Midterm 3 - Version 048/AADAA midterm 03 Turner (58220) 1...

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Unformatted text preview: Version 048/AADAA midterm 03 Turner (58220) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A potters 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 , = 65 rev / min , and t = 6 . 3 s . The angular acceleration is = f t = t since f = 0. The applied friction is f = N , so the magnitude of the net torque is = | I | f R = vextendsingle vextendsingle vextendsingle I parenleftBig t parenrightBigvextendsingle vextendsingle vextendsingle RR = I t = I N Rt = (65 rev / min)(1 . 036 kg m 2 ) (100 N)(0 . 16 m)(6 . 3 s) 2 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 left-hand face at a height of 4 3 a . The bullet gets embedded in the cube. 2 a M mvectorv min 4 3 a M g Find the minimum value of v min required to tip the cube so that it falls its right-hand 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 U + 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 2 . (1) From conservation of angular momentum 4 a 3 mv = parenleftbigg 8 M a 2 3 parenrightbigg = 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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Midterm 3 - Version 048/AADAA midterm 03 Turner (58220) 1...

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