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HW8 - boland(kbb544 HW8 mackie(10611 This print-out should...

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boland (kbb544) – HW8 – mackie – (10611) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A wheel rotating with a constant angular ac- celeration turns through 20 revolutions during a 3 s time interval. Its angular velocity at the end of this interval is 13 rad / s. What is the angular acceleration of the wheel? Note that the initial angular veloc- ity is not zero. Correct answer: 19 . 2586 rad / s 2 . Explanation: Let : N = 20 , t = 3 s , and ω = 13 rad / s . From kinematics α t = ω f ω 0 ω 0 = ω f α t and Δ θ = N 2 ( π ) , so Δ θ = ω 0 t + 1 2 α t 2 2 N π = ( ω f α t ) t + 1 2 α t 2 = ω f t 1 2 α t 2 α = 2 ω f t 2 N π t 2 = 2 (13 rad / s) (3 s) 2 (20) π (3 s) 2 = 19 . 2586 rad / s 2 . keywords: 002 10.0 points A copper block rests 21.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the surface is 0.50. The turntable starts from rest and rotates with a constant angular acceleration of 0.60 rad/s 2 . After what time interval will the block start to slip on the turntable? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 8 . 05487 s. Explanation: Let : r = 21 . 0 cm , μ s = 0 . 50 , α = 0 . 60 rad / s 2 , and g = 9 . 81 m / s 2 . From kinematics ω f = ω i + α Δ t = α Δ t since ω i = 0 rad/s, F c = m a c = m ( r ω 2 f ) when the block starts to slip, and F s = μ s F n = μ s m g , so F c = F s m r ω 2 f = μ s m g r ( α Δ t ) 2 = μ s g Δ t = radicalbigg μ s g r α 2 = radicalBigg 0 . 5 (9 . 81 m / s 2 ) (0 . 21 m) (0 . 6 rad / s 2 ) 2 = 8 . 05487 s . 003 10.0 points A rotating wheel requires 3 . 9 s to rotate through 43 . 1 rev. Its angular speed at the end of the 3 . 9 s interval is 108 . 8 rad / s. What is its constant angular acceleration? Assume the angular acceleration has the same sign as the angular velocity.
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boland (kbb544) – HW8 – mackie – (10611) 2 Correct answer: 20 . 186 rad / s 2 . Explanation: Let : t = 3 . 9 s , Δ θ = 43 . 1 rev , and ω f = 108 . 8 rad / s . From kinematics ω f = ω 0 + α t ω 0 = ω f α t , so Δ θ = ω 0 t + 1 2 α t 2 = ( ω f α t ) t + 1 2 α t 2 = ω t 1 2 α t 2 α = 2 ( ω t Δ θ ) t 2 Since ω t Δ θ = (108 . 8 rad / s) (3 . 9 s) (43 . 1 rev) 2 π 1 rev = 153 . 515 rad , α = 2 (153 . 515 rad) (3 . 9 s) 2 = 20 . 186 rad / s 2 . 004 10.0 points A rotating bicycle wheel has an angular speed of 74 / s at 5 s and a constant angular acceleration of 35 / s 2 . With the center of the wheel at the origin, the valve stem is on the positive x -axis (horizontal) at t 0 = 1 . 5 s.
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