# Rate of change of ##L## in a rotating coordinate system

• I
• Kashmir
In summary, we discussed the relationship between the time rates of a vector in an inertial frame and a co-located rotating frame. We found that there exists a relationship between the two rates, with the addition of the angular velocity vector. However, this derivation only applies when the vector is independent of the coordinate system, which is not the case for the angular momentum vector. Therefore, we cannot use the same formula for the time rate of angular momentum.

#### Kashmir

* We've a vector ##\mathbf{A}## lying in space, changing according to some rule.

* We introduce an inertial frame and find ##\left(\frac{d}{d t}
\mathbf{A} \right)_{i n}## in it.

* We also introduce a co located frame rotating with ##\mathbf{\omega}##. In this rotating frame I find ##\left(\frac{d}{d t}
\mathbf{A} \right)_{rot}##

* There exists a relationship between the two time rates as
##\left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf A##

* In all of this derivation it was assumed that the vector ##\mathbf{A}## was independent of the coordinate system. We merely observed the vector in two frames. The vector is independent of the coordinate system.

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* Can we use the above equation on angular momentum vector ##\mathbf{L}## i.e ##\left ( \frac{d\mathbf L}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf L}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf L##?

I think no we can't.

In the derivation of the ##\left ( \frac{d\mathbf A}{dt} \right )_\text{inertial} = \left ( \frac{d\mathbf A}{dt} \right )_\text{rot} + \boldsymbol \omega \times \mathbf A## we assumed that the vector ##\mathbf{A}## was independent of the coordinate system, its lengths and direction in space is independent of the coordinate system.

However for ##\mathbf{L}## that isn't the case.
##\mathbf{L}## has a different length in a stationary frame than in a rotating one. So the derivation doesn't apply.

$$L=r \times p$$
$$\frac{dL}{dt}=\frac{dr}{dt} \times p + r \ \times \frac{dp}{dt}$$
Why don't you use this relation to check the result ?

vanhees71
Of course you can use the formula. The formula refers to the time derivative of polar or axial vector components (not vectors!).

## 1. What is the rate of change of ##L## in a rotating coordinate system?

The rate of change of ##L## in a rotating coordinate system refers to the change in angular momentum with respect to time in a rotating coordinate system. This measures how quickly the angular momentum of an object is changing as it rotates.

## 2. How is the rate of change of ##L## calculated in a rotating coordinate system?

The rate of change of ##L## in a rotating coordinate system can be calculated using the equation ##\frac{dL}{dt} = I\frac{d\omega}{dt}##, where ##I## is the moment of inertia and ##\omega## is the angular velocity.

## 3. What factors affect the rate of change of ##L## in a rotating coordinate system?

The rate of change of ##L## in a rotating coordinate system is affected by the moment of inertia, the angular velocity, and any external torques acting on the system. Changes in any of these factors can result in a change in the rate of change of ##L##.

## 4. How does the rate of change of ##L## in a rotating coordinate system impact the motion of an object?

The rate of change of ##L## in a rotating coordinate system is directly related to the angular acceleration of an object. It determines how quickly the object's angular velocity is changing, which in turn affects its rotational motion.

## 5. Can the rate of change of ##L## be negative in a rotating coordinate system?

Yes, the rate of change of ##L## can be negative in a rotating coordinate system. This means that the angular momentum is decreasing over time, which can happen if there is an external torque acting in the opposite direction of the rotation.