Kaplunovsky - hw 10

Kaplunovsky - hw 10 - homework 10 Due: Apr 4 2007, 4:00 am...

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homework 10 – – Due: Apr 4 2007, 4:00 am 1 Question 1, chap 12, sect 5. part 1 of 1 6 points The rigid object shown is rotated about an axis perpendicular to the paper and through center point O . O ω 34 kg 34 kg 68 kg 68 kg 16 m 16 m 8 m 8 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: 26112 kg m 2 (tolerance ± 1 %). Explanation: The moment of inertia of a rigid body whose mass is concentrated in several points is given by I = s i m i r 2 i . For the body at hand, the mass is concen- trated in the four small balls. Two of them have masses m 1 , 2 = 34 kg and are located at distances r 1 , 2 = 16 m from the rotation axis O ; the other two balls have m 3 , 4 = 68 kg each and are located at distances r 3 , 4 = 8 m from the axis. Therefore, I = 4 s i =1 m i r 2 i = 2 × (34 kg)(16 m) 2 + 2 × (68 kg)(8 m) 2 = 26112 kg m 2 . Question 2, chap 12, sect 5. part 1 of 3 6 points Four particles with masses 5 kg, 5 kg, 5 kg, and 5 kg are connected by rigid rods of negli- gible mass at the corners of a rectangle. The origin is at the center of the ±gure. x y O ω 6 m 8 m 5 kg 5 kg 5 kg 5 kg What is the moment of inertia about the y -axis? Correct answer: 320 kg m 2 (tolerance ± 1 %). Explanation: Let : m 1 = 5 kg , top right m 2 = 5 kg , top left m 3 = 5 kg , bottom left m 4 = 5 kg , bottom right w = 8 m , = 6 m , x 0 = 4 m , and y 0 = 3 m . The distance r i in the expression for the moment of inertia is simply the distance of each particle to the y axis. i.e. r i = x 0 . Hence I y = s m i r 2 i = x 2 0 s m i = (4 m) 2 (20 kg) = 320 kg m 2 . Question 3, chap 12, sect 5. part 2 of 3 6 points What is the moment of inertia about the z -axis? Correct answer: 500 kg m 2 (tolerance ± 1 %). Explanation:

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homework 10 – – Due: Apr 4 2007, 4:00 am 2 Moment of inertia about the z -axis: I z = s m i r 2 i r i is the same r 0 for each mass r 2 i = r 2 0 = x 2 0 + y 2 0 So I z = r 2 0 s m i = (5 m) 2 × [(5 kg) + (5 kg) + (5 kg) + (5 kg)] = (5 m) 2 (20 kg) = 500 kg m 2 . Question 4, chap 12, sect 5. part 3 of 3 6 points The system rotates around the z -axis at an angular velocity of 8 rad / s. What is the kinetic energy of the system? Correct answer: 16000 J (tolerance ± 1 %). Explanation: Kinetic energy is given by K = 1 2 I z ω 2 = (500 kg m 2 ) (8 rad / s) 2 2 = 16000 J . Question 5, chap 12, sect 5. part 1 of 1 6 points Consider three objects of equal masses but diFerent shapes: a solid disk, a thin ring, and a thin hollow square. The sizes of the three objects are related according to the following picture disk R ring R square 2 R Note: The ring and the square are hollow and their perimeters carry all the mass, but the disk is solid and has uniform mass density over its whole area.
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This note was uploaded on 01/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Kaplunovsky - hw 10 - homework 10 Due: Apr 4 2007, 4:00 am...

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