Answer:
1) I_ pendulum = 2.3159 kg m²
, 2) I_pendulum = 24.683 kg m²
Explanation:
In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum
Sphere
They indicate the density of the sphere roh = 37800 kg / m³ and its radius
r = 5 cm = 0.05 m
we use the definition of density
ρ = M / V
M = ρ V
the volume of a sphere is
V = 4/3 π r³
we substitute
M = ρ 4/3 π r³
we calculate
M = 37800 4/3 π 0.05³
M = 19,792 kg
Bar
the density is ρ = 32800 kg / m³ and its dimensions are 1 m,
0.8 cm = 0.0008 m and 4cm = 0.04 m
The volume of the bar is
V = l w h
m = ρ l w h
we calculate
m = 32800 (1 0.008 0.04)
m = 10.496 kg
Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is
r_cm = 1 / M (M r_sphere + m r_bar)
M = 19,792 + 10,496 = 30,288 kg
r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))
r_cm = 1 / 30,288 (5,248 + 20,7816)
r_cm = 0.859 m
This is the center of mass of the pendulum.
1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:
Sphere I = 2/5 M r2
Bar I = 1/12 m L2
parallel axes theorem
I = I_cm + m D²
where m is the mass of the body and D is the distance from the body to the axis of rotation
Sphere
m = 19,792 ka
the distance D is
D = 1.05 -0.85
D = 0.2 m
we calculate
I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168
I_sphere = 0.811472 kg m²
Bar
m = 10.496 kg
distance D
D = 0.85 - 0.5
D = 0.35 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576
I_bar = 1.5044 kg m²
The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts
I_pendulum = I_sphere + I_bar
I_pendulum = 0.811472 +1.5044
I_ pendulum = 2.3159 kg m²
this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m
2) The moment is requested with respect to the pivot point at r = 0 m
Sphere
D = 1.05 m
I_sphere = 2/5 M r2 + M D2
I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82
I_sphere = 21.84 kg m²
Bar
D = 0.5 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624
I_bar = 2,84266 kg m 2
The pendulum moment of inertia is
I_pendulum = 21.84 +2.843
I_pendulum = 24.683 kg m²
This moment of inertia is about the turning point at r = 0 m
