PHYS 1120 Angular Momentum Solutions.pdf

# PHYS 1120 Angular Momentum Solutions.pdf -...

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Questions: 1 2 3 4 5 6 7 8 9 10 Physics 1120: Angular Momentum Solutions 1. Determine the direction of the angular momentum for the following cases: Angular momentum is defined as the cross product of position and momentum, L = r × p . The direction of the angular momentum is perpendicular to the plane formed by the position and momentum vectors. For this problem that means either into the paper, denoted by ×, or out of the paper, (·) . To find the direction, we sweep our right hand through the smallest angle formed by the vector. The way the thumb points indicates the direction of the angular momentum. 2. Calculate the angular momentum for the following particles. Find the angle between the position and the momentum vectors. (a) r = (4, ­5, 3) and p = (1, 4, ­2) (b) r = (1, ­2, 3) and p = (7, ­1, 1) (c) r = (0, 2, 0) and p = (1, 0, 0) The angular momentum can be found either by evaluating the determinant or by using L = rpsinφ. We will use the first method to find L . We can find the angle between the momentum and position vectors using φ = sin ­1 (L/rp). We find the magnitudeof the vectors using the 3D form of the Pythagorean Theorem. (a) L = [(­2) 2 + (11) 2 + (21) 2 ] ½ = 23.791 kg­m 2 /s

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r = [(4) 2 + (­5) 2 + (3) 2 ] ½ = 7.0711 m p = [(1) 2 + (4) 2 + (­2) 2 ] ½ = 4.5826 kg­m/s φ = sin ­1 (L/rp) = sin ­1 [ 23.791 / (7.0711 × 4.5826)] = 47.2° (b) L = [(1) 2 + (20) 2 + (13) 2 ] ½ = 23.875 kg­m 2 /s r = [(1) 2 + (­2) 2 + (3) 2 ] ½ = 3.7417 m p = [(7) 2 + (­1) 2 + (1) 2 ] ½ = 7.1414 kg­m/s φ = sin ­1 (L/rp) = sin ­1 [ 23.875 / (3.7417 × 7.1414)] = 63.3° (c) L = [(0) 2 + (0) 2 + (­2) 2 ] ½ = 2 kg­m 2 /s r = [(0) 2 + (2) 2 + (0) 2 ] ½ = 2 m p = [(1) 2 + (0) 2 + (0) 2 ] ½ = 1 kg­m/s φ = sin ­1 (L/rp) = sin ­1 [ 2 / (2 × 1)] = 90°
3. Calculate the angular momentum of a phonograph record (LP) rotating at 33 1 / 3 rev/min. An LP has a radius of 15 cm and a mass of 150 g. A typical phonograph can accelerate an LP from rest to its final speed in 0.35 s, what average torque would be exerted on the LP? The angular momentum of a rotating body is L = IΩ. An LP is a solid disk. Consulting a table of moments of inertia, we find I = ½MR 2 . The angular velocity must be converted to rad/s Ω = 100/3 rev/min × 2π rad / rev × 1 min / 60 s = 3.4907 rad/s . Thus we find the angular momentum of the LP to be L = IΩ = ½MR 2 Ω = ½(0.15 kg)(0.15 m) 2 (3.4907 rad/s) = 5.8905 × 10 ­3 kg­m 2 /s . Torque is equal to the change in angular momentum with time τ = ΔL / Δt = (L f ­ L i ) / Δt = (5.8905 × 10 ­3 kg­m 2 /s ­ 0) / 0.35 s = 1.68 × 10 ­2 N­m .

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