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 6.10 m long, his tangential speed at the bottom of the swing is 9.70 m/s. (a) What is the centripetal acceleration of the child at the bottom of the swing? m/s₂ (b) What provides the centripetal force that keeps the child moving in an arc? the weight of the child the tension in the rope the height of the tree gravitational acceleratio

Answer 1

(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.