A stone tied to the end of a string 1m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, what is the magnitude and direction of acceleration of the stone

Options

A. `pi^(2)//4 ms^(-2)` and direction along the radius towards the centre
B. `pi^(2) ms^(-2)` and direction along the radius away from the centre
C. `pi^(2) ms^(-2)` and direction along the radius towards the centre
D. `pi^(2) ms^(-2)` and direction along the tangent to the circle

Explanation

To solve the problem of a stone tied to a string and whirled in a horizontal circle, we will follow these steps:

### Step 1: Identify the given data
- Length of the string (radius of the circle, R) = 1 m
- Number of revolutions = 22
- Time taken = 44 seconds

### Step 2: Calculate the frequency of the motion
Frequency (F) is the number of revolutions per second.

\[ F = \frac{\text{Number of revolutions}}{\text{Time in seconds}} = \frac{22}{44} = 0.5 \text{ revolutions per second} \]

### Step 3: Convert frequency to angular velocity
Angular velocity (ω) can be calculated using the formula:

\[ \omega = 2\pi F \]

Substituting the value of frequency:

\[ \omega = 2\pi \times 0.5 = \pi \text{ radians per second} \]

### Step 4: Calculate the centripetal acceleration
Centripetal acceleration (a_c) is given by the formula:

\[ a_c = R \omega^2 \]

Substituting the values of R and ω:

\[ a_c = 1 \times (\pi)^2 = \pi^2 \text{ m/s}^2 \]

### Step 5: Determine the direction of the acceleration
The direction of centripetal acceleration is always directed towards the center of the circular path.

### Final Answer
The magnitude of the acceleration of the stone is \( \pi^2 \text{ m/s}^2 \) and its direction is towards the center of the circle.

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