Math Problem Statement
Solution
The problem states:
"A 1000 g (1 kg) ball is tied to a 40 cm (0.4 m) long string and swings in a vertical circle. We are asked to calculate the minimum velocity the ball must have at the highest point of its trajectory for it to maintain circular motion."
Step-by-step solution:
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Concept: At the highest point of the vertical circle, the ball must maintain enough velocity so that the centripetal force is provided by both the tension in the string and the gravitational force. The minimum velocity occurs when the tension in the string is zero, meaning the centripetal force is provided solely by gravity.
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Centripetal Force Requirement: At the highest point, the gravitational force provides the necessary centripetal force. Therefore, we can write: where:
- is the mass of the ball (1 kg),
- is the velocity at the highest point,
- is the radius of the circle (0.4 m),
- is the acceleration due to gravity (9.8 m/s²).
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Solve for the minimum velocity: Cancel from both sides: Now, solve for :
Thus, the minimum velocity at the highest point of the trajectory is approximately 2 m/s.
Do you have any questions, or would you like further details on this problem?
Here are 5 related questions to deepen your understanding:
- How would the velocity change if the mass of the ball were increased?
- What would happen if the length of the string were doubled?
- How does the minimum velocity depend on the gravitational constant?
- What is the velocity at the lowest point of the trajectory?
- How is the tension in the string affected at different points in the circular motion?
Tip: The minimum velocity at the highest point ensures the object remains in circular motion without the string going slack. At this point, all centripetal force comes from gravity.
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Math Problem Analysis
Mathematical Concepts
Circular Motion
Centripetal Force
Gravitational Force
Formulas
Centripetal force formula: F_c = (mv^2)/r
Gravitational force: F_g = mg
Velocity formula: v = sqrt(gr)
Theorems
Newton's Second Law of Motion
Suitable Grade Level
Grades 9-12
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