A 25-kg child is placed on a swing with a chain length of 3.5-m. Her father pull

Question

A 25-kg child is placed on a swing with a chain length of 3.5-m. Her father pulls the swing back until it touches his chest, so that the chain is at 5.9-degrees to the vertical. He gives the swing a small push, so that it is travelling at 0.54-m/s. When the child reaches the apex of her motion, what is the new angle of the chain to the vertical? degrees Check How far back should the parent step so that the swing would just avoid hitting him when it returns (ignore air resistance and assume that the child does no work)? cm Check

Explanation / Answer

A) Let the angle be theta,

mg R (cos 5.9 degree - cos theta) = 0.5mv^2

0.9947 - cos theta = 0.5v^2/(gR) = 0.5*0.54^2/(9.8*3.5)

= 0.004251

theta = arccos (0.9947-0.004251)

= 7.925 degree answer

B) x = 3.5*( sin 7.925 degree - sin 5.9 degree)

= 0.123 m answer

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