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Unformatted text preview: oldquiz 03 – GUTHMANN, MATTHEW R – Due: Apr 3 2006, 1:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:5. Question 1 Part 1 of 1. 10 points. Consider three objects of equal masses but different 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. Compare the three objects’ moments of in- ertia when rotated around their respective centers of mass. Which of the following conditions is cor- rect? 1. I cm disk > I cm square > I cm ring 2. I cm square > I cm ring > I cm disk correct 3. I cm ring > I cm disk > I cm square 4. I cm disk > I cm ring > I cm square 5. I cm ring > I cm square > I cm disk 6. I cm square > I cm disk > I cm ring Explanation: The moment of inertia I = Z r 2 dm can be though as I = M × r 2 , where M is the object’s net mass and r 2 is the average distance 2 of massive points making up the object from the rotation axis. For a hoop or a thin ring, all massive point are at the same distance R from the axis, hence r 2 = R 2 = ⇒ I cm ring = M R 2 . For the solid disk, the distance ranges from zero in the center to R at the perimeter, hence the average distance is less than R , r 2 < R 2 = ⇒ I cm disk < M R 2 . On the other hand, on the perimeter of the hollow square, the distance from the center ranges from R to R √ 2, so the average distance is greater than R , r 2 > R 2 = ⇒ I cm square < M R 2 . Therefore, without any calculations we can say that I cm square > I cm ring > I cm disk . If you want actual values of the three mo- ments of inertia, you can find two of them in the textbook, namely I cm ring = M R 2 and I cm disk = 1 2 M R 2 . For the square, we need to calculate: A hollow square is a sum of four thin rods, each having mass M 4 and length 2 R , and the center of each rod is located at the distance R from the square’s center. Hence, using the parallel axis theorem I rod = M rod × R 2 + I cm rod = M 4 R 2 + 1 12 M 4 (2 R ) 2 = 1 3 M R 2 , and therefore I cm square = 4 I rod = 4 3 M R 2 . And this confirms than indeed I cm square > I cm ring > I cm disk . Question 2 Part 1 of 1. 10 points. oldquiz 03 – GUTHMANN, MATTHEW R – Due: Apr 3 2006, 1:00 am 2 Given: A rotating bicycle wheel has an angular speed of 44 ◦ / s at t 1 = 3 . 4 s and a constant angular acceleration of 38 ◦ / s 2 . Given: With the center of the wheel at the origin, the valve stem is on the positive x-axis (horizontal) at t = 1 . 4 s. Using 0 ◦ ≤ θ ≤ 360 ◦ , what angle θ does the valve stem make with its original direction at t 2 = 8 s?...
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This note was uploaded on 04/15/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas.
- Spring '10