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`        Why in circular motion a Bob of mass get fallen down in the upper half still having a component of mg?`
2 years ago

## Answers : (1)

jitender
114 Points
```							Bob need to have a min of (5gl)^1/2 m/s at the lowest point To complete the vertical circular motion.If it has velocity less than that it will either oscillate about the lowest position. If it has v= (2gl)^1/2 at lowest point  then velocity and tension in the string will be zero at perpendicular distance from the lowest point.If it has velocity in between ( 2gl)^1/2 and (5gl)^1/2 at lowest point Then it first completes a semi circle or then when the tension become 0 above perpendicular distance ( T = mgcos¤ + (mv^2)/l ) then it will separate from circular path and folloes projectile motion because of remaining velocity ( V = u^2 - 2gl [1-cos¤] )u= velocity at lowest point.And gravitaion field is a conservative field bon is under the constant acceleration due to gravity g.
```
2 years ago
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• Test paper with Video Solution
• Mind Map
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• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions