Question

Below is a picture of the setup of the experiment done in class,  For each of the three locations of the smartphone on the bar, you need to determine the center of mass of the combined bar and smartphone as well as the mass moment of inertia of the bar and smartphone about the axis of rotation of the pendulum. To do these computations, you need to have information on the mass and dimensions of your smartphone (width and length). Note that the bar is made of 6061 aluminum (density 2700 kg/m3, has a width of 1.25 inches, a length of 12 inches, and a thickness of 0.25 inches.

The three distances of the phone are 4,6, and 8 inches

Answer 1

Mass of the bar:

$Mass = \rho * Volume$

$\rho = 2700 kg/m^3 = 2.7g/cm^3$

COM of Bar is at the mid length at 6 inches = 6*2.54 = 15.24

Moment of Inertia of bar about an axis parallel to axis of pendulum and passing through its COM is:

$I_G = \frac{M*(1.25^2+12^2)}{12}$

$I_G = \frac{166 *(1.25^2+12^2)}{12}$

Moment of Inertia of bar about axis of pendulum is:

Iphone: We assume that the COM of the phone is in its geometric center. And that given distances(4,6,8) are from axis to Center of the phone.

Moment of Inertia of Phone about an axis parallel to axis of pendulum and passing through its COM is:

$I_G = \frac{M*(14.36^2+7.09^2)}{12}$

$I_G = \frac {174*(14.36^2+7.09^2)}{12}$

Location 1: 4 inches = 2.54*4 = 10.16

$COM = \frac{M_{phone}*x_{phone}+M_{bar}*x_{bar}}{M_{phone}+M_{bar}}$

$COM = \frac{174*10.16+166*15.24}{174+166}$

Moment of Inertia of phone about axis of pendulum is:

Total Moment of Inertia is:

Location 2: 6 inches = 2.54*6 = 15.24

$COM = \frac{M_{phone}*x_{phone}+M_{bar}*x_{bar}}{M_{phone}+M_{bar}}$

$COM = \frac{174*15.24+166*15.24}{174+166}$

Moment of Inertia of phone about axis of pendulum is:

Total Moment of Inertia is:

Location 1: 8 inches = 2.54*8 = 20.32

$COM = \frac{M_{phone}*x_{phone}+M_{bar}*x_{bar}}{M_{phone}+M_{bar}}$

$COM = \frac{174*20.32+166*15.24}{174+166}$

Moment of Inertia of phone about axis of pendulum is:

Total Moment of Inertia is: