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A simple pendulum of bob of mass m and l...

A simple pendulum of bob of mass `m` and length `L` has one of its ends fixed at the centre `O` of a vertical circle, as shown in the figure. If `0=60^(@)` at the point `P`, the minimum speed `u` that should be given to the bob so that it completes vetical circle is

A

`sqrt(gL)`

B

`sqrt(2gL)`

C

`sqrt(5gL)/(2)`

D

`sqrt(3gL)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Energy conservation at heighest point
`{v_("top")=sqrt(gR)}`
`(1)/(2)m u^(2)=(mgR)/(2)+(1)/(2)m(sqrt(gR))^(2)rArru=sqrt(2gR)`
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Knowledge Check

  • A simple pendulum bob of mass M and length l has one of its end fixed at the centre of a vertical circle as shown in figure. If theta = 60^(@) at the point P, the minimum speed u that should be given to the bob so that it completes vertical circle is :

    A
    `sqrt(gl)`
    B
    `sqrt(2 gl)`
    C
    `sqrt((5 gl)/(2))`
    D
    `sqrt(1.5 gl)`

  • A simple pendulum of effective length 'l' is kept in equilibrium in vertical position . What horizontal velocity should be given to its bob , so that it just completes a vertical circular motion ?

    A
    `sqrt(5gl) `
    B
    `sqrt(3gl)`
    C
    `sqrt(gl)`
    D
    `sqrt(7gl)`

  • A simple pendulum of mass m and length l stands in equilibrium in vertical position .The maximum horizontal velocity that should be given to the bob at the bottom so that it completes on revolution is

    A
    `sqrt(lg)`
    B
    `sqrt(2 l g)`
    C
    `sqrt(3 l g )`
    D
    `sqrt (5 l g)`

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