The WaiterPROBLEMPROBLEM SKILLSBaCKGROUND INFORMATIONproblem solvingOBSERVATIONSIntroductory Mechanics Problems Laboratory1The WaiterGoals: Calculate the center of mass of an individual object and a collection of objects. Understand the role of the center of mass for stability and equilibrium.PROBLEM We have all watched as a waiter carries an amazing numebr of items on a tray without dropping any. Perhaps, you’ve even had the chance to wait or bus tables as well. The act of balancing a number of different items on a large tray requires balance by the waiter, but also requires placement of items based on physics.Our waiter has two types of items: plates and glassware. The plates can be treated as uniform thin disks with a mass of 900 g and a 28 cm diameter. The glasses have a mass of 300 g and are cylinders with a radius of 4 cm and a height of 16 cm. The mass is not uniform in a glass. Two thirds of the mass is in the base which is 3 cm tall.2 The WaiterThe waiter has a choice of two trays. One is a narrow rectangular tray that measures 21 cm by 88 cm and has a mass of 500 g. On the narrow tray, items are lined up in a row without overlapping. The other is a large circular tray with a diameter of 76 cm and mass of 1000 g. Items on this tray can be placed in any position if they don’t overhang the edge and don’t overlap.The waiter carries his tray supported on one hand that spans a diameter of 20 cm. What arrangements of plates and glasses can be balanced on a tray?PROBLEM SKILLS We are used to problems with variables to stand for quantities that may change during a the course of a problem. Problems that have a changing number of items with the same types of variables use a shorthand notation to represent a set of variables. In this short-hand the same variable name is given to all variables of a certain type. A subscript is then used to distinguish between the individual variables.Let’s consider a set of objects that each have a mass. For three objects the masses could be represented as m1, m2 and m3. Just as easily we can write mi, where we understand that i = 1, 2, or 3. The advantage of writing mi is that the number of objects can be an arbitrary number N. The same notation can apply to the x and y position coordinates for each object, writing them as xi and yi.If we wish to find the total mass M in the problem we write that using summation nota-tion as in EQ 1.(EQ 1)This shorthand means add all the values mi, from i = 1 to N. If the sum is assumed to be for all items EQ 1 is shortened to EQ 2.(EQ 2)Using summation notation, the average mass can be written as in EQ 3.(EQ 3)BACKGROUND INFORMATIONSimple kinematics and dynamics problems treat objects as single points with all their mass at the one point. Real objects have mass at many points, and there may be a num-ber of objects all moving together. Rather than compute the dynamics of each separate mass, the center of mass can be used with the external forces as if there was a single object.The center of mass is sometimes called the center of gravity. This is because gravity, as an external force, act on the set of objects as if they were a single object with the total Mmii 1=N∑=Mmii∑=M1N----mii 1=N∑=The Waiter 3mass at the point of the center of gravity. If a set of objects is to be supported without tipping over, the center of mass must be directly over a point anywhere between the sup-ports.The center of mass of a system of objects is a weighted average of the positions of the individual objects. A weighted average is like a regular average, but assigns each of the values a “weight”. For the center of mass the position is weighted by the mass of the object. Using summation notation the center of mass (xCM) of a set of objects with posi-tion xi in one dimension is given by EQ 4.(EQ 4)For each of the objects in EQ 4 the position is taken to be the center of mass of the indi-vidual object.The center of mass can also be found in more than one dimension. In this case the posi-tion is a vector and the form of the center of mass is (EQ 5)Like any vector equation this one can be separated into separate equations for each com-ponent. The component equation is then exactly the form of EQ 4.For uniform, symmetrical objects the center of mass is just the midpoint of the object. If a single object is not uniform it is often possible to find the center of mass with a simple approximation. If the ojbject can be pictured as a finite number of uniformly shaped pieces the center of mass of the complete object is just the center of mass of those pieces.PROBLEM SOLVING Part A. Center of Mass in One Dimension1. Find the center of mass of a plate and a glass. Assume that the glass is made up of two uniform cylinders stacked on top of each other.2. Set up a diagram of the rectangular tray as seen sideways looking at the long axis. Place the origin at the center of the tray.3. Assume that the waiter holds the rectangular tray under the center of mass of the empty tray. Find the range of points above the waiter’s hand that can support a center of mass.4. Find the center of mass of the tray and one glass as a function of the position of the glass using EQ 4.xCMmixii∑mii∑-----------------mixii∑M-----------------==rCMmirii∑mii∑-----------------mirii∑M-----------------==4 The Waiter5. Use the result of step 4 to find the farthest out from the center a glass can be on the tray.6. Repeat steps 4 and 5 with the glass resting on its side.7. Repeat steps 4 and 5 with one plate and one glass, and determine how far out the glass can be now.8. Share your results with the other groups and discuss any differences between the groups.Part B. Center of Mass in Two Dimensions9. Set up a diagram of the round tray as seen from above. Place the origin at the center of the tray.10. Assume that the waiter holds the round tray under the center of mass of the empty tray. Find the range of points above the waiter’s hand that can support a center of mass.11. Use EQ 5 and find a formula for the center of mass of two plates and two glasses resting on the tray.12. Use the result of steps 10 and 11 to find an arrangement of glasses and plates that keeps them as far apart as possible. Keep in mind that the plates and glasses cannot overlap on the tray.13. Share your results of part B with the other groups and discuss any differences between the groups.OBSERVATIONS For each observation below write a short paragraph to explain your thinking.What