Finding moment of inertia of a grapefruit and metal ring

[badaboom]

Homework Statement

Suppose we rotate a grapefruit and a metal ring
on a dining table with a smooth surface. Grapefruit has a volume of
575 cm3 and density of 0.4 g/cm3, while the ring has mass of 57 g,
length of 2 cm and area of 38.5 cm2. Determine
A. Moment of inertia of each of the objects
B. Which of the two objects will be harder to rotate through the surface if
neglecting friction?

Homework Equations

area of circle= (pi) r²
volume of a sphere = (4(pi)r³) / 3
moment of inertia of a ring = m * r² / 2
moment of inertia of a sphere = (2mr²) / 5

The Attempt at a Solution

We found the mass of the grapefruit to be 230 g. The radio of the grapefruit was 5.16 cm and we used this to calculate the moment of inertia of the grapefruit (2,449.56). After finishing with the grapefruit, I calculated the radius of the ring (3.5cm). How do I calculate the radius of the inner ring?

 

[rock.freak667]

The Attempt at a Solution

We found the mass of the grapefruit to be 230 g. The radio of the grapefruit was 5.16 cm and we used this to calculate the moment of inertia of the grapefruit (2,449.56). After finishing with the grapefruit, I calculated the radius of the ring (3.5cm). How do I calculate the radius of the inner ring?

You don't need the inner ring radius, you just need the radius of the ring which you found.

I_ring=½mr²

 

[badaboom]

then the moment of inertia of the ring is 349.125. Are the other values I got correct?
thank you

 

[rock.freak667]

then the moment of inertia of the ring is 349.125. Are the other values I got correct?
thank you

you need to convert the grams to kilograms and centimeters to meters. Remember the unit of the mass moment of inertia is kgm²