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The bob of a stationary pendulum is give...

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of `sqrt(3gl)`. Find the angle rotated by the string before it becomes slack.

Text Solution

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Let `theta ` be the angle rotated by the string of pendulum before it becomes slack as shown in figure and let v be the speed of bob at this position
Applying circular dynamics we can write the following equation :
`T + mg cos (pi-theta ) = (mv^(2))/l `
Applying conservation of mechanical energy betweeen bottom most point and the point where string become slack , we can write following equation :
Loss in K.E = Gain in gravitational potential energy
`1/2 m (sqrt(3gl))^(2) - 1/2 mv^(2) = mg [ l +l cos ( pi - theta )] `
Substituting from equation (i) we get
`1/2 ,(sqrt(3gl))^(2) - 1/2 mgl cos ( pi - theta) = mg [ l+l cos (pi - theta ) ] `
` rArr " " 3/2 mgl+1/2mgl cos theta = mg [ l - l cos theta ] `
`rArr " " 3/2 +1/2 cos theta =1 - cos theta `
`rArr " " 3/2 +1/2 cos theta = 1 - cos theta `
`rArr " " cos theta = - 1/ 3 `
`rArr " " theta = cos^(-1) (-1/3)`
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