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A particle is hanging from a fixed point O by means of a string of length a.there is a small nail Q in the same horizontal line with O at a distance between(b=a/3) from O.find the minimum velocity with which the particle should be projected so that it may make a complete revolution around the nail without being slackened.
A particle is hanging from a fixed point O by means of a string of length a.there is a small nail Q in the same horizontal line with O at a distance between(b=a/3) from O.find the minimum velocity with which the particle should be projected so that it may make a complete revolution around the nail without being slackened.


2 years ago

Arun
25768 Points
							Dear student  Here we have to find the velocity of particle at the lowest position required to complete the constrained motion arround the nailFirstly we know that the velocity required to complete 90° is sqrt(2ga) Total kinetic energy decrease to complete 90 of the circle of radius a K1 = 1/2mv2 now the radius of particle is decreased to 2a/3 so , we know that the velocity required to complete a circle with radius R is /{sqrt{5gr}}so here R =2a/3  v atLets take it as a completely rotational motion about point O and Q.by energy conservation(wb representa angular velocity at the bottom (ZPL) and wT represents the velocity at the top)(1/2)I1wb2 = mg(a+2a/3) + (1/2)I2wT2  ...(1)Simplify it to get  (1/2)(ma2)wb2 = mg(a+2a/3) + (1/2)(m(2a/3)2)wT2 put values I1 = ma2 ; I2 =m(2a/3)2 .here we need to find wb , so we will eleminate wT :at the top the Tension just tends to 0Therefore, mwT2(2a/3) = mgNow, substitute the value of mwT(2a/3) in equation 1 :(1/2)(ma2)wb2 = mg(5a/3)+(1/2)(mg(2a/3)​)Solve this to get :$\fn_cm Wb=2{\sqrt{g/a}}$So the velocity at bottom will be $\fn_cm Vb = 2{\sqrt{ga}}$

2 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