A physical pendulum is covered from a sheet of aluminum. The sheet, which is 1/4 inch thick, is cut down to square having a side length of 30.0 cm. A small hole of diameter 5mm is drilled near one corner, a small rod is passed through the hole, and the pendulum swings about an axis through the center of the rod. If the center of the hole for the rod is 3 cm from the corner,
A) find the moment of inertia of the pendulum relative to the axis of rotation.
B) Assuming a small angle of oscillation, determine the theoretical period of its oscillations.
C) Another hole is drilled along the top to bottom diagonal, a distance of 18cm from the corner nearest the axis of rotation. if the diameter of this hole is 10cm, determine the new period of the pendulum\'s oscillations.
Show all work.
A physical pendulum is covered from a sheet of aluminum. The sheet, which is 1/4 inch thick, is cut down to square having a side length of 30.0 cm. A small hole of diameter 5mm is drilled near one corner, a small rod is passed through the hole, and the pendulum swings about an axis through the center of the rod. If the center of the hole for the rod is 3 cm from the corner,
A) find the moment of inertia of the pendulum relative to the axis of rotation.
B) Assuming a small angle of oscillation, determine the theoretical period of its oscillations.
C) Another hole is drilled along the top to bottom diagonal, a distance of 18cm from the corner nearest the axis of rotation. if the diameter of this hole is 10cm, determine the new period of the pendulum\'s oscillations.
Show all work.
A) find the moment of inertia of the pendulum relative to the axis of rotation.
B) Assuming a small angle of oscillation, determine the theoretical period of its oscillations.
C) Another hole is drilled along the top to bottom diagonal, a distance of 18cm from the corner nearest the axis of rotation. if the diameter of this hole is 10cm, determine the new period of the pendulum\'s oscillations.
Show all work.
density of aluminum = 2.7 gms/cc
volume of the sheet v = 30x30x2.54/4 = 571.5cc
mass of the sheet = 571.5*2.7 gms = 1.543 kg
MI of the plate about an axis through the center and perpendicular to the plane = ML2/6
= 1.543*0.32 /6 = 0.28 kg-m2
the point of rotation is 3 cm from the corner
diagonal length = 42.43 cm
axis of rotation from the center = 21.22-3.0 = 18.22 cm
= 0.1822 m
using parallel axis theorem, MI about the axis of rotation
= I0 + Md2
= 0.28 + 1.543*(0.1822)2 = 0.33 kg-m2
period of the pendulum is given by T = 2(I/mgl)
I = 0.33, m = 1.543 kg, g =9.8 m/s2
distance from COM l = 0.1822m
T = 2*3.14*sqrt(0.33/1.543*9.8*0.18)
= 2.19 s
dia of the new hole = 10cm
mass of the cut plate = *52*(2.54/4)*2.7 gms
= 0.135 kg
mass of the remaining plate = 1.543- 0.135
= 1.408 kg
MI of the cut plate about its central axis
= Mr2/2 = 0.135*0.052/2
= 0.00017 kg-m2
center of this hole is 18cm from the corner
distance from the axis of rotation = 18-3 = 15 cm
MI of the cut plate about the axis of rotation
=0.00017 + 0.135*0.152
=0.0032 kg-m2
MI of the plate after the hole is cut = 0.33-0.0032
= 0.327
COM shits after the hole is cut
let new COM is at x from the corner then
(18*0.135 + x*1.408)/1.543 = 21.22
x= 21.53 cm from the corner
distance of COM = 21.53 -3 = 18.53
period of the hole is cut T = 2*3.14*sqrt(0.327/1.408*9.8*0.185)
=2.25 s
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