Homework 09 Solutions

Homework 09 Solutions - homework 09 – GADHIA TEJAS –...

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Unformatted text preview: homework 09 – GADHIA, TEJAS – Due: Mar 28 2007, 4:00 am 1 Question 1 part 1 of 3 10 points A ball at the end of a string of length 1 . 7 m rotates at a constant speed in a horizontal circle. It makes 5 . 2 rev / s. What is the period of the ball’s motion? Correct answer: 0 . 192308 s (tolerance ± 1 %). Explanation: The period T is the time for one complete revolution and in this case the ball makes 5 . 2 rev / s complete revolutions per second, so T = 1 rev 5 . 2 rev / s = 0 . 192308 s . Question 2 part 2 of 3 10 points What is the frequency of the motion? Correct answer: 5 . 2 Hz (tolerance ± 1 %). Explanation: f = 1 . 192308 s = 5 . 2 Hz . Question 3 part 3 of 3 10 points What is the ball’s angular velocity? Correct answer: 32 . 6726 rad / s (tolerance ± 1 %). Explanation: ω = 2 π rad T = 2 π . 192308 s = 32 . 6726 rad / s . Question 4 part 1 of 1 10 points A girl ties a toy airplane to the end of a string and swings it around her head. The plane’s average angular speed is 2.5 rad/s. In what time interval will the plane move through an angular displacement of 3.4 rad? Correct answer: 1 . 36 s (tolerance ± 1 %). Explanation: Basic Concept: ω avg = Δ θ Δ t Given: ω avg = 2 . 5 rad / s Δ θ = 3 . 4 rad Solution: Δ t = Δ θ ω avg = 3 . 4 rad 2 . 5 rad / s = 1 . 36 s Question 5 part 1 of 2 10 points A grinding wheel, initially at rest, is ro- tated with constant angular acceleration of 2 . 1 rad / s 2 for 9 . 04 s. The wheel is then brought to rest, with uniform deceleration, in 13 . 1 rev. Find the magnitude of the angular deceler- ation required to bring the wheel to rest. Correct answer: 2 . 18924 rad / s 2 (tolerance ± 1 %). Explanation: Let : α 1 = 2 . 1 rad / s 2 , t = 9 . 04 s , and θ 1 = 82 . 3098 rad . Basic Concepts ω = ω + αt ω 2 = ω 2 + 2 αθ θ = ¯ ω t. homework 09 – GADHIA, TEJAS – Due: Mar 28 2007, 4:00 am 2 Solution: First find the speed attained before the wheel begins to slow down. With ω = 0 rad / s , we have ω 1 = α 1 t 1 = (2 . 1 rad / s 2 ) (9 . 04 s) = 18 . 984 rad / s . Next, consider the wheel as it comes to rest. Now ω 1 = 18 . 984 rad / s, ω 2 = 0 rad / s , so ω 2 2 = 0 = ω 2 1 + 2 α 2 θ 1 α 2 = − ω 2 1 2 θ 1 (1) = − (18 . 984 rad / s) 2 2 (82 . 3098 rad) = − 2 . 18924 rad / s 2 bardbl vectorα 2 bardbl = 2 . 18924 rad / s 2 . Question 6 part 2 of 2 10 points Determine the time needed to bring the wheel to rest. Correct answer: 8 . 67149 s (tolerance ± 1 %). Explanation: Since ω 1 = 18 . 984 rad / s , ω 2 = 0 rad / s , and α 2 from Eq. 1, we have ω 2 = ω 1 + α 2 t t = − ω 1 α 2 = 2 θ 1 ω 1 = 2 (82 . 3098 rad) (18 . 984 rad / s) = 8 . 67149 s . Question 7 part 1 of 1 10 points A wheel rotates with a constant angular acceleration of 2 . 05 rad / s 2 . Suppose that at t = 0, the angular speed of the wheel is 2 . 83 rad / s....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework 09 Solutions - homework 09 – GADHIA TEJAS –...

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