EXAM3 - Nguyen, Phan – Midterm 3 – Due: Apr 7 2004,...

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Unformatted text preview: Nguyen, Phan – Midterm 3 – Due: Apr 7 2004, 10:00 pm – Inst: V Kaplunovsky 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 8 points An abstract sculpture consists of a ball (radius R = 55 cm) resting on top of a cube (each side L = 130 cm long). The ball and the cube are made of the same material of uniform density; there are no hollow spaces inside them. The bottom face of the cube rests on a horizontal floor. How high is the sculpture’s center of mass above the floor? Correct answer: 93 . 8983 cm. Explanation: Basic Principle: If a body comprises several parts and we know the location of each part’s center of mass, the the whole body’s center of mass is located at ~ R cm whole = X parts M part M whole ~ R cm part . (1) For the problem at hand, we have two parts — the cube and the ball — and we are inter- ested only in the vertical coordinate Z cm of the center of mass, thus Z cm sculpture = M cube M cube + M ball Z cm cube + M ball M cube + M ball Z cm ball . (2) Each part here has its center of mass in its geometric center, hence for the cube of size L × L × L resting on the floor Z cm cube = L 2 (3) while for the ball of radius R resting on top of the cube Z cm ball = L + R. (4) Next, the masses follow from volumes accord- ing to M cube = V cube ρ = L 3 ρ, (5) M ball = V ball ρ = 4 π 3 R 3 ρ, (6) hence M cube M cube + M ball = 3 L 3 3 L 3 + 4 πR 3 , (7) M ball M cube + M ball = 4 πR 3 3 L 3 + 4 πR 3 , (8) and the density ρ of the scuplture’s material cancels out of these formulae. Finally, we substitute eqs.(3–4) and (7–8) into eq. (2) and obtain Z cm sculpture = 3 L 3 3 L 3 + 4 πR 3 × L 2 + 4 πR 3 3 L 3 + 4 πR 3 × ( L + R ) = 93 . 8983 cm . (9) 002 (part 1 of 1) 10 points A triangular wedge of height H = 0 . 846 m, base length L = 1 . 41 m and mass M = 9 . 98 kg is placed on a frictionless table. A small block of mass m = 2 . 54 kg is placed on top of the wedge as shown on the picture below: M m All surfaces are frictionless, so the block slides down the wedge while the wedge slides side- wise on the table. By the time the block slides all the way down to the bottom of the wedge, how far does the wedge slide to the right? Nguyen, Phan – Midterm 3 – Due: Apr 7 2004, 10:00 pm – Inst: V Kaplunovsky 2 Correct answer: 0 . 286054 m. Explanation: Consider the wedge and the block as a two- body system. The external forces acting on this system — the weight of the wedge, the weight of the block and the normal force from the table — are all vertical, hence the net hor- izontal momentum of the system is conserved, P wedge x + P block x = const ....
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This note was uploaded on 03/23/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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EXAM3 - Nguyen, Phan – Midterm 3 – Due: Apr 7 2004,...

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