A bob attached to a light string of length 10m is given an initial velocity of 20m/s at the bottom-most point to perform vertical circular motion. What will be the velocity of bob just before slacking of string?
(Take g=10m/s2)
A
0m/s
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B
10√23m/s
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C
20√3m/s
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D
10√3m/s
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Solution
The correct option is B10√23m/s The FBD of the bob at the position where the slackness happen is shown in the figure
The slackness of the string happens when the tension in the string becomes zero.
Hence the component of mg along the string (mgcosθ) will balance the centrifugal force which is mv2R in the radially outward direction.
⇒mgcosθ=mv2R...(i),
Height of the bob, H=R+Rcosθ
Now by the conservation of mechanical energy we get, 12mu2+0=12mv2+mg(R+Rcosθ)...(ii), where u is the initial velocity given to the bob at the bottom most point and v is the velocity at the point where the slackness happen.
Putting the value of , cosθ=v2Rg from equation (i) in the equation (ii) we get