The Waiter Physics Lab

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Zach Hicks (Z1603274)
Karen Richards
Section F (0900 - 1150)
11/30/12

The Waiter

Problem Solving
Part A: Center of Mass in One Direction
1. Since the mass of the plate is a uniform, the center of mass is in the center of the plate. The plate has diameter d = 28 cm, thus the center of mass is 14 cm from the edge of the plate. In the lab, two scenarios will be analyzed; 1) when the glass is standing up, and 2) when the glass is laying on its side. When the glass is standing up, its mass is uniform about the axis of symmetry. Therefore, the horizontal center of mass will be at the exact midpoint between the two sides of the glass. The radius of the glass is 4 cm, thus the horizontal center of mass, when the glass is standing up, is 4 cm from the edge of the glass. On the other hand, when on its side, because the mass of the glass is not uniform, that is two-thirds of the mass is in the base, the vertical center of mass has to be calculated using the equation xcm=miximi, where mi is the mass of the ith component and xi is the distance away from the origin the center of mass of the ith component is. For a glass 8 cm in diameter, 16 cm tall, of mass 300 grams (two-thirds of which is in the base), whose base is 3 cm tall and is at the origin, the vertical center of mass is xcm=miximi=200g1.5cm+100g(6.5cm)(300g)=4.17cm. The vertical center of mass is located 4.17 cm from the bottom of the glass. 3. Because the waiter's hand has a diameter d = 20 cm, the range of points above the waiter's hand that can support a center of mass is 10 cm. 4. The center of mass of the tray plus one glass--with the origin at the center of the tray--given as a function of time, is xcm=mtxt+mgxgmt+mg=500g0+300gxg500g+300g=300xg800=38xg. 5. The farthest from the center of the tray a glass--when standing up--can be is 38xg=384 cm=1.5cm. Because the waiter's hand can support a center of mass up to 10 cm from the center of the tray, the distance a glass can be from the...
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