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`        A particle of mass m tied with a string of length l is released from a horizontal position . Find the velocity at lowest position `
11 months ago

Arun
24739 Points
```							A particle of mass m attached to end of the string of length l is released from horizontal position. then, energy at horizontal position of particle = energy at lowest position of particle.Let at lowest position speed of particle = vnow, mgl = 1/2 mv² v = √2gl now we should find speed of particle at highest position . Let v' be the speed of particle at highest position .so, mg(2l) + 1/2mv'² = 1/2 mv² 2mgl + 1/2mv² = 1/2mg(2gl) = mgl 1/2mv² = -mgl v² = -2gl , here v is imaginary number .means this situation is not possible . if we released particle from horizontal position , particle doesn't complete circle
```
11 months ago
Harsh
47 Points
```							As the particle is at horizontal position,let its height from the lowest position be h.Applying conservation of energy on the system we getKe at horizontal position+pr wt horizontal position=Ke at lower position+ pe at lower position.Hence,0+mgh=1/2mv²+0mgh=1/2mv²gh=1/2mv²V=√2gh.
```
11 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions