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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' The equation of motion about A in the small angle approximation (sin ) is given by 1. I d 2 dt 2 = mg R . 2. I d 2 dt 2 = 2 mg R . 3. I d 2 dt 2 = mg R . 4. I A d 2 dt 2 = mg R . 5. I A d 2 dt 2 = 1 2 mg R . 6. I A d 2 dt 2 = mg R . correct 7. I A d 2 dt 2 = 2 mg R . 8. I d 2 dt 2 = 1 2 mg R . 9. I d 2 dt 2 = 2 mg R . 10. I A d 2 dt 2 = 2 mg R . Explanation: The torque on the hoop is provided by grav ity, and its magnitude is given by = mg R sin mg R , 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 dt 2 = mg R , 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, = radicalBigg mg R I A = radicalbigg mg R 2 mR 2 = radicalbigg g 2 R and the period is T = 2 = 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 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 = I 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 , we have summationdisplay = parenleftbigg mg L 2 + k y L parenrightbigg cos = parenleftbigg mg L 2 + k L sin L parenrightbigg cos , (2) where y = L sin is the displacement from equilibrium. Using I from Eq. (1), and for small , cos = 1 and sin = , we have mg L 2 k L 2 = 1 3 mL 2 d 2 dt 2 . (3) Substitute =  mg 2 k L and d 2 dt 2 = d 2 dt 2 , we have d 2 dt 2 + 3 k m = 0 , (4) where the coefficient of is 2 = 3 k m . There fore T 2 = 2 radicalbigg...
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 Spring '08
 Turner
 Inertia, Mass

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