Rotational Inertia Seesaw Picot

[premedonna89]

CORRECTION ON TITLE: it should be "Rotational Inertia Seesaw Pivot"

Homework Statement

Alice (20kg) and Bonnie (25kg) sit on seesaw (10m long, 12 kg mass). Pivot of seesaw is in the middle. In order for the seesaw to stay level we determined (in the previous step of the multi-part problem) that Bonnie has to sit 4 m away from the pivot/center.

What is the rotational inertia of this system about the seesaw pivot?

Homework Equations

Inet = Ia + Ig + Ib

I = mass x radius^2^

Ig = 1/12 x mass x length^2^

The Attempt at a Solution

I actually have the answer already (this is a practice exam). But I don't understand how he got it.

Inet = Ia + Ig + Ib

Inet = mass,a x radius,a^2^ + 1/12 x mass x length^2^ + mass,b x radius,b^2^

Inet = 500 + 100 + 25

Inet = 625 kg/m^2^

I don't understand how:

mass,b x radius,b^2^ = 25

Shouldn't it be: 25 x 4^2^ = 400 ?

I imagine radius, b = 1, therefore 1 square is just 1. Since Bonnie is sitting 4 meters away from the center, and the edge is 5 meters away from the center, so 5 minus 4 = 1. Then multiply that by 25 kg to get 25. However, I can't find anywhere in the textbook section for inertia that says to measure radius from the end of the seesaw (?) rather than from the center/pivot.

Thank you!

 

[clamtrox]

You can of course calculate the moment of inertia with respect to any point (and the result is different for different points) but you're right about this making no sense at all. The distance in your formulas is indeed the distance from the point where you wish to calculate I.

 

[kerimek]

Hello,

Thank you for sharing your attempt at solving this problem. Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to rotational motion. It depends on both the mass of the object and its distribution of mass around the axis of rotation.

In this case, the seesaw is a long thin object with most of its mass concentrated at the ends where Alice and Bonnie are sitting. The rotational inertia of this system is calculated by adding the individual moments of inertia of each object, Alice (Ia), Bonnie (Ib), and the seesaw itself (Ig).

The moment of inertia of a point mass (I = mr^2) is calculated using the distance from the axis of rotation (r) and the mass (m) of the object. In this case, the distance from the seesaw pivot to Alice (radius,a) is 5 meters, and the distance from the pivot to Bonnie (radius,b) is 4 meters. So, the moment of inertia for Alice (Ia) is calculated as:

Ia = 20kg x (5m)^2 = 500 kgm^2

Similarly, the moment of inertia for Bonnie (Ib) is calculated as:

Ib = 25kg x (4m)^2 = 400 kgm^2

Lastly, the moment of inertia for the seesaw itself (Ig) is calculated using the parallel axis theorem, which takes into account the distribution of mass along the length of the object. This is where the 1/12 factor comes in. The equation is Ig = 1/12 x mass x length^2. In this case, the mass of the seesaw is 12 kg and the length is 10m, so Ig = 100 kgm^2.

Now, we can add all three moments of inertia together to get the total rotational inertia of the system:

Inet = Ia + Ig + Ib = 500 kgm^2 + 100 kgm^2 + 400 kgm^2 = 1000 kgm^2

However, this is the rotational inertia of the system about the center of mass, which in this case is the seesaw pivot. To find the rotational inertia about the seesaw pivot, we need to use the parallel axis theorem again, but this time we will use the distance from the center of mass to the pivot (which is half the length of