A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children swings from this rope that is 5.70 m long, his tangential speed at the bottom of the swing is 9.10 m/s.What is the centripetal acceleration, in m/s2, of the child at the bottom of the swing?

Respuesta :

Given,

The length is r=5.70 m

The tangential speed is r=9.10 m/s

The centripetal acceleration is:

[tex]\begin{gathered} a=\frac{v^2}{r} \\ \Rightarrow a=\frac{9.10^2}{5.70} \\ \Rightarrow a=\frac{14.52m}{s^2} \end{gathered}[/tex]

The acceleration is:

[tex]a=14.52m/s^2[/tex]