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A ball of mass M at one end of a string of length L rotates in a vertical circle just fast enough to prevent the string from going slack the speed of the ball at the bottom of the circle is
A ball of mass M at one end of a string of length L rotates in a vertical circle just fast enough to prevent the string from going slack the speed of the ball at the bottom of the circle is

```
3 years ago

Arun
25768 Points
```							I think the speed is not meant tp be constant; as the mass falls it loses potential energy and gains kinetic energy. At the top its speed is given by mv^2/L = mg .as you say. So its kinetic energy at the top is (1/2)mv^2 = mgL/2 At the bottom it has dropped a distance 2L, so its potential energy has decreased 2mgL. Its kinetic energy has increased by 2mgL. Total kinetic energy at bottom = mgL/2 + 2mgL = 5mgL/2 At the bottom (1/2)mv^2 = 5mgL/2 v = sqrt (5gL)
```
3 years ago
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions