11 - obaidi (aao476) – OH-11 – Bradley – (191) 1 This...

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Unformatted text preview: obaidi (aao476) – OH-11 – Bradley – (191) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 11. It is due before 06:00 AM, Tuesday 11 November (Abu Dhabi Time). Note: OH-11 is due 11-11! Also note that THIS TIME the solutions will be available at 06:05 on Tuesday morning! 001 10.0 points In a circus performance, a large 4 kg hoop of radius 3 . 3 m rolls without slipping. If the hoop is given an angular speed of 2 . 7 rad / s while rolling on the horizontal and is allowed to roll up a ramp inclined at 22 ◦ with the horizontal, how far (measured along the incline) does the hoop roll? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 21 . 6249 m. Explanation: Let : m = 4 kg , r = 3 . 3 m , ω i = 2 . 7 rad / s , θ = 22 ◦ , and g = 9 . 8 m / s 2 . By conservation of mechanical energy, U f = K i since v f = 0 m/s and h i = 0 m. U f = K trans,i + K rot,i mgh = 1 2 mv 2 i + 1 2 ( mr 2 ) ω 2 i mg d sin θ = 1 2 m ( r ω i ) 2 + 1 2 ( mr 2 ) ω 2 i = 1 2 mr 2 ω 2 i + 1 2 mr 2 ω 2 i g d sin θ = r 2 ω 2 i d = r 2 ω 2 i g sin θ = (3 . 3 m) 2 (2 . 7 rad / s) 2 (9 . 8 m / s 2 ) sin 22 ◦ = 21 . 6249 m . keywords: 002 (part 1 of 3) 4.0 points A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 1 . 3 m above the ground. Find the final linear velocity of the thin cylindrical shell.The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 56931 m / s. Explanation: Let : H = 1 . 3 m , and g = 9 . 8 m / s 2 . H S h 1 2 S Because there is no slipping, v 1 = ω 1 R and the rotational inertia of the cylindrical shell is I 1 = mR 2 . Thus, from conservation of energy K rot + K trans = H 1 2 I 1 ω 2 1 + 1 2 m ( ω 1 R ) 2 = mg H 1 2 ( mR 2 ) parenleftBig v 1 R parenrightBig 2 + 1 2 mv 2 1 = mg H , so v 1 = radicalbig g H = radicalBig (9 . 8 m / s 2 ) (1 . 3 m) = 3 . 56931 m / s . 003 (part 2 of 3) 3.0 points Find the final linear velocity of the solid cylin- der. Correct answer: 4 . 12149 m / s. obaidi (aao476) – OH-11 – Bradley – (191) 2 Explanation: Because there is no slipping, v 2 = ω 2 R and the rotational inertia of the solid cylinder is I 2 = 1 2 mR 2 , so K rot + K trans = H 1 2 I 2 ω 2 2 + 1 2 m ( ω 2 R ) 2 = mg H 1 2 parenleftbigg 1 2 mR 2 parenrightbigg ω 2 2 + 1 2 mv 2 2 = mg H v 2 = radicalbigg 4 3 g H v 2 = radicalbigg 4 3 (9 . 8 m / s 2 ) (1 . 3 m) = 4 . 12149 m / s . 004 (part 3 of 3) 3.0 points When the first object reaches the bottom, what is the height above the ground of the other object?...
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This note was uploaded on 02/16/2009 for the course PYHS 191 taught by Professor Curtise during the Spring '08 term at Abilene Christian University.

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11 - obaidi (aao476) – OH-11 – Bradley – (191) 1 This...

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