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solution23_pdf - arellano(aa39398 Practice Homework Chapter...

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arellano (aa39398) – Practice Homework Chapter 10 Solutions – weathers – (17101) 1 This print-out should have 53 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A girl sitting on a merry-go-round moves counterclockwise through an arc length of 2.57 m. If the girl’s angular displacement is 2.27 rad, how far is she from the center of the merry-go-round? Correct answer: 1 . 13216 m. Explanation: Let : Δ s = 2 . 57 m and Δ θ = 2 . 27 rad . Δ s = r Δ θ r = Δ s Δ θ = 2 . 57 m 2 . 27 rad = 1 . 13216 m . keywords: 002 10.0 points A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3 . 9 times? Correct answer: 0 . 84498 s. Explanation: Let : ω avg = 29 rad / s and Δ θ = 3 . 9 rev . ω avg = Δ θ Δ t Δ t = Δ θ ω avg = 3 . 9 rev 29 rad / s · 2 π rad 1 rev = 0 . 84498 s . 003 (part 1 of 2) 10.0 points A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 15 rad / s in 3 . 05 s. Find the magnitude of the angular acceler- ation of the wheel. Correct answer: 4 . 91803 rad / s 2 . Explanation: Let : Δ ω = 15 rad / s and Δ t = 3 . 05 s . Angular acceleration is α = Δ w Δ t = 15 rad / s 3 . 05 s = 4 . 91803 rad / s 2 . 004 (part 2 of 2) 10.0 points Find the angle in radians through which it rotates in this time. Correct answer: 22 . 875 rad. Explanation: The angle through which the wheel rotates during this time interval is θ = 1 2 α t 2 = 1 2 (4 . 91803 rad / s 2 ) (3 . 05 s) 2 = 22 . 875 rad . keywords: 005 (part 1 of 2) 10.0 points A grinding wheel, initially at rest, is ro- tated with constant angular acceleration of 5 . 4 rad / s 2 for 3 . 9 s. The wheel is then brought to rest with uniform deceleration in 8 . 75 rev.
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arellano (aa39398) – Practice Homework Chapter 10 Solutions – weathers – (17101) 2 Find the angular acceleration required to bring the wheel to rest. Note that an in- crease in angular velocity is consistent with a positive angular acceleration. Correct answer: - 4 . 03365 rad / s 2 . Explanation: Let : ω 0 = 0 rad / s , α 1 = 5 . 4 rad / s 2 , t 1 = 3 . 9 s , and θ 2 = 8 . 75 rev . From kinematics ω f = ω 0 + α t . First find the speed attained before the wheel begins to slow down. Since ω 0 = 0 rad / s for the acceleration, the final speed ω 1 = α 1 t 1 for the acceleration is also the initial speed for the deceleration. Considering the wheel as it comes to rest, ω 2 2 = ω 1 2 + 2 α 2 θ 2 = 0 α 2 = - ω 2 1 2 θ 2 = - α 2 1 t 2 1 2 θ 2 = - (5 . 4 rad / s 2 ) 2 (3 . 9 s) 2 2 (8 . 75 rev) · 1 rev 2 π rad = - 4 . 03365 rad / s 2 . 006 (part 2 of 2) 10.0 points Determine the time needed to bring the wheel to rest. Correct answer: 5 . 22107 s. Explanation: The initial speed for the deceleration is ω 1 = α 1 t 1 , so from kinematics, ω f = ω 1 + α 2 t 2 = 0 t 2 = - ω 1 α 2 = - α 1 t 1 α 2 = - (5 . 4 rad / s 2 ) (3 . 9 s) - 4 . 03365 rad / s 2 = 5 . 22107 s . 007 (part 1 of 2) 10.0 points A remote-controlled car’s wheel accelerates at 22.7 rad/s 2 . If the wheel begins with an angular speed of 10.3 rad/s, what is the wheel’s angular speed after exactly seventeen full turns?
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