A stone tied at the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude of acceleration of the stone ?

Options

A. 1. `9.0 m s^(-2)`
B. 2. `12.0 m s^(-2)`
C. 3. `11 m s^(-2)`
D. 4. `9.9 m s^(-2)`

Answer

To find the magnitude of acceleration of the stone being whirled in a horizontal circle, we can follow these steps:

### Step 1: Identify the given values
- Length of the string (radius, \( R \)) = 80 cm = 0.8 m (conversion to SI units)
- Number of revolutions = 14
- Time taken = 25 s

### Step 2: Calculate the frequency (\( \nu \))
Frequency is the number of revolutions per second. We can calculate it using the formula:
\[ \nu = \frac{\text{Number of revolutions}}{\text{Time in seconds}} = \frac{14}{25} \text{ revolutions per second} \]

### Step 3: Calculate the angular velocity (\( \omega \))
Angular velocity can be calculated using the formula:
\[ \omega = 2\pi \nu \]
Substituting the value of \( \nu \):
\[ \omega = 2\pi \left(\frac{14}{25}\right) \]

### Step 4: Simplify the angular velocity
Calculating \( \omega \):
\[ \omega = 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25} \text{ radians per second} \]
Calculating this gives:
\[ \omega \approx 3.52 \text{ radians per second} \]

### Step 5: Calculate the centripetal acceleration (\( a \))
The formula for centripetal acceleration is:
\[ a = R \omega^2 \]
Substituting the values of \( R \) and \( \omega \):
\[ a = 0.8 \times (3.52)^2 \]

### Step 6: Calculate \( \omega^2 \)
Calculating \( \omega^2 \):
\[ (3.52)^2 \approx 12.3904 \]

### Step 7: Calculate the acceleration
Now substituting back to find \( a \):
\[ a = 0.8 \times 12.3904 \approx 9.91232 \text{ m/s}^2 \]

### Final Answer
The magnitude of acceleration of the stone is approximately:
\[ a \approx 9.91 \text{ m/s}^2 \]