# 3 - homework 20 – FISHER CHRISTIAN – Due Apr 8 2008...

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Unformatted text preview: homework 20 – FISHER, CHRISTIAN – Due: Apr 8 2008, 11:00 pm 1 Question 1, chap 13, sect 4. part 1 of 2 10 points A 3 kg bicycle wheel rotating at a 2684 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 . 4 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 6 8 4 r e v / m i n ω . 4 m radius Find the change in the precession angle (in degree) after a 1 . 9 s time interval. Correct answer: 14 . 2339 ◦ (tolerance ± 1 %). Explanation: Let : m = 3 kg , ω = 2684 rev / min = 2 π (2684 rev / min) (60 s / min) = 281 . 068 rad / s , b = 0 . 6 m , and R = 0 . 4 m . Basic Concepts: vector τ = d vector 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, vector τ = vector b × mvectorg . 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 . 9 s) (0 . 4 m) 2 (281 . 068 rad / s) = 0 . 248428 rad = 14 . 2339 ◦ . Question 2, chap 13, sect 4. part 2 of 2 10 points The direction of precession as viewed from the top is 1. clockwise. correct 2. along the direction of rotation of the wheel. 3. counter-clockwise. 4. opposite to the direction of rotation of the wheel. 5. static, since angular momentum is con- served. Explanation: vector τ = vector b × mvectorg , where vector b is along the positive y-axis and vectorg is into the page, in the figure below. Therefore the direction of the torque vector τ is along the positive y-axis. homework 20 – FISHER, CHRISTIAN – Due: Apr 8 2008, 11:00 pm 2 + 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 vector τ = −→ Δ L Δ t , that is Δ L is in the direction of the torque τ and is producing clockwise motion ( vector L × vector τ produces clockwise motion). Question 3, chap 13, sect 99. part 1 of 2 10 points A thin hollow cylindrical rod has length 1 . 4 m, mass 0 . 09 kg, and diameter 0 . 3 cm . The thin cylindrical rod is free to rotate about a vertical axis that passes through its center and is perpendicular to the cylinder’s axis. Inside the cylinder are two masses of 0 . 7 kg each, attached to springs of spring constant k and unstretched lengths 0 . 3 m . The springs have negligible mass. The inside walls of the cylinder are frictionless....
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3 - homework 20 – FISHER CHRISTIAN – Due Apr 8 2008...

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