A stone is tied to one end of a string 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s , what is the magnitude of acceleration of the stone

Options

A. `493 cm//s`
B. `720 cm//s`
C. `860 cm//s`
D. `990 cm//s`

Answer

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

### Step 1: Identify the given data
- Length of the string (radius, r) = 50 cm = 0.5 m (convert to meters)
- Number of revolutions (n) = 10
- Time taken (t) = 20 s

### Step 2: Calculate the angular velocity (ω)
The angular velocity can be calculated using the formula:
\[ \omega = \frac{2\pi n}{t} \]
Substituting the values:
\[ \omega = \frac{2\pi \times 10}{20} = \frac{20\pi}{20} = \pi \, \text{rad/s} \]

### Step 3: Calculate the centripetal acceleration (a_c)
The centripetal acceleration is given by the formula:
\[ a_c = r \omega^2 \]
Substituting the values:
\[ a_c = 0.5 \times (\pi)^2 \]
Calculating \((\pi)^2\):
\[ \pi^2 \approx 9.87 \]
Thus,
\[ a_c = 0.5 \times 9.87 \approx 4.935 \, \text{m/s}^2 \]

### Step 4: Final Result
The magnitude of the acceleration of the stone is approximately:
\[ \boxed{4.94 \, \text{m/s}^2} \]