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Unformatted text preview: practice 12 – GADHIA, TEJAS – Due: Apr 11 2007, 4:00 am 1 Question 1 part 1 of 2 10 points A hoop has a radius R and mass m . It is pivoted at A , a point on the circumference of the hoop. Denote the moment of inertia about the center by I and about the pivot point A by I A . The acceleration of gravity is 9 . 8 m / s 2 . A O O' θ The equation of motion about A in the small angle approximation (sin θ ≈ θ ) is given by 1. I d 2 θ dt 2 = mg Rθ . 2. I d 2 θ dt 2 = 2 mg Rθ . 3. I d 2 θ dt 2 = mg Rθ . 4. I A d 2 θ dt 2 = mg Rθ . 5. I A d 2 θ dt 2 = 1 2 mg Rθ . 6. I A d 2 θ dt 2 = mg Rθ . correct 7. I A d 2 θ dt 2 = 2 mg Rθ . 8. I d 2 θ dt 2 = 1 2 mg R θ . 9. I d 2 θ dt 2 = 2 mg Rθ . 10. I A d 2 θ dt 2 = 2 mg Rθ . Explanation: The torque on the hoop is provided by grav ity, and its magnitude is given by τ = mg R sin θ ≈ mg Rθ , where we used the small angle approximation in the last step. Note: The distance from the pivot A to the center of mass of the hoop is R . Finally, the equation of motion reads I A d 2 θ dt 2 = mg Rθ , where the minus sign indicates a restoring torque. Question 2 part 2 of 2 10 points Let m = 1 kg, R = 0 . 5 m, g = 9 . 8 m / s 2 , I = mR 2 . Determine the period of small oscillation of the hoop. Correct answer: 2 . 00709 s (tolerance ± 1 %). Explanation: To find the moment of inertia about the pivot A , use the parallel axis theorem. I A = I + mR 2 = 2 mR 2 . From the equation of motion, ω = radicalBigg mg R I A = radicalbigg mg R 2 mR 2 = radicalbigg g 2 R and the period is T = 2 π ω = 2 . 00709 s . Question 3 part 1 of 1 10 points A uniform plank of mass 0 . 542 kg and length 32 . 2 cm is pivoted at one end, and the opposite end is attached to a spring of force constant 404 N / m. The height of the pivot has been adjusted so that the plank will be in equilibrium when it is horizontally ori ented as in the figure. practice 12 – GADHIA, TEJAS – Due: Apr 11 2007, 4:00 am 2 3 2 . 2 cm . 5 4 2 k g θ 404N / m Find the period of small oscillation about the equilibrium point. Correct answer: 0 . 13287 s (tolerance ± 1 %). Explanation: Let : L = 32 . 2 cm , m = 0 . 542 kg , and k = 404 N / m . summationdisplay τ = I α Using the parallel axis theorem, the rotational inertia of the plank about the pivot point is I = 1 12 mL 2 + m parenleftbigg L 2 parenrightbigg 2 = 1 3 mL 2 . (1) From the freebody diagram, after displace ment by a small angle θ , we have summationdisplay τ = parenleftbigg mg L 2 + k y L parenrightbigg cos θ = parenleftbigg mg L 2 + k L sin θ L parenrightbigg cos θ , (2) where y = L sin θ is the displacement from equilibrium. Using I from Eq. (1), and for small θ , cos θ = 1 and sin θ = θ , we have mg L 2 k L 2 θ = 1 3 mL 2 d 2 θ dt 2 . (3) Substitute θ ′ = θ mg 2 k L and d 2 θ ′ dt 2 = d 2 θ dt 2 , we have d 2 θ ′ dt 2 + 3 k m θ ′ = 0 , (4) where the coefficient of θ ′ is ω 2 = 3 k m . There fore T ≡ 2 π ω = 2 π radicalbigg...
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 Spring '08
 Turner
 Energy, Inertia, Mass, Simple Harmonic Motion, Correct Answer

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