Question

A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point, is (if \( \pi^2 = 9.8 \) and \( g = 9.8 \, \text{m/s}^2 \))

• A. 97 N
• B. 9.8 N
• C. 8.82 N
• D. 17.8 N

Answer 1

The Correct Option is B

To find the tension in the string when the stone is at the lowest point, we need to analyze the forces acting on the stone in circular motion.

First, let's outline the necessary parameters: 

• Mass of stone, \(m = \frac{900}{1000} \, \text{kg} = 0.9 \, \text{kg}\)
• Radius of the circle, \(r = 1 \, \text{m}\)
• Revolutions per minute, \(N = 10 \, \text{rpm}\)
• Gravitational acceleration, \(g = 9.8 \, \text{m/s}^{2}\)

At the lowest point of the circle, the forces acting on the stone are the tension \( T \) in the string and the gravitational force. The net centripetal force needed to keep the stone moving in a circle is provided by the tension in the string minus the gravitational force:

\(T - mg = m \omega^2 r\)

Where \(\omega\) is the angular velocity in radians per second. We need to convert rpm to radians per second using the formula:

\(\omega = \frac{2\pi N}{60} \, \text{rad/s}\)

\(\omega = \frac{2 \times 3.14 \times 10}{60} = \frac{62.8}{60} \approx 1.047 \, \text{rad/s}\)

Now substitute \(\omega\) back into the centripetal force formula:

\(T = mg + m \omega^2 r\)

Let's calculate each component:

• \(mg = 0.9 \times 9.8 = 8.82 \, \text{N}\)
• \(m \omega^2 r = 0.9 \times (1.047)^2 \times 1 \approx 0.99 \, \text{N}\)

Add these to find the total tension:

\(T = 8.82 + 0.99 = 9.81 \, \text{N}\)

After rounding off based on significant figures provided in the options, the closest value is:

Correct Answer: 9.8 N

Answer 2

Given:  
- Mass of the stone, \( m = 900 \, \text{g} = 0.9 \, \text{kg} \)  
- Radius of the circle, \( r = 1 \, \text{m} \)  
- Angular velocity in rpm, \( \omega = 10 \, \text{rpm} \)

Step 1. Convert rpm to rad/s:  
  
  \(\omega = 10 \times \frac{2\pi}{60} = \frac{\pi}{3} \, \text{rad/s}\)
  \]

Step 2. Calculate the centripetal force at the lowest point:  
  The centripetal force \( F_c = m\omega^2r \):  
 
  \(F_c = 0.9 \times \left(\frac{\pi}{3}\right)^2 \times 1 = 0.9 \times \frac{\pi^2}{9} = 0.9 \times \frac{9.8}{9} = 0.98 \, \text{N}\)
 
Step 3. Calculate the tension \( T \) at the lowest point:  
  At the lowest point, the tension \( T \) in the string must support both the gravitational force and the centripetal force. Thus:  
  \(T = mg + F_c\)
 \(T = (0.9 \times 9.8) + 0.98 = 8.82 + 0.98 = 9.8 \, \text{N}\)
Thus, the tension in the string at the lowest point is 9.8 N.

The Correct Answer is: 9.8 N