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oldhw24_s - oldhomewk 24 YOO HEE Due Apr 1 2008 4:00 am we...

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oldhomewk 24 – YOO, HEE – Due: Apr 1 2008, 4:00 am 1 Question 1, chap -1, sect -1. part 1 of 2 10 points Four particles with masses 2 kg, 2 kg, 2 kg, and 9 kg are connected by rigid rods of negli- gible mass as shown. x y O ω 4 m 4 m 2 kg 2 kg 2 kg 9 kg The origin is at the center of the rectangle, which is 4 m wide and 4 m long. If the system rotates in the xy plane about the z axis (origin, O ) with an angular speed of 9 rad / s, calculate the moment of inertia of the system about the z axis. Correct answer: 120 kg m 2 (tolerance ± 1 %). Explanation: Let : m 1 = 2 kg , top left m 2 = 2 kg , bottom left m 3 = 2 kg , bottom right m 4 = 9 kg , top right w = 4 m , and = 4 m , From I = summationdisplay j m j r 2 j , where in this case all distances are equal to r = radicalBigg bracketleftBig w 2 bracketrightBig 2 + bracketleftbigg 2 bracketrightbigg 2 = radicalBigg bracketleftbigg (4 m) 2 bracketrightbigg 2 + bracketleftbigg (4 m) 2 bracketrightbigg 2 = 2 . 82843 m , we obtain I z = [ m 1 + m 2 + m 3 + m 4 ] r 2 = [(2 kg) + (2 kg) + (2 kg) + (9 kg)] × (2 . 82843 m) 2 = 120 kg m 2 . Question 2, chap -1, sect -1. part 2 of 2 10 points Find the rotational energy of the system. Correct answer: 4860 J (tolerance ± 1 %). Explanation: The rotational energy of the system is K = 1 2 I ω 2 = 1 2 (120 kg m 2 ) (9 rad / s) 2 = 4860 J . Question 3, chap 12, sect 5. part 1 of 2 10 points The rigid object shown is rotated about an axis perpendicular to the paper and through center point O . The total kinetic energy of the object as it rotates is 4 . 7 J. O ω 11 kg 11 kg 22 kg 22 kg 7 m 7 m 3 . 5 m 3 . 5 m What is the moment of inertia of the ob- ject? Neglect the mass of the connecting rods and treat the masses as point masses. Correct answer: 1617 kg m 2 (tolerance ± 1 %). Explanation:
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oldhomewk 24 – YOO, HEE – Due: Apr 1 2008, 4:00 am 2 Let : M = 11 kg , L = 3 . 5 m , and K = 4 . 7 J . Basic Concepts: I = summationdisplay m i r 2 i Solution: The moment of inertia of the rigid body is given by I = summationdisplay m i r 2 i = 2 bracketleftBig 2 M ( L ) 2 bracketrightBig + 2 bracketleftbig M (2 L ) 2 bracketrightbig = 12 M L 2 = 12 (11 kg) (3 . 5 m) 2 = 1617 kg m 2 . Question 4, chap 12, sect 5. part 2 of 2 10 points What is the angular velocity of the object? Correct answer: 0 . 0762446 rad / s (tolerance ± 1 %). Explanation: Since the rotational kinetic energy is given by K R = 1 2 I ω 2 , we may then solve for ω . ω = radicalbigg 2 K R I = radicalBigg 2 (4 . 7 J) (1617 kg m 2 ) = 0 . 0762446 rad / s . Question 5, chap 12, sect 5. part 1 of 1 10 points A uniform rod of mass 3 . 3 kg is 15 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 15 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 48 with the horizontal. The rod is released from rest at an angle of 48 with the horizontal, as shown in the figure below The acceleration of gravity is 9 . 8 m / s 2 . Hint: The moment of inertia of the rod about its center-of-mass is I cm = 1 12 m ℓ 2 .
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