Answer:
(a) To find the centripetal acceleration of the child at the bottom of the swing, we can use the formula:
(a_c = \\frac{v^2}{r})
where (v) is the tangential speed and (r) is the radius of the swing.
Given that (v = 9.70 , \\text{m/s}) and (r = 6.10 , \\text{m}), we can substitute these values into the formula:
(a_c = \\frac{(9.70 , \\text{m/s})^2}{6.10 , \\text{m}})
Solving this equation gives us:
(a_c = 15.40 , \\text{m/s}^2)
Therefore, the centripetal acceleration of the child at the bottom of the swing is (15.40 , \\text{m/s}^2).
(b) The centripetal force that keeps the child moving in an arc is provided by the tension in the rope. In this case, the tension in the rope acts as the inward force required to maintain the circular motion of the child. The tension force is directed towards the center of the circular path and is responsible for providing the centripetal force necessary to keep the child swinging.