An object is formed by attaching a uniform, thin rod with a mass of mr = 7.42 kg and length L = 4.88 m to a uniform sphere with mass ms = 37.1 kg and radius R = 1.22 m. Note ms = 5mr and L = 4R.
3) What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)
To find the moment of inertia of the object about an axis at the center of mass, we need to calculate the individual moments of inertia of the rod and the sphere, and then use the parallel axis theorem.
First, let's find the moment of inertia of the rod. The moment of inertia of a thin rod rotating about one end is given by the formula:
I_rod = (1/3) * m * L^2,
where m is the mass of the rod and L is its length.
Substituting the given values, we have:
I_rod = (1/3) * (7.42 kg) * (4.88 m)^2.
Simplifying this expression, we get:
I_rod ≈ 38.13 kg·m^2.
Next, let's find the moment of inertia of the sphere. The moment of inertia of a solid sphere rotating about its diameter axis can be calculated using the formula:
I_sphere = (2/5) * m * R^2,
where m is the mass of the sphere and R is its radius.
Substituting the given values, we have:
I_sphere = (2/5) * (37.1 kg) * (1.22 m)^2.
Simplifying this expression, we get:
I_sphere ≈ 16.72 kg·m^2.
Now, we can use the parallel axis theorem to find the moment of inertia of the combined object about an axis at its center of mass. The parallel axis theorem states that the moment of inertia about an axis parallel to and a distance d away from an axis passing through the center of mass is given by:
I_total = I_rod + I_sphere + M * d^2,
where M is the total mass of the object and d is the distance between the two axes.
We are given that the center of mass is located at a point halfway between the center of the sphere and the left edge of the sphere, so the distance between the two axes is L/2. Also, the mass of the entire object is given by M = ms + mr.
Substituting the given values, we have:
I_total = (38.13 kg·m^2) + (16.72 kg·m^2) + (5 * 7.42 kg + 37.1 kg) * (4.88 m / 2)^2.
Simplifying this expression, we get:
I_total ≈ 192.03 kg·m^2.
Therefore, the moment of inertia of the object about an axis at the center of mass is approximately 192.03 kg·m^2.