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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

Here, ` r= 80 cm =0.8 , v= 14// 25 s^(-1)`. :.` omega = 2 pi = 2xx (22) / 7 xx (14)/(25) = (88)/(25) rad. S^(-10`.
The centripetal acceleration , ` a = omega^2 r = ( (88)/( 25))^2 xx 0.80 = 9. 90 ms^(-2)`
The direction of centripetal acceleration is along the string directed towards the centre of circular path.
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Knowledge Check

  • 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 s, the magnitude of acceleration is :

    A
    `20 ms^(-2)`
    B
    `12 m//s^(-2)`
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    D
    `8 ms^(-2)`

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    A
    `(pi^(2))/4 ms^(-2)` and direction along the radius towards the centre
    B
    `pi^(2)ms^(-2)` and direction along the radius away from centre
    C
    `pi^(2)ms^(-2)` and direction along the radius towards the centre
    D
    `pi^(2)ms^(-2)` and direction along the tangent to the circle.

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

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    C
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