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A stone tied to the end of a string 80 c...

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?

Text Solution

Verified by Experts

The acceleration will be directed towards the centre of the circular loop
angular velocity ` omega = 2 pi f = 2 xx3 .14 xx (14)/(25 ) , omega rad //s`
Centripetal accerelation `=r omega ^2 =( 0.8 xx (88 )^2 )/( (25)^2) , a _(c ) = 9.91 m//s^2`
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Knowledge Check

  • If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms^(-1) , then the acceleration of the stone is

    A
    `12 ms^(-2)`
    B
    `36 ms^(-2)`
    C
    `2pi^2 ms^(-2)`
    D
    `72 ms^(-2)`

  • When a stone tied to the end of a string whireled in a circular path the centripetal force is provided by the

    A
    weight of the stone
    B
    centrifugal force
    C
    tension in the string
    D
    weight of the string

  • When a stone tied to the end of a string whireled in a circular path the centripetal forcwe is provided by the

    A
    weight of the stone centrufugal force q
    B
    centrufugal force
    C
    tension in the string
    D
    weight of the string

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