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

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 centre
C. `pi^(2)ms^(-2)` and direction aloing the radius towards the centre
D. `pi^(2)ms^(-2)` and direction along the tangent to the circle

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the frequency of the stone's motion.
The stone makes 22 revolutions in 44 seconds. To find the frequency (f), we use the formula:
\[ f = \frac{\text{Number of revolutions}}{\text{Time in seconds}} = \frac{22}{44} = 0.5 \text{ Hz} \]

### Step 2: Calculate the angular velocity (ω).
The angular velocity (ω) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 0.5 = \pi \text{ rad/s} \]

### Step 3: Calculate the centripetal acceleration (a_c).
Centripetal acceleration can be calculated using the formula:
\[ a_c = \omega^2 r \]
Where \( r \) is the radius of the circular path (1 meter in this case). Substituting the values:
\[ a_c = \pi^2 \times 1 = \pi^2 \text{ m/s}^2 \]

### Step 4: Determine the direction of the acceleration.
The direction of centripetal acceleration is always directed towards the center of the circular path along the radius.

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

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