In each of the following cases, the mass is 7.99 kg and the radius is 7.50 cm.

(a) What is the moment of inertia (kg/m2
) of a solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder?

(b) What is the moment of inertia (kg/m2
) of a solid sphere rotating about an axis tangent to its surface?

The wording in (a) is confusing me. I tried but wasn't even close on both.

Using the parallel axis theorem, we know that I = I_cm + mr^2, where r is the amount of displacement from the symmetry axis.

And we know that for a solid cylinder, I_cm = 1/2(mr^2); Further, for a solid sphere I_cm = 2/5(mr^2)

(a)
==> I = 1/2(mr^2) + mr^2 = 3/2(mr^2) = 3/2(7.99)(.075^2) = 6.7416E-2 kg

(b)
==> I = 2/5(mr^2) + mr^2 = 7/5(mr^2) = 7/5(7.99)(.075^2) = 6.2921E-2 kg

(a) To calculate the moment of inertia of a solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder, we can use the formula:

I = ½ * M * R²

Where:
I = Moment of inertia
M = Mass
R = Radius

Given that the mass (M) is 7.99 kg and the radius (R) is 7.50 cm (which is 0.075 m), we substitute these values into the formula:

I = ½ * 7.99 kg * (0.075 m)²

Calculating this gives us:

I = 0.5 * 7.99 kg * 0.005625 m²
I = 0.0225159375 kg·m²

Therefore, the moment of inertia of the solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder is approximately 0.0225 kg·m².

(b) To calculate the moment of inertia of a solid sphere rotating about an axis tangent to its surface, we can use the formula:

I = ⅖ * M * R²

Where:
I = Moment of inertia
M = Mass
R = Radius

Given that the mass (M) is 7.99 kg and the radius (R) is 7.50 cm (which is 0.075 m), we substitute these values into the formula:

I = ⅖ * 7.99 kg * (0.075 m)²

Calculating this gives us:

I = 0.4 * 7.99 kg * 0.005625 m²
I = 0.017997 kg·m²

Therefore, the moment of inertia of the solid sphere rotating about an axis tangent to its surface is approximately 0.018 kg·m².

To calculate the moment of inertia, you need to use the mass and geometrical properties of the object. Let's solve the problems one by one:

(a) For a solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder, you can calculate the moment of inertia using the formula:

I = (1/2) * m * r^2,

where:
I is the moment of inertia,
m is the mass of the cylinder, and
r is the radius of the cylinder.

In this case, the mass (m) is given as 7.99 kg, and the radius (r) is given as 7.50 cm. However, it's important to convert the radius to meters before plugging the values into the formula. 1 cm is equal to 0.01 m, so the radius in meters would be:

r = 7.50 cm * 0.01 m/cm = 0.075 m.

Now you can substitute the values into the formula:

I = (1/2) * 7.99 kg * (0.075 m)^2.

I = (1/2) * 7.99 kg * (0.075 m)^2
I = 0.2248875 kg*m².

Therefore, the moment of inertia of the solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder is approximately 0.2249 kg*m².

(b) For a solid sphere rotating about an axis tangent to its surface, you can use the formula for the moment of inertia of a solid sphere:

I = (2/5) * m * r^2,

where:
I is the moment of inertia,
m is the mass of the sphere, and
r is the radius of the sphere.

In this case, the mass (m) is given as 7.99 kg, and the radius (r) is given as 7.50 cm. Again, convert the radius to meters:

r = 7.50 cm * 0.01 m/cm = 0.075 m.

Now substitute the values into the formula:

I = (2/5) * 7.99 kg * (0.075 m)^2.

I = (2/5) * 7.99 kg * (0.075 m)^2
I = 0.2248875 kg*m².

Therefore, the moment of inertia of the solid sphere rotating about an axis tangent to its surface is approximately 0.2249 kg*m².

Remember to always pay attention to units and make appropriate conversions to ensure accurate calculations.