A student ties a rock to a string then spins the rock in a vertical circular motion at constant speed. The tension in the string:

Question

BlackTom AI · Accepted Answer

Centripetal force required for circular motion is m v^2 / r, and gravity adds or subtracts from the string’s required inward force depending on position.

1) Relevant concepts
- Centripetal acceleration a_c = v^2 / r toward the circle’s center.  
- Tension provides the net radial (centripetal) force; weight mg acts downward.  
- In vertical circular motion at constant speed, the magnitude v is the same at all positions, so m v^2 / r is the required net inward force at every point.

2) Step-by-step reasoning
- At the bottom of the circle the center is upward. Upward tension T_bottom must overcome the downward weight mg and still provide the upward centripetal force: T_bottom − mg = m v^2 / r.
- Solve: T_bottom = m v^2 / r + mg.
- At the top the center is downward. Both tension (toward center) and weight point downward, so their sum provides the required centripetal force: T_top + mg = m v^2 / r.
- Solve: T_top = m v^2 / r − mg.

3) Intermediate calculation example
- Let m v^2 / r = A (some positive number). Then T_bottom = A + mg and T_top = A − mg.
- Difference: T_bottom − T_top = (A + mg) − (A − mg) = 2 mg, so the bottom tension is larger by 2 mg.
- Numerical example: m = 0.5 kg, mg ≈ 4.9 N, and A = 10 N → T_bottom = 14.9 N, T_top = 5.1 N.

4) Why the correct answer is correct
- Because gravity adds to the inward force requirement at the bottom (it acts opposite the center direction) and subtracts at the top (it acts toward the center), the tension is maximal at the bottom: T_bottom = m v^2 / r + mg.

5) Why the other options are incorrect
- b (greatest at top): false — tension at top is m v^2 / r − mg, which is smaller.  
- c (constant): false — gravity’s direction relative to the center changes with position, so T varies.  
- d (halfway between): false — at the side positions the radial component of weight is zero, so tension there would be T = m v^2 / r (assuming external tangential forces maintain constant speed), which is between the top and bottom values but not the greatest.

6) Conclusion
Therefore the tension is greatest at the bottom of the circular path, confirming answer a.

A student ties a rock to a string then spins the rock in a vertical circular motion at constant speed. The tension in the string:单项选择题

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