# Thin rods (H) rotating about an axis

In summary, a rigid body consisting of three identical thin rods, each with length L=0.600m and arranged in an H shape, is allowed to fall from rest with the plane of the H in a horizontal position. The body is free to rotate about an axis running along the length of one of the legs of the H. To find the angular speed of the body when the plane of the H is vertical, the moment of inertia of each rod must be calculated separately, using the perpendicular and parallel axis theorems. The moment of inertia for a rod pivoted about one end can be looked up or determined, and the third rod will act like a point mass at the end of a rod. The potential energy can then be

## 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?

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.

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?

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/

## 1. What is the definition of rotational motion?

Rotational motion is the movement of an object around an axis or center point, where each point on the object moves in a circular path with the same angular velocity.

## 2. How is angular velocity calculated for thin rods rotating about an axis?

The angular velocity of a thin rod rotating about an axis is calculated by dividing the angular displacement (change in angle) by the time taken for that displacement to occur. It is represented by the symbol omega (ω) and is measured in radians per second.

## 3. What is the moment of inertia for a thin rod rotating about an axis?

The moment of inertia for a thin rod rotating about an axis is a measure of the object's resistance to rotational motion. It is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation.

## 4. How does the rotational energy of a thin rod change as it rotates about an axis?

The rotational energy of a thin rod increases as it rotates about an axis, due to the work done by the applied torque. The rotational energy is directly proportional to the moment of inertia and the square of the angular velocity.

## 5. What factors affect the period of rotation for a thin rod about an axis?

The period of rotation for a thin rod about an axis is affected by the length and mass of the rod, as well as the distance from the axis of rotation. It is also influenced by any external forces acting on the rod, such as friction or air resistance.

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