Yo-Yo Question - Rotational motion/angular momentum

[admanrich]

Moment of inertia center of mass...

Homework Statement

Assume: Friction is negligible.
Given g= 9.81 m/s^2. The density of this large yo-yo like solid is uniform throughout. The yo-yo like solid has a mass of 2.8 kg. A cord is wrapped around the stem of the yo-yo like solid and attached to the ceiling. The radius of the stem is 5 m, and the radius of the disk is 6 m. Calculate the moment of inertia about the center of mass (axis of rotation). Answer in units of kg m^2.

Homework Equations

I = 1/2 M R^2

The Attempt at a Solution

I tried doing I = 1/2MR^2, which is wrong, I also thought of summing the I's of the two cylinders and the stem, but I that won't give you the I center of mass, and they don't give you the masses of the individual parts. I also tried MR^2 of the stem radius, also wrong. I also tried doing 1/2M(r^2 + R^2) which is also wrong. There are four parts to this question, but I think i can manage it as long as I get the inertia because I'm going to need it for the rest of the questions. Can anyone help me, I'm getting so frustrated with this problem!

 

[admanrich]

Bump. Anyone?

 

[m2003]

I would suggest approaching this problem by first understanding the concept of moment of inertia and how it is affected by the distribution of mass and the axis of rotation. The moment of inertia is a measure of an object's resistance to rotational motion, and it is affected by both the mass and the distance of the mass from the axis of rotation.

In this case, we have a yo-yo like solid with a stem and a disk, and we are asked to calculate the moment of inertia about the center of mass (axis of rotation). This means that we need to consider the mass and distribution of mass of the entire object, rather than just individual parts.

To calculate the moment of inertia about the center of mass, we can use the parallel axis theorem, which states that the moment of inertia about an axis parallel to the axis passing through the center of mass is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes.

In this case, we can use the moment of inertia for a solid cylinder (I = 1/2MR^2) for both the stem and the disk, and then add the moment of inertia for the stem and the disk using the parallel axis theorem. The moment of inertia for the stem is given by I = 1/2MR^2, where M is the mass of the stem and R is the radius of the stem. Similarly, the moment of inertia for the disk is given by I = 1/2MR^2, where M is the mass of the disk and R is the radius of the disk.

To calculate the moment of inertia about the center of mass, we first need to find the total mass of the yo-yo like solid. We are given that the density of the solid is uniform, so we can use the formula for density (ρ = M/V) to find the mass (M) of the solid. The volume (V) of the solid can be calculated by subtracting the volume of the stem (Vstem = πR^2h) from the volume of the disk (Vdisk = πR^2h), where h is the height of the solid. We are not given the height, but we can calculate it using the Pythagorean theorem (h = √(R^2 + r^2)).

Once we have the mass of the solid, we can use the parallel axis