Circular Motion Physics Example 5

Follow the full solution, then compare it with the other examples linked below.

Example 5

hard
A ball on a 0.8 m0.8 \text{ m} string is swung in a vertical circle. What minimum speed must it have at the top of the circle so the string remains taut? Use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution

  1. 1
    At the top, both gravity and tension point downward (toward center). For minimum speed, tension T=0T = 0.
  2. 2
    At minimum speed: mg=mvmin⁡2rmg = \frac{mv_{\min}^2}{r}, so vmin⁡2=grv_{\min}^2 = gr.
  3. 3
    vmin⁡=gr=9.8×0.8=7.84=2.8 m/sv_{\min} = \sqrt{gr} = \sqrt{9.8 \times 0.8} = \sqrt{7.84} = 2.8 \text{ m/s}

Answer

vmin⁡=2.8 m/sv_{\min} = 2.8 \text{ m/s}
At the top of a vertical circle, gravity alone must provide sufficient centripetal force. If the speed is too low, the required centripetal force exceeds gravity and the string goes slack.

About Circular Motion

Motion of an object along a circular path where the speed may be constant but the velocity is continuously changing direction, requiring a centripetal acceleration.

Learn more about Circular Motion →

More Circular Motion Examples

Example 1 easy

A car drives around a circular track of radius [formula] at a constant speed of [formula]. What is t

Example 2 medium

A satellite orbits Earth at a radius of [formula] with a period of [formula]. What is its orbital sp

Example 3 medium

A car travels around a circular track of radius 50 m at 20 m/s. Find the centripetal acceleration.

Example 4 medium

A point on the rim of a bicycle wheel ([formula]) moves at [formula]. What is the period of rotation

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