{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

328_Physics ProblemsTechnical Physics

# 328_Physics ProblemsTechnical Physics - 330 Angular...

This preview shows page 1. Sign up to view the full content.

330 Angular Momentum P11.14 F ma x x = T mv r sin θ = 2 F ma y y = T mg cos θ = So sin cos θ θ = v rg 2 v rg = sin cos θ θ L rmv L rm rg L m gr r L m g = ° = = = = sin . sin cos sin cos sin sin cos 90 0 2 3 2 3 4 θ θ θ θ θ θ θ A A , so θ m l FIG. P11.14 P11.15 The angular displacement of the particle around the circle is θ ω = = t vt R . The vector from the center of the circle to the mass is then R R cos ± sin ± θ θ i j + . The vector from point P to the mass is r i i j r i j = + + = + F H G I K J F H G I K J + F H G I K J L N M O Q P R R R R vt R vt R ± cos ± sin ± cos ± sin ± θ θ 1 The velocity is v r i j = = − F H G I K J + F H G I K J d dt v vt R v vt R sin ± cos ± So L r v = × m L i j i j L k = + + × − + = F H G I K J + L N M O Q P mvR t t t t mvR vt R 1 1 cos ± sin ± sin ± cos ± ± cos ω ω ω ω a f x y θ m R P Q v FIG. P11.15
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}