Question

A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s what is the acceleration of the stone?

• A. $\left(\frac{88}{25}\right)^{2} m s ^{-2}$
• B. $\left(\frac{25}{88}\right)^{2} m s ^{-2}$
• C. $\left(\frac{88}{25}\right) m s ^{-2}$
• D. $\left(\frac{25}{88}\right) m s ^{-2}$

Answer 1

Answer: (A)

Here, $r=100\, cm =1 m , v =\frac{14}{25} s ^{-1}$
$\therefore \omega=2 \pi v=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} rad s ^{-2}$
Centripetal acceleration,
$a_{c}=\omega^{2} r=\left(\frac{88}{25}\right)^{2} \times 1.0$
$=\left(\frac{88}{25}\right)^{2} m s ^{-2}$