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Unformatted text preview: Version 048/AADAA – midterm 03 – Turner – (58220) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A potter’s wheel, having a radius of 0 . 16 m and a moment of inertia of 1 . 036 kg · m 2 , is freely rotating at 65 rev / min. The potter can stop the wheel in 6 . 3 s by pressing a wet rag against the rim and exerting a radially inward force of 100 N. Find the effective coefficient of kinetic fric tion between the wheel and the wet rag. 1. 0.0683028 2. 0.338833 3. 0.100109 4. 0.119946 5. 0.0860267 6. 0.0945969 7. 0.0699586 8. 0.290063 9. 0.104973 10. 0.271477 Correct answer: 0 . 0699586. Explanation: Let : I = 1 . 036 kg · m 2 , R = 0 . 16 m , N = 100 N , ω = 65 rev / min , and t = 6 . 3 s . The angular acceleration is α = ω f − ω t = − ω t since ω f = 0. The applied friction is f = μN , so the magnitude of the net torque is τ =  I α  f R = vextendsingle vextendsingle vextendsingle I parenleftBig − ω t parenrightBigvextendsingle vextendsingle vextendsingle μRR = I ω t μ = ω I N Rt = (65 rev / min)(1 . 036 kg · m 2 ) (100 N)(0 . 16 m)(6 . 3 s) × 2 π 1 rev · 1 min 60 s = . 0699586 002 10.0 points Assume: A bullet of mass m and cube of mass M undergo an inelastic collision, where m ≪ M . Note: The moment of inertia of this cube (with edges of length 2 a and mass M ) about an axis along one of its edges is 8 M a 2 3 . A solid cube is resting on a horizontal sur face. The cube is constrained to rotate about an axis at its bottom left edge (due to a small obstacle on the table). A bullet with speed v min is shot at the lefthand face at a height of 4 3 a . The bullet gets embedded in the cube. 2 a M mvectorv min 4 3 a ω M g Find the minimum value of v min required to tip the cube so that it falls its righthand face. 1. v min = m M radicalbigg 3 g a parenleftBig √ 5 − 1 parenrightBig 2. v min = m M radicalbigg 2 g a parenleftBig √ 2 − 1 parenrightBig 3. v min = m M radicalbigg 3 g a parenleftBig √ 3 − 1 parenrightBig Version 048/AADAA – midterm 03 – Turner – (58220) 2 4. v min = m M radicalbigg 5 g a parenleftBig √ 2 − 1 parenrightBig 5. v min = M m radicalbigg 3 g a parenleftBig √ 2 − 1 parenrightBig correct Explanation: Basic Concepts: summationdisplay vector L = const Δ U + Δ K = 0 For the cube to tip over the center of mass (CM) must rise so that it is over the axis of rotation AB . To do this the CM must be raised a distance of a parenleftBig √ 2 − 1 parenrightBig . From conservation of energy M g a parenleftBig √ 2 − 1 parenrightBig = 1 2 I cube ω 2 . (1) From conservation of angular momentum 4 a 3 mv = parenleftbigg 8 M a 2 3 parenrightbigg ω ω = parenleftBig mv 2 M a parenrightBig . (2) Thus, substituting Eq. 2 into 1, we have 1 2 parenleftbigg 8 M a 2 3 parenrightbiggparenleftbigg m 2 v 2 4 M 2 a 2 parenrightbigg = M g a parenleftBig √ 2 − 1 parenrightBig...
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 Spring '08
 Turner
 Physics, Angular Momentum, Moment Of Inertia, Correct Answer, Irod

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