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hw1_MechanicsI - 1 Mechanics I | Homework#1 1-10 2 sin cos...

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Unformatted text preview: 1 Mechanics I | Homework #1 1-10. 2 sin cos b t b t Z Z £ r i j a) 2 2 2 c os sin 2 s in cos b t b t b t b t Z Z Z Z 2 Z Z Z Z Z ¤ ¤ ¤ ¤ v r i j a v i j r ¢ ¢ 1 2 2 2 2 2 2 2 1 2 2 2 speed 4 cos sin 4 cos sin b t b b t t Z Z Z Z Z Z Z t ª º £ ¬ ¼ ª º £ ¬ ¼ v 1 2 2 speed 3 cos 1 b t Z Z ª º £ ¬ ¼ b) At 2 t S Z , sin 1 t Z , cos t Z So, at this time, b Z ¤ v j , 2 2 b Z ¤ a i So, 90 T ¡ q 2 1-25. e r e φ e θ θ φ The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by ¡ ¢ ¡ ¢ ¡ ¢ cos cos , cos sin , sin sin , cos , 0 sin cos , sin sin , cos r T I T I T I T I I T I T I T º £ » » £ » » » ¼ e e e (1) Thus, ¡ ¢ cos sin sin cos , cos cos sin sin , cos T I T I T T I I T I T T I T T £ £ £ £ e ¡ ¡ ¡ ¡ ¡ ¡ cos r I T I T ¤ e ¡ ¡ £ (2) e Similarly, ¡ ¢ cos , sin , 0 I I I I I £ £ e ¡ ¡ ¡ cos sin r T I T I T £ e ¡ ¡ £ (3) e sin r I T I T T ¤ e e ¡ ¡ ¡ e (4) Now, let any position vector be x . Then, r r x e (5) ¡ ¢ sin sin r r r r r r r r r r I T I T I T T I T T ¤ ¤ ¤ ¤ ¤ x e e e e e e e e ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ r (6) ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ 2 2 2 2 sin cos sin sin 2 sin 2 cos sin sin 2 s in cos r r r r r r r r r r r r r r r r r r r r I I T T I T I T TI T I T I T T T T I T TI T I T I T T T T I T T ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ £ £ ¤ ¤ £ x e e e e e e ¡ ¡ ¡¡ ¡ ¡ ¡¡ ¡ ¡ ¡ ¡¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ r e e e ¡ ¡¡ ¡ (7) or, ¡ ¢ ¡ ¢ 2 2 2 2 2 2 2 1 sin sin cos 1 sin sin r d r r r r r r dt d r r d t T I T I T T I T T I T T ª º ª º £ £ ¤ £ ¬ ¼ « » ¬ ¼ ª º ¤ « » ¬ ¼ x a e e e ¡ ¡ ¡ ¡¡ ¡ ¡¡ ¡ ( 8 ) 3 2-12. The equation of motion for the upward motion is 2 2 2 d x m m kv dt ¡ ¡ mg (1) Using the relation 2 2 d x dv dv dx dv v dt dt dx dt dx (2) we can rewrite (1) as 2 v dv dx kv g ¡ ¤ (3) Integrating (3), we find ¢ £ 2 1 log 2 kv g x C k ¤ ¡ ¤ (4) where the constant C can be computed by using the initial condition that when x = 0: v v ¢ £ 2 1 log 2 C kv k g ¤ (5) Therefore, 2 2 1 log 2 kv g x k k v g ¤ ¤ (6) Now, the equation of downward motion is 2 2 2 d x m m kv dt ¡ ¤ mg (7) This can be rewritten as 2 v dv dx kv g ¡ ¤ (8) Integrating (8) and using the initial condition that x = 0 at v = 0 (w take the highest point as the origin for the downward motion), we find 4 2 1 log 2 g x k g kv ¡ (9) At the highest point the velocity of the particle must be zero. So we find the highest point by substituting v = 0 in (6): 2 1 log 2 h kv g x k g ¢ (10) Then, substituting (10) into (9), 2 2 1 1 log log 2 2 kv g g k g k g k ¢ ¡ v (11) Solving for v...
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