Question

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?

Answer 1

Given,

The length is r=5.70 m

The tangential speed is r=9.10 m/s

The centripetal acceleration is:

`\begin{gathered} a=(v^2)/(r) \\ \Rightarrow a=(9.10^2)/(5.70) \\ \Rightarrow a=(14.52m)/(s^2) \end{gathered}`

The acceleration is:

`a=14.52m/s^2`