# Total angular momentum of a translating and rotating pancake

• I
• Rikudo
In summary: The origin and CM frame of reference are chosen as the reference frames. Then, the angular momentum is also in the CM frame of reference.
Rikudo
I have read Classical Mechanics book by David Morin, and there are some parts that I do not understand from its derivation.

Note :
## V## and ##v## is respectively the velocity of CM and a particle of the body relative to the fixed origin , while ##v'## is velocity of the particle relative to CM
From this, we can get : $$v = V + v'$$

In the 3rd step, It is written that ## \int r' \times V\, dm ## vanised because the position of CM in CM frame is zero.
But, isn't the position of CM should be calculated relative to the origin? (Since we are calculating the angular momentum relative to the origin, not CM)

Well, by definition if you are in the CM frame of reference the CM is chosen as the origin of your reference frame. Then also the reference point of the angular momentum is the CM.

Rikudo said:
and is respectively the velocity of CM and a particle of the body relative to the fixed origin , while is velocity of the particle relative to CM
it is a very bad phrase
there can not be a velocity of one point relative other point
there can be velocity of a point relative a frame

vanhees71 and ergospherical
wrobel said:
it is a very bad phrase
there can not be a velocity of one point relative other point
there can be velocity of a point relative a frame
Ah...I see.Thanks for telling me!
vanhees71 said:
Well, by definition if you are in the CM frame of reference the CM is chosen as the origin of your reference frame. Then also the reference point of the angular momentum is the CM.
Alright. So, ##\int V \times r' \,dm ## use CM as its origin point, i suppose?

Figure 8.4 says it all imho. ##r'## is the position vector in the frame of reference of CM, so when you calculate those terms (the ones that vanish) it is like you are in the frame of reference of CM, regardless if the whole calculation is done in the frame of reference of another point that is taken as origin , different than CM.
so in a nutshell it is
$$\int \vec{r'} dm=\vec{0}$$
$$\int \vec{r} dm=M\vec{R}$$

Last edited:
vanhees71
Ok I might not have explained it very well but here is a better (i think) explanation.

From figure 8.4 it is ##\vec{r'}=\vec{r}-\vec{R}## hence integrating both sides we get $$\int\vec{r'}dm=\int\vec{r}dm-\int\vec{R}dm=M\vec{R}-\vec{R}M=0$$

vanhees71
It's nice to meet you again!
Delta2 said:
Ok I might not have explained it very well but here is a better (i think) explanation.

From figure 8.4 it is ##\vec{r'}=\vec{r}-\vec{R}## hence integrating both sides we get $$\int\vec{r'}dm=\int\vec{r}dm-\int\vec{R}dm=M\vec{R}-\vec{R}M=0$$
But, isn't ##\int V \times r' \,dm = V\int r'\, sin\alpha ## since it is a cross product? (With ##\alpha## is the angle between vector## V ##and vector ##r'##).
Then, to find ##\int r'\, sin\alpha ##, we need to multiple both sides by ##sin \alpha## before integrating it.
$$\int\vec{r'} sin (\alpha)dm=\int\vec{r}sin (\alpha) dm-\int\vec{R}sin (\alpha) dm$$

No , you confuse $$\int\vec{V}\times\vec{r'} dm=\vec{V}\times\int\vec{r'}dm$$ with $$\int|\vec{V}\times\vec{r'}|dm=|\vec{V}|\int|\vec{r'}|\sin a dm$$, the first above gives a vector as result(where in the integration opposite vectors cancel out), while the second gives a number greater than zero (where in the integration opposite vectors don't cancel out, because we take their magnitude).

vanhees71
Delta2 said:
No , you confuse $$\int\vec{V}\times\vec{r'} dm=\vec{V}\times\int\vec{r'}dm$$ with $$\int|\vec{V}\times\vec{r'}|dm=|\vec{V}|\int|\vec{r'}|\sin a dm$$, the first above gives a vector as result(where in the integration opposite vectors cancel out), while the second gives a number greater than zero (where in the integration opposite vectors don't cancel out, because we take their magnitude).
OK

Last edited:
Delta2 said:
it is like you are in the frame of reference of CM, regardless if the whole calculation is done in the frame of reference of another point that is taken as origin
So, we use the origin and CM frame of reference. but, how is it possible to use two different frames at the same time?

## 1. What is total angular momentum?

Total angular momentum is the measure of the rotational motion of an object around a fixed point. It takes into account both the mass and distribution of the object's mass, as well as its rotational speed.

## 2. How is total angular momentum calculated?

Total angular momentum is calculated by multiplying the moment of inertia (a measure of an object's resistance to rotational motion) by the angular velocity (the rate at which an object rotates around a fixed point).

## 3. What is the significance of a translating and rotating pancake?

A translating and rotating pancake is a common example used in physics to demonstrate the concept of total angular momentum. It is significant because it shows how an object can have both linear and rotational motion, and how these two types of motion affect each other.

## 4. How does the shape of the pancake affect its total angular momentum?

The shape of the pancake affects its total angular momentum by changing its moment of inertia. A pancake with a larger radius will have a greater moment of inertia, and therefore a greater total angular momentum, compared to a pancake with a smaller radius.

## 5. Can total angular momentum be conserved in a system?

Yes, total angular momentum can be conserved in a system as long as there are no external torques acting on the system. This means that the total angular momentum of the system will remain constant, even if individual objects within the system are rotating or translating.

• Mechanics
Replies
1
Views
400
• Mechanics
Replies
3
Views
1K
• Mechanics
Replies
14
Views
2K
• Mechanics
Replies
17
Views
2K
• Mechanics
Replies
3
Views
895
• Mechanics
Replies
13
Views
2K
• Mechanics
Replies
7
Views
2K
• Mechanics
Replies
19
Views
1K
• Mechanics
Replies
42
Views
4K
• Mechanics
Replies
2
Views
1K