A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in figure. If the string gets slack at some point P making an angle from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v0 is: [figure]
- ( 2 + 3 ) 12
- ( 2 + 3 ) 12
- ( ) 12
- ( 12 + 3 ) 12
Answer
(2) ( 2 + 3 ) 12
Solution
At P the string becomes slack, so T = 0. The radial component of gravity alone gives the centripetal force:
mg = mv2l v2 = gl
Height of P above the lowest point: h = l + l
Energy conservation: v02 = v2 + 2gl(1 + )
v02 = gl + 2gl + 2gl = gl(2 + 3 )
vv0 = ( 2 + 3 ) 12
Answer: (2)