Apply the principle of conservation of angular momentum
Since there is no external torque acting on the system, the angular momentum before and after the change in radius must be equal. The angular momentum (L) is given by:
\[ L = I \cdot \omega \]
where \(I\) is the moment of inertia.

The moment of inertia for a point mass (the ball) is:
\[ I = mR^2 \]
Thus, for the initial and final states:
\[ L_{\text{initial}} = mR_{\text{initial}}^2 \cdot \omega_{\text{initial}} \]
\[ L_{\text{final}} = mR_{\text{final}}^2 \cdot \omega_{\text{final}} \]

Setting these equal gives:
\[ mR_{\text{initial}}^2 \cdot \omega_{\text{initial}} = m\left(\frac{R_{\text{initial}}}{2}\right)^2 \cdot \omega_{\text{final}} \]

### Step 5: Simplify the equation
Canceling \(m\) from both sides and substituting \(R_{\text{final}} = \frac{R_{\text{initial}}}{2}\):
\[ R_{\text{initial}}^2 \cdot \omega_{\text{initial}} = \frac{R_{\text{initial}}^2}{4} \cdot \omega_{\text{final}} \]
\[ \omega_{\text{final}} = 4 \cdot \omega_{\text{initial}} \]

### Step 6: Calculate the final angular velocity
Substituting the value of \(\omega_{\text{initial}}\):
\[ \omega_{\text{final}} = 4 \cdot \frac{\pi}{2} = 2\pi \text{ radians/second} \]

### Step 7: Calculate the time for one revolution
The time for one revolution (T_final) is the reciprocal of the angular velocity in revolutions per second:
\[ T_{\text{final}} = \frac{1 \text{ revolution}}{\omega_{\text{final}}} = \frac{1 \text{ revolution}}{1 \text{ revolution/second}} = 1 \text{ second} \]

### Final Answer
The ball will take **1 second** to complete one revolution after the radius is halved.

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