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Sudheesh Singanamalla Grade: 12
        

A srone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is in the lowest position , and has a speed u. The magnitude of the change in its velocity as it reaches a position where the sting is horizontal is :


a) root (u^2 - 2gL)


b) root (2gL)


c) root (u^2 - gL)


d) root (2(u^2 - gL))


 


please help by answering this Q and the method to do this Question

7 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

applying energy conservation


 1/2mu^2=mgl+1/2mv^2


  we get v=root(u^2-2gl)


  in vector form we can express as v vector=root(u^2-2gl)j and u vector =ui


  difference is v-u =root(u^2+v^2)=root(2(u^2-gl))

7 years ago
Sudheesh Singanamalla
114 Points
										

Can some1 explain this more clearly !

6 years ago
Raiwat Bapat
13 Points
										student-name Abhiram M V asked in PhysicsA stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in the velocity as it reaches a position where the string is horizontal is (A) u2−2gL−−−−−−−√                 (B) 2gL−−−√                     (C) u2−gL−−−−−−√              (D) 2(u2−gL)−−−−−−−−−√ 7   Follow 0student-name A K Daya Sir answered this2879 helpful votes in Physics, Class XI-ScienceLet the direction of revolution be counter-clockwiseInitial velocity = u i^ (at bottom)i^ ( i-cap suggests that the direction of initial velocity is towards positive x-axis)--When the string is horizontal, the body has covered a quarter of revolution, its velocity is v j^j^ (j-cap suggests that the direction of its velocity is towards positive y-axis when the string is horizontal)-Change in velocity = vj^ - ui^magnitude of change in velocity = √ v2 + u2-Using conservation of energyKinetic Energy of body when it is at bottom = 1/2 mu2When it goes to the quarter position, it does some work against gravity. So a part of its kinetic energy is spent in doing work equal to mgh and remaining kinetic energy is 1/2 mv21/2 mu2 = 1/2 mv2 + mghhere h = l1/2 mu2 = 1/2 mv2 + mglv2 - u2 = -2glv2 - u2 + 2u2 = 2u2 - 2glv2 + u2 = 2 (u2 - gl)Therefore, change in magnitude of velocity = √ v2 + u2 = √2 (u2 - gl)
										
16 days ago
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