IIT JEE , AIEEE Forum - Physics , Mechanics

perin

HCV part 1 page 135 Q.49
a small heavy block is attached to the lower end of a light rod of length l which can be rotated about its clamped upper end. what minimum horizontal velocity should the block be given so that it moves in a complete circle?

rtiit

root(5gl)

rtiit

hey is my ans rite????

tell me i'll post the solution

perin

@ tanmay

yur answer is wrong.

answer is 2

gl

rtiit

oops i didnt read ques...

it's min horizontal velocity

rtiit

wait i'll try to solve

varun.tinkle

if the particle moves in a citcle then the radius is 2l
therefore the work done against gravity =2mgl
2mgl=1/2mv^2
2gl=1/2v^2
rooot 4gl
=2 root gl

rtiit

oh how can the radius be 2l????

varun.tinkle

sorry the radius is l
but at the highest position the block is at a dis tance 2l
so work done against gravity=2mgl

priyesh

the radius is l and not 2l

now to complete a full circle the block must have enough energy to reach the topmost point

so 1/2mv^2 = mg(2l) (distance of topmost and initial bottommost point is 2l)

=> v = 2root(gl)

anchitsaini

probably your doubt is that why the answer is not

see first of all lets consider the case of block attached to a thread

there at the topmost point

T + mg = mv^2 / l

now for minimum velocity

T = 0

hence

hence at bottommost point

v_1^2 = v^2 + 2g2l = 5gl

NOW coming to the case given -->

at the topmost point the weight is already balanced because it is placed on the rod unlike in the string ,

hence

T = mv^2/ l

for minimum velocity

T = 0 , hence- v = 0

thus at bottommost point

v_1^2 = 4gl

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