Lab 11 omments of Inertia In lab, we will learm how obtain the ent of Inertia of

Question

Lab 11 omments of Inertia In lab, we will learm how obtain the ent of Inertia ofany object this to I-Ir dm. whether it has constant or variable density. We will use the definition, Equipments solid Rectangular Plate solid cylinder, sphere, Various Triangles Theory: Moment of Inertia is the counterpart of mass. Mass (m) measures rotational therefore it is the and it is to make an object line, how difficult best measure of inertia. On the other hand, Moment of inertia object it is to object Therefore, it the rotational inertia of the is, make a The Definition for Moment of Inertia and hence the term, "Moment from a string at distance Jr ahr. In the case of a point obiect of mass "m" rotating a "d", this definition reduces to, I m The case for extended objects is more They Principle Moments of Inertias depending on what complicated. axis you are trying is the Moment of Inertia if you Moment Inertia if you rotate around the xaxis. Ly proven the rotate it around similar for La. In higher level physics, it is that medium value the y-axis and unstable principle moment of inertia yields a chaotic and Here are the different cases we will work with. Case 1: Rectangular Plate Lou-You Figure it out by Logic Same as Above Case 2: Solid Cylinder MR22 Case 3: Solid Sphere 46

Explanation / Answer

Case I: Reactangular plate

M = 0.113 kg , L = 0.119 m, W = 0.103 m , H = 0.0195 m

Izz = M(L2 + W2 )/12 = 0.113(0.1192 +0.1032)/12 = 2.33e-4 kg-m2

Ifront or back edge = Izz + M(L/2)2 = 2.33e-4 + 0.113* (0.119/2)2

                          = 6.33e-4 kg-m2

Ixx = M(H2 + W2 )/12 = 0.113(0.0.01952 +0.1032)/12 = 1.035e-4 kg-m2

Ileft or right edge = Ixx + Md2   --- (d2 = (W/2)2 + (H/2)2 = 0.0027 )

= 1.035e-4 + 0.113* 0.0027

                          = 4.05e-4 kg-m2

Iyy = M(L2 + H2 )/12 = 0.113(0.1192 +0.0.01952)/12 = 1.37e-4 kg-m2

Itop or bottom edge = Iyy + Md2   -- ( d2 = (L/2)2 + (H/2)2 = 0.0036 )

                          = 2.27e-4 kg-m2

Any corner axis ( paralle to z-axis)

d2 = (L/2)2 +(W/2)2 = 0.0062

I = Izz + Md2 = 2.33e-4 + 0.0523*0.0062 = 5.57e-4 kg-m2

Ifront or back vertical mid point axis , it is somewhat confusing, it is assumed to be parallel to y-axis in the x-y plane at the edge

then d= L/2 = 0.119/2 = 0.0595

I = Iyy + Md2/2 = 1.37e-4 + 1.85e-4 = 3.22 kg-m2

Case 2: Cylinder M = 0.523 kg, R = 0.012 m, L = 0.01735

Izz = MR2/2 = 0.0523 * 0.0122/2 = 3.766e-6 kg-m2

Iside-edge = Izz + Md2 = 3.766e-6 + 0.0523*(0.012)2

                    = 1.13 e-5 kg-m2

Ixx = Iyy = ML2 /12 + Izz/2 ( not sure why the factor Izz/2 is included here , it is not required)

            = 0.0523*0.017352/12 + 0.565e-5 = 1.312e-6 + 0.565e-5 ( correct one is 1.312e-6)

             = 6.962 e-6 kg-m2

Itop or bottom edge = ML2 /3 = 0.0523*0.017352/3 = 5.248 e-6 kg-m2

From the fromula given for Ixx

we get

Itop or bottom edge = Ixx + Md2 = 6.962e-6 + 0.0523*(0.01735/2)2

                        = 10.9 e-6 kg-m2

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