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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 s, what is the magnitude and direction of acceleration of the stone ?

Text Solution

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If the stone makes n revolutions `(n=14)` in time t (t=25s), then its angular velocity is :
`omega=(2pin)/(t)=(2xx22//7xx14)/(25)`
`=3.52" rad/s"`
The radius of the horizontal circle is 80 cm = `0.80` m. Therefore the magnitude of the acceleration of the stone,
`a=omega^(2).r=0.80xx3.52^(2)=9.9" m/s"^(2)`
The acceleration is always direcly along the radius. Pointing towards the centre of the circle.
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