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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 ω 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 ℓ = 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 ℓ 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 ω 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 ω 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 ω 2 , we may then solve for ω . ω = 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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 Spring '09
 KLEINMAN
 Physics, Mass, Moment Of Inertia, Rigid Body, kg, MATT –

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