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  • 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?

Q. 4.17 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?

Answers (1)

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Frequency is given by :

                                              Frequency\ =\ \frac{No.\ of\ revolutions}{Total\ time\ taken}

                                                                         =\ \frac{14}{25}\ Hz

And, the angular frequency is given by :

                                                     \omega \ =\ 2 \Pi f

Thus,                                                   =\ 2 \times\ \frac{22}{7}\times\frac{14}{25}

                                                            =\ \frac{88}{25}\ rad/s

Hence the acceleration is given by  :

                                                    a\ =\ \omega ^2 r

or                                                      =\ \left ( \frac{88}{25} \right ) ^2 \times 0.8

or                                                a\ =\ 9.91\ m/s

Posted by

Devendra Khairwa

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