A stone is tied to a string and whirled in a vertical circle of radius 1 m. What is the minimum speed at the top of the circle to keep the stone in circular motion?
Practice Questions
1 question
Q1
A stone is tied to a string and whirled in a vertical circle of radius 1 m. What is the minimum speed at the top of the circle to keep the stone in circular motion?
1 m/s
2 m/s
3 m/s
4 m/s
At the top, centripetal force = weight. mv²/r = mg. v² = rg. v = √(1*9.8) ≈ 3.13 m/s.
Questions & Step-by-step Solutions
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Q
Q: A stone is tied to a string and whirled in a vertical circle of radius 1 m. What is the minimum speed at the top of the circle to keep the stone in circular motion?
Solution: At the top, centripetal force = weight. mv²/r = mg. v² = rg. v = √(1*9.8) ≈ 3.13 m/s.
Steps: 11
Step 1: Understand that the stone is moving in a vertical circle and we need to find the minimum speed at the top of the circle.
Step 2: At the top of the circle, the forces acting on the stone are its weight (downward) and the tension in the string (also downward).
Step 3: For the stone to stay in circular motion at the top, the centripetal force needed to keep it moving in a circle must be provided by its weight.
Step 4: Write the equation for centripetal force: centripetal force = (mass * speed²) / radius.
Step 5: Set the centripetal force equal to the weight of the stone: (mv²/r) = mg.
Step 6: Cancel the mass (m) from both sides of the equation since it appears in both terms: v²/r = g.
Step 7: Rearrange the equation to solve for speed (v): v² = rg.
Step 8: Substitute the values: radius (r) = 1 m and acceleration due to gravity (g) = 9.8 m/s² into the equation: v² = 1 * 9.8.
Step 9: Calculate v²: v² = 9.8.
Step 10: Take the square root of both sides to find v: v = √9.8.
