OLDHW24 - oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008,...

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Unformatted text preview: oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am 1 Question 1, chap -1, sect -1. part 1 of 2 10 points Four particles with masses 3 kg, 3 kg, 3 kg, and 8 kg are connected by rigid rods of negli- gible mass as shown. x y O 6m 4 m 3 kg 3 kg 3 kg 8 kg The origin is at the center of the rectangle, which is 4 m wide and 6 m long. If the system rotates in the xy plane about the z axis (origin, O ) with an angular speed of 8 rad / s, calculate the moment of inertia of the system about the z axis. Correct answer: 221 kg m 2 (tolerance 1 %). Explanation: Let : m 1 = 3 kg , top left m 2 = 3 kg , bottom left m 3 = 3 kg , bottom right m 4 = 8 kg , top right w = 4 m , and = 6 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 (6 m) 2 bracketrightbigg 2 = 3 . 60555 m , we obtain I z = [ m 1 + m 2 + m 3 + m 4 ] r 2 = [(3 kg) + (3 kg) + (3 kg) + (8 kg)] (3 . 60555 m) 2 = 221 kg m 2 . Question 2, chap -1, sect -1. part 2 of 2 10 points Find the rotational energy of the system. Correct answer: 7072 J (tolerance 1 %). Explanation: The rotational energy of the system is K = 1 2 I 2 = 1 2 (221 kg m 2 ) (8 rad / s) 2 = 7072 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 6 . 2 J. O 30 kg 30 kg 60 kg 60 kg 8 m 8 m 4 m 4 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: 5760 kg m 2 (tolerance 1 %). Explanation: oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am 2 Let : M = 30 kg , L = 4 m , and K = 6 . 2 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 (30 kg) (4 m) 2 = 5760 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 . 046398 rad / s (tolerance 1 %). Explanation: Since the rotational kinetic energy is given by K R = 1 2 I 2 , we may then solve for . = radicalbigg 2 K R I = radicalBigg 2 (6 . 2 J) (5760 kg m 2 ) = . 046398 rad / s . Question 5, chap 12, sect 5. part 1 of 1 10 points A uniform rod of mass 1 . 4 kg is 2 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 2 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 59 with the horizontal. The rod is released from rest at an angle of 59 with the horizontal, as shown in the figure below The acceleration of gravity is 9...
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.

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OLDHW24 - oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008,...

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