# Solve Rigid Body Motion: Find Angle Between Axis & Invariable Line

• schuksj
In summary, the problem at hand involves a rigid body rotating freely about a fixed point with an axis of symmetry. Using Euler equations and the angle alpha between the axis of symmetry and the instantaneous axis of rotation, the angle between the axis of rotation (omega) and the invariable line (L) can be calculated using the formula tan^-1((Is-I)*tan(alpha)/(Is+ I*tan(alpha)^2)). It is also mentioned that there may be another angle, beta, between the symmetry axis and the angular momentum, which could be used to find the magnitude of L and the cross product L x omega to ultimately find the angle theta.
schuksj
I am having problems figuring out this problem. A rigid body having an axis of symmetry rotates freely about a fixed point under no torques. If alpha is the angle nbetween the axis of symmetry and hte instantous axis of rotaiton, show that he angle between the axis of rotaiton(omega) and the invariable line(L) is tan^-1((Is-I)*tan(alpha)/(Is+ I*tan(alpha)^2)).

Using Euler equations:

omega y'=omega*sin(alpha)
omega z'=omega*cos(alpha)
Ly'=I*omega*sin(alpha)
Lz'=Is*omega*cos(alpha)

and Ly'/Lz'=tan(theta)=I/Is=tan(alpha)

Would the cross product of omega and L give me the angle between the axis of rotation and the invariable line? I am not sure how to begin this problem or if this is the right start.

What exactly are I and Is? I see the possibility of another angle in the problem. If alpha is the angle between the symmetry axis and omega (it looks like you have used z as the symmetry axis), then there is some angle beta between the symmetry axis and the angular momentum. That would make

Ly'=L*sin(beta)
Lz'=L*cos(beta)

The angle theta would be (alpha - beta). The components of L are related to the components of ω by the moment of inertial tensor. If you know the components of L in terms of ω then you can find the magnitude of L and cross product L x ω as you suggested to find sin(theta) and theta.

## 1. What is rigid body motion?

Rigid body motion is the movement of an object where all points of the object move in a parallel manner, maintaining their relative distances and orientations to each other.

## 2. How is the angle between axis and invariable line calculated in rigid body motion?

The angle between axis and invariable line is calculated using the cross product of the two vectors. The magnitude of the cross product is divided by the product of the magnitudes of the two vectors, and the inverse cosine of this value gives the angle between the two vectors.

## 3. Why is it important to solve for the angle between axis and invariable line in rigid body motion?

Knowing the angle between axis and invariable line is important in understanding the orientation and movement of a rigid body. It can also be used to calculate the angular velocity and acceleration of the body.

## 4. What are the applications of solving for the angle between axis and invariable line in rigid body motion?

This calculation is commonly used in engineering and physics, especially in fields such as robotics, mechanics, and aerospace. It can also be applied in computer graphics and animation to accurately represent the movement of objects.

## 5. Are there any limitations to solving for the angle between axis and invariable line in rigid body motion?

This calculation assumes that the rigid body is moving in a two-dimensional plane. It may not be accurate for bodies that are moving in three-dimensional space. Additionally, it assumes that the body is rigid and does not take into account any deformations or flexibility.

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