Thin rods (H) rotating about an axis

[deadpenguins]

Homework Statement

A rigid body is made of three identical thin rods, each with length L=0.600m, fastened together in the form of a letter H. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?

Homework Equations

Not really sure

The Attempt at a Solution

I am not really sure where to start since all that is given is L. I think that each rod must have its moment of inertia calculated separately about the axis using the perpendicular and parallel axis theorems, but we have not reviewed these subjects extensively and I am not sure of how I would go about doing this.

 

[LowlyPion]

Yes you are sensing the right solution.

As I read it the plane of the H rotates about one leg. Since one leg is already the pivot it can be disregarded. (r=0 after all.)

So that means then you have two more moments to account for.

Do you know the moment of inertia for a rod pivoted about 1 end? It's easily looked up or determined.

Then you just need to figure the moment of the 3rd rod. But since it is just falling and not rotating won't it simply act like a point mass at the end of a rod?

After you have the moment of inertia you know how to figure the potential energy to rotational kinetic energy don't you?

 

[nlightner]

First, it is important to understand the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass, shape, and distribution of mass of the object. In this case, we can assume that the thin rods are uniform in mass and shape.

To solve this problem, we can use the principle of conservation of energy. When the body is released from rest, it has only potential energy. As it falls, this potential energy is converted into kinetic energy. At the bottom of the fall, all the potential energy is converted into kinetic energy, and the body will have a maximum velocity.

Using this concept, we can write the equation:

PE = KE

Where PE is the potential energy and KE is the kinetic energy.

The potential energy of the body is given by PE = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the body.

The kinetic energy of the body can be written as KE = 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular velocity.

Since the body is initially at rest, the initial potential energy is zero. At the bottom of the fall, the potential energy is converted into kinetic energy, giving us the equation:

mgh = 1/2 * I * ω^2

We know the mass of the body (since it is made of three identical rods), the acceleration due to gravity, and the height of the fall (which is the length of the horizontal leg of the H, L).

We can also calculate the moment of inertia of the body. Since the body is made of three identical rods, we can use the parallel axis theorem to calculate the moment of inertia about the axis of rotation. The moment of inertia of a thin rod about an axis perpendicular to its length is 1/12 * m * L^2. Using the parallel axis theorem, we can add the moments of inertia of the three rods to get the moment of inertia of the entire body.

Putting all of this together, we can rearrange the equation to solve for the angular velocity:

ω = √(2 * mgh / I)

Substituting the values we know, we get:

ω = √(2 * 3m * 9.8m/s^2 * 0.6m / (1/