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

Motion in a Plane

Question : 17 of 32

Marks: +1, -0

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 make 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone?

Solution:

Here,

r=80\,\text{cm}=0.8\,\text{m} ; \upsilon=\frac{14}{25}\,\text{s}^{-1}

\omega=2\pi\upsilon=2\times\frac{22}{7}\times\frac{14}{25}

=\frac{88}{25}\,\text{rad}\,\text{s}^{-1}

The centripetal acceleration,

a=\omega^{2}r=\left(\frac{88}{25}\right)^{2}\times 0.80=9.90\,\text{ms}^{-2}

The direction of centripetal acceleration is along the string directed towards the centre of circular path.

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