notes_10

672 so a b 1487 sin k ie sin 1487 the solution

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Unformatted text preview: ˆ ) + 2( ˆ × ˆ ) − 12( ˆ × ˆ ) + 3( ˆ × ˆ ) ii ij ji jj 0 0 ˆ ˆ ˆ = 2 k − 12( − k ) = 10 k . rr rr ˆ (b) We know A × B = A B sin θ k , r but A = 22 + 32 = 13 r and B = ( −4)2 + 12 = 17 rr 10 ˆ = 0.672 . So A × B = 14.87 sin θ k , i.e., sin θ = 14.87 The solution θ = 42.2o is clearly not appropriate, but we remember that sin α ≡ sin(180 − α ). ∴ θ = 180o − 42.2o = 137.8o. Another important concept in rotation about an axis is angular momentum ... y ˆ j r v r r O φ m r r ( p = mv ) ˆ i x r L is outward The angular momentum about O is defined as: rrr L = r × p, r r where p = mv is the linear momentum of the object. The magnitude of the angular momentum is; L = mvr sin φ = pr sin φ , r and the direction of L is given by the right-hand-rule and r r is perpendicular to r and v . Note: a particle does not have to be rotating to have angular momentum. Dimension: [M][L]2 [ T] Units: kg ⋅ m2 /s (or J.s) Question 2: An object of mass 3kg is moving with a r ˆ velocity v = (3.0m/s)i along the line z = 0, y = 5.3m. (a) What is the angular momentum of the object relative to the origin at the point x = 0, y = 5.3m? (b) What is the angular momentum of the object relative to the origin at r i the point x = 12 m, y = 5.3m? (c) If a force F = ( −3N)ˆ is applied to the object, what is the torque of this force relative to the origin? ˆ j y (0, 5.3m) r 5.3m r1 r r2 (12m, 5.3m) r v = (3m/s) ˆ i ˆ k x O Take a closer look at a similar scenario ... ˆ j Origin ˆ i d ˆ i r rr (a) Angular momentum about O is L1 = r1 × p, r j where the position vector is: r1 = (5.3ˆ ) m and the r r momentum is p = mv . r ˆ ∴ L = 5.3ˆ × 9ˆ = ( −47.7 k ) kg ⋅ m2 /s . ji 1 (Note the direction) r i j (b) For the position vector r2 = (12ˆ + 5.3ˆ ) m, r rr ˆ L2 = r2 × p = (12ˆ + 5.3ˆ ) × 9ˆ = ( −47.7 k ) kg ⋅ m2 /s, i j i i.e., the same ... very interesting! rrr (c) The torque about O is: τ = r × F ˆ = (12ˆ + 5.3ˆ ) × ( −3ˆ ) = ( +15.9 k ) N ⋅ m. i j i (Note the direction) r r1 r r r2 p θ1 r p r r3 θ2 r p θ3 The angular momentum of the object about the origin is r rr ˆ Li = ri × p = ri p sin θi k (outward) r ∴ Li = Li = ri p sin θi r But ri sin θi = d , which is constant for all pairs of ri , θi, r providing the direction of p is unchanged! r So, the angular momentum Li is constant provided there r are no external forces to...
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