89.1 Momentum and Newtons second law of motion Example 1: P420 56(b) y 3m Soluti 2 mx0+2ml+3ml/2 7l CM 6m 12 m×0+2m×0+3msin60°3√31 J CM 6m 12 89.1 Momentum and Newton's second law of motion Example 2: a thin strip of material is bent into the shape of a semicircle of radius r Find its center of mass Solution: dm since the symmetry, we obtained R C 0 w=∫om= M (Rsin)-dφ R 2R SIn =0.637R 元 元4 Example 1: P420 56(b) l l l m 2m 3m x y 12 3 3 6 0 2 0 3 sin60 12 7 6 0 2 3 / 2 l m m m ml y l m m ml ml x CM CM = × + × + = = × + + = o Solution: §9.1 Momentum and Newton’s second law of motion Example 2: A thin strip of material is bent into the shape of a semicircle of radius R. Find its center of mass. Solution: since the symmetry, we obtained R R R M R M y m M y x CM CM 0.637 2 sin d ( sin ) d 1 d 1 0 0 0 = = = = = = ∫ ∫ ∫ π π π φ φ π φ π φ x y R φ dm dφ §9.1 Momentum and Newton’s second law of motion