A rod with a diameter of 1.8 cm is used to make the letter A in an 18 by 12 cm space. If the number were to spin about the left side of the space calculate the moment of inertia. The rod has a linear mass density of 4.6 kg/m
First compute the moment of inertia about a vertical axis of symmetry. Then apply the parallel axis theorem. The answer will depend upon where the "horizontal bar" of the letter "A" is placed, and whether the 18 cm or the 12 cm is the vertical dimension.
To calculate the moment of inertia of the spinning rod, we need to know the mass of the rod and the radius of gyration. The radius of gyration can be calculated using the formula:
k = √(I/m)
Where:
k = Radius of gyration
I = Moment of inertia
m = Mass
First, let's calculate the mass of the rod. We can find the volume of the rod using the formula for the volume of a cylinder:
V = πr²h
Where:
V = Volume
r = Radius (diameter/2)
h = Height (length of the rod)
Given that the diameter of the rod is 1.8 cm, the radius would be 0.9 cm (0.009 m). The height can be calculated as the width of the space, which is 18 cm (0.18 m). Plugging these values into the formula:
V = π(0.009)²(0.18) ≈ 0.000115084 m³
Next, the mass of the rod can be calculated using the linear mass density:
m = linear mass density × height
Given that the linear mass density is 4.6 kg/m and the height is 0.18 m:
m = 4.6 × 0.18 = 0.828 kg
Now we can calculate the radius of gyration using the formula mentioned earlier:
k = √(I/m)
Rearranging the formula, we get:
I = k² × m
To find the radius of gyration k, we need to know the moment of inertia. Let's assume the rod is a solid cylinder. The moment of inertia of a solid cylinder about an axis perpendicular to its length is given by the formula:
I = (1/4) × m × r²
Plugging in the values:
I = (1/4) × 0.828 × (0.009)² ≈ 0.000002815 kg·m²
Now, let's calculate the radius of gyration k:
k = √(0.000002815/0.828) ≈ 0.00337 m
Finally, to calculate the moment of inertia when the number spins about the left side of the space, we can use the formula:
I_spinning = m × (k + h)²
Plugging in the values:
I_spinning = 0.828 × (0.00337 + 0.18)² ≈ 0.0005603 kg·m²
Therefore, the moment of inertia when the number spins about the left side of the space is approximately 0.0005603 kg·m².