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A stone is tied to a string and swung in a vertical circle. At the highest point

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Question: A stone is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 5 N and the weight of the stone is 10 N. What is the speed of the stone at the highest point if the radius of the circle is 2 m?

Options:

  1. 2 m/s
  2. 3 m/s
  3. 4 m/s
  4. 5 m/s

Correct Answer: 4 m/s

Solution: 

At the highest point, T + mg = mv²/r. 5 + 10 = (m*v²)/2. Solving gives v = 4 m/s.

A stone is tied to a string and swung in a vertical circle. At the highest point

Practice Questions

Q1
A stone is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 5 N and the weight of the stone is 10 N. What is the speed of the stone at the highest point if the radius of the circle is 2 m?
  1. 2 m/s
  2. 3 m/s
  3. 4 m/s
  4. 5 m/s

Questions & Step-by-Step Solutions

A stone is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 5 N and the weight of the stone is 10 N. What is the speed of the stone at the highest point if the radius of the circle is 2 m?
  • Step 1: Identify the forces acting on the stone at the highest point. The tension in the string (T) is 5 N and the weight of the stone (mg) is 10 N.
  • Step 2: Write down the equation for the forces at the highest point. The equation is T + mg = mv²/r, where m is the mass of the stone, v is the speed, and r is the radius of the circle.
  • Step 3: Substitute the known values into the equation. We have T = 5 N, mg = 10 N, and r = 2 m. So, the equation becomes 5 + 10 = (m * v²) / 2.
  • Step 4: Combine the forces on the left side of the equation. This gives us 15 = (m * v²) / 2.
  • Step 5: Rearrange the equation to solve for v². Multiply both sides by 2 to get 30 = m * v².
  • Step 6: We need to find the mass (m). Since the weight (mg) is 10 N, we can find m by dividing the weight by g (acceleration due to gravity, approximately 10 m/s²). So, m = 10 N / 10 m/s² = 1 kg.
  • Step 7: Substitute the mass back into the equation. Now we have 30 = 1 * v², which simplifies to 30 = v².
  • Step 8: Take the square root of both sides to find v. So, v = √30, which is approximately 5.48 m/s.
  • Step 9: However, the short solution provided states v = 4 m/s. This discrepancy suggests a need to double-check the calculations or assumptions made in the problem.
  • Centripetal Force – Understanding the forces acting on an object in circular motion, particularly at the highest point where both tension and weight contribute to the centripetal force.
  • Newton's Second Law – Applying Newton's second law to relate the net force acting on the stone to its mass and acceleration.
  • Circular Motion – Calculating the speed of an object in circular motion using the relationship between radius, mass, and forces.
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