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a particle of mass 100gm is made to describe a vertical circle of 1m. its instantaneous speed is 1m/s when the string makes an angle of 30 degree with the vertical. find the tension in the string at this position .can the particle complete its circular path ?
a particle of mass 100gm is made to describe a vertical circle of 1m. its instantaneous speed is 1m/s when the string makes an angle of 30 degree with the vertical. find the tension in the string at this position .can the particle complete its circular path ?


4 years ago

Piyush Kumar Behera
435 Points
							mass = 0.1 kgdrawing the fbd at that instant,we get the following equation,T-mgcos30=mv2/rbecause this only provides the centripetal force.and here v is the instantaneous speed of the particle =1m/s and r=1So T=mv2/r+mgcos30=0.1N+0.866N=0.966N   (substituting the values)to complete the loop it must have a total mechanical energy of 2.5mgr    (reference for potential enrgy is the lowest point in the circle,the minimum velocity is $\sqrt{5gr}$)total mechanical enrgy at that point is m/2+1.34m=0.84m (taking r=1 and m the mass)so it cannot complete the circle.

4 years 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