old-midterm3 - Castaneda (ic4477) – oldmidterm 03 –...

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Unformatted text preview: Castaneda (ic4477) – oldmidterm 03 – Turner – (58220) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rotating bicycle wheel has an angular speed of 62 ◦ / s at 4 . 4 s and a constant angular acceleration of 37 ◦ / s 2 . With the center of the wheel at the origin, the valve stem is on the positive x-axis (horizontal) at t = 1 . 4 s. Through what angle has the valve stem turned between 1 . 4 s and 7 . 6 s? Correct answer: 407 . 34 ◦ . Explanation: Let : ω 1 = 62 ◦ / s , t 1 = 4 . 4 s , α = 37 ◦ / s 2 , θ = 0 ◦ , t = 1 . 4 s , and t = 7 . 6 s . The initial angular velocity can be deter- mined from ω 1 = ω + α ( t 1- t ) ω = ω 1- α ( t 1- t ) = 62 ◦ / s- (37 ◦ / s 2 ) (4 . 4 s- 1 . 4 s) =- 49 ◦ / s . This θ 2 = ω ( t 2- t ) + 1 2 α ( t 2- t ) 2 = (- 49 ◦ / s) (7 . 6 s- 1 . 4 s) + 1 2 (37 ◦ / s 2 ) (7 . 6 s- 1 . 4 s) 2 = 407 . 34 ◦ . 002 (part 1 of 3) 10.0 points A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendulum is 153 cm long. If the beetle rides through a swing of 33 ◦ , how far has he traveled along the path of the pendulum? Correct answer: 88 . 1217 cm. Explanation: Let : r = 153 cm , and θ = 33 ◦ . Arc length is defined as s = r θ = (153 cm)(33 ◦ ) · π 180 ◦ = 88 . 1217 cm . 003 (part 2 of 3) 10.0 points What is the displacement experienced by the beetle while moving theough the same angle 33 ◦ ? Correct answer: 86 . 9087 cm. Explanation: Using the law of cosines, we have , R = r 2 + r 2- 2 r r cos θ = r 2 (1- cos θ ) = (153 cm) 2(1- cos 33 ◦ ) = 86 . 9087 cm . Alternative Solution: Divide the isosce- les triangle in half. Then , R = 2 . r sin θ 2 = 2 . 0 (153 cm) sin 16 . 5 ◦ = 86 . 9087 cm . 004 (part 3 of 3) 10.0 points If the pendulum at some instant is swinging at 1 . 2 rad / s, how fast is the beetle traveling? Correct answer: 183 . 6 cm / s. Explanation: Castaneda (ic4477) – oldmidterm 03 – Turner – (58220) 2 Let : ω = 1 . 2 rad / s . Linear and angular velocity are related by v = r ω = (153 cm)(1 . 2 rad / s) = 183 . 6 cm / s . keywords: 005 10.0 points At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0 . 82 rad / s 2 has an angular velocity of 2 . 8 rad / s and an angular position of 6 . 3 rad. What is the angular position of the wheel after 5 s? Correct answer: 10 . 05 rad. Explanation: Let : α =- . 82 rad / s 2 , ω = 2 . 8 rad / s , θ = 6 . 3 rad , and t = 5 s . The angular position is θ f = θ + ω t + 1 2 α t 2 = 6 . 3 rad + (2 . 8 rad / s) (5 s) + 1 2 (- . 82 rad / s 2 ) (5 s) 2 = 10 . 05 rad ....
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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old-midterm3 - Castaneda (ic4477) – oldmidterm 03 –...

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