Question

A stone tied to the end of a string which is $80cm$ long is whirled in a horizontal circle with a constant speed. If the stone makes $14$ revolutions in $25s$, what is the magnitude of centripetal acceleration? (in $m/s^2$)

• A. $9.91$
  
• B. $2.36$
  
• C. $10.36$
  
• D. $12.69$
  

Answer 1

The correct option is A $9.91$

Period of one revolution $=\frac{25}{14}sec$

As we know $\omega =\frac{2\pi }{periodofonerevolution\left(sec\right)}$$=\frac{2\pi }{25}\times 14=3.52rad/s$

Radius of the string $r=\frac{80}{100}cm$

Centripetal acceleration $={\omega }^{2}r=(3.52{)}^2\frac{80}{100}=9.91m/s^2$