IIT JEE , AIEEE Forum - Physics , Mechanics

varun.tinkle

FIND THE MINIMIMUM VSUE OF X/L FOR WHICH THE BOB GOES IN A COMPLETE CORCLE ABOUT THE PEG WHEN THE PENDULUM IS RELEASED FROM @=90

Conjurer

Assume the lowest point of the motion of the bob to be the PE level. So initial energy = mgL.

At the top most point, in circular motion , tension can be just zero to perform circular motion.

At the top most point of the circle , the equations of forces are mg+T = mv^2/R

Here R = L-x

Put T =0 to get v^2 = gR = g(L-x)

Now at this point energy is 1/2 mv^2 +2mg(L-x) = 1/2 m g(L-x) + 2mg(L-x)

This should be equal to inital energy i.e mgL. Just equate to get answer.

varun.tinkle

how is radius l-x

varun.tinkle

at the top most point
why is p.e=2mg(L-X)

varun.tinkle

HOW IS THE RADIUS L-X
PLSSSS EXPLN

harshthakar017

is the answer 3/5

varun.tinkle

YEP
BUT HOW IS THE RADIUS
L-X
IF IT IS SO THEN THE DIAMETER IS COMING AS 2L-X
WHICH SHOULD BE 2L
PLS EPLN

harshthakar017

Actually 2(l-x) < l and radius is l-x because when the string just winds around the nail, the pt. at the nail becomes fixed and it behaves as if the bob is suspended by by a string of length l-x

varun.tinkle

IF IT IS WINDING AROUND THE PEG THEN THE RADIUS AT THE BOTTOM MI=OST POINT WILL BE L-X
AND AT THE HIGHEST POINT WILL BE L+X
PLS EXPLN

harshthakar017

NO Varun not at all. When it strikes the nail, x length of the string is stuck to the nail. You can even consider the string to be fixed at the nail.

varun.tinkle

harshthakar017

Anytime

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