This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Overton, Mays – Midterm 3 – Due: Apr 6 2005, 10:00 pm – Inst: Turner 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two objects, of masses 25 kg and 10 kg, are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown. 25 kg 10 kg A B C D E F G 10 20 30 40 50 60 If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord? 1. F 2. D 3. E 4. A correct 5. B 6. C Explanation: Let : ‘ = 70 cm , m 1 = 25 kg , m 2 = 10 kg . For static equilibrium, τ net = 0. Denote x the distance from the left end point of the stick to the point where the cord is attached. m 1 g x- m 2 g [ ‘- x ] = τ = 0 ( m 1- m 2 ) x = m 2 ‘ x = m 2 ‘ m 1- m 2 = (10 kg)(70 cm) 25 kg- 10 kg = 20 cm . Therefore the point should be point A . 002 (part 1 of 1) 10 points A 3 kg bicycle wheel rotating at a 2368 rev / min angular velocity has its shaft supported on one side, as shown in the figure. When viewing from the left (from the positive x-axes), one sees that the wheel is rotating in a clockwise manner. The distance from the center of the wheel to the pivot point is 0 . 6 m. The wheel is a hoop of radius 0 . 5 m, and its shaft is horizontal. Assume all of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9 . 8 m / s 2 . y z x . 6 m mg ω 2 3 6 8 r e v / m i n ω . 5 m radius Find the change in the precession angle after a 1 . 6 s time interval. Correct answer: 8 . 695 ◦ . Explanation: Let : m = 3 kg , ω = 2368 rev / min = 2 π (2368 rev / min) (60 s / min) = 247 . 976 rad / s , b = 0 . 6 m , and R = 0 . 5 m . Overton, Mays – Midterm 3 – Due: Apr 6 2005, 10:00 pm – Inst: Turner 2 Basic Concepts: ~ τ = d ~ L dt Solution: The magnitude of the angular momentum I of the wheel is L = I ω = mR 2 ω , and is along the negative x-axis. From the figure in Part 2, we get Δ φ = Δ L L . Using the relation, Δ L = τ Δ t, where τ is the torque, ~ τ = ~ b × m~g . The precession angle Δ φ is Δ φ = Δ L L = τ Δ t L = mg b Δ t mR 2 ω = g b Δ t R 2 ω = (9 . 8 m / s 2 )(0 . 6 m)(1 . 6 s) (0 . 5 m) 2 (247 . 976 rad / s) = 0 . 151756 rad = 8 . 695 ◦ . ~ τ = ~ b × m~g , where ~ b is along the positive y-axis and ~g is into the page, in the figure below. Therefore the direction of the torque ~ τ is along the positive y-axis. + x- x L + Δ L Δ L Δ φ Δ L τ wheel Viewed from Above + y L We can see the direction of precession is clockwise, due to ~ τ =-→ Δ L Δ t , that is Δ L is in the direction of the torque τ and is producing clockwise motion ( ~ L × ~ τ produces clockwise motion)....
View Full Document
This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
- Spring '08