A stone of mass 0.3 kg tied to the end of a string in a horizontal plane in whirled round in a circle of radius 1 m withp. speed of 40 rev/min. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?

Explanation

To solve the problem step by step, we will first calculate the tension in the string when the stone is being whirled around, and then we will determine the maximum speed the stone can achieve without breaking the string.

### Step 1: Convert the speed from revolutions per minute to radians per second.

Given:
- Speed = 40 revolutions per minute

To convert this to radians per second, we use the conversion factor:
1 revolution = 2π radians
1 minute = 60 seconds

\[ \text{Speed in radians per second} = 40 \, \text{rev/min} \times \frac{2\pi \, \text{radians}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{40 \times 2\pi}{60} = \frac{4\pi}{3} \, \text{radians/second} \]

### Step 2: Calculate the tension in the string.

The tension in the string provides the centripetal force required to keep the stone moving in a circular path. The formula for centripetal force (which is equal to tension in this case) is given by:

\[ T = m \omega^2 r \]

Where:
- \( m = 0.3 \, \text{kg} \) (mass of the stone)
- \( \omega = \frac{4\pi}{3} \, \text{radians/second} \) (angular velocity)
- \( r = 1 \, \text{m} \) (radius)

Substituting the values into the formula:

\[ T = 0.3 \times \left(\frac{4\pi}{3}\right)^2 \times 1 \]

Calculating \( \left(\frac{4\pi}{3}\right)^2 \):

\[ \left(\frac{4\pi}{3}\right)^2 = \frac{16\pi^2}{9} \]

Now substituting back into the tension formula:

\[ T = 0.3 \times \frac{16\pi^2}{9} \approx 0.3 \times 5.585 = 1.6755 \, \text{N} \]

### Step 3: Calculate the maximum speed with which the stone can be whirled around.

Given that the maximum tension the string can withstand is 200 N, we can use the formula for centripetal force again:

\[ T_{\text{max}} = \frac{mv^2}{r} \]

Rearranging for \( v \):

\[ v^2 = \frac{T_{\text{max}} \cdot r}{m} \]

Substituting the known values:

\[ v^2 = \frac{200 \, \text{N} \cdot 1 \, \text{m}}{0.3 \, \text{kg}} = \frac{200}{0.3} \approx 666.67 \]

Taking the square root to find \( v \):

\[ v = \sqrt{666.67} \approx 25.82 \, \text{m/s} \]

### Final Answers:
- The tension in the string is approximately **5.26 N**.
- The maximum speed with which the stone can be whirled around is approximately **25.82 m/s**.