Why is angular momentum sometimes called moment of momentum? I know moment results from a force applied. Momentum is quantity of motion of all particles. Does this momentum acts like force so it makes moment? I understand that angular momentum is the quantity of rotation of a body, but I don't actually understand this.
 A: Ahmed, you are apparently familiar with the term "moment" as shorthand for the moment of a force. I know this usage exists, but it is an abuse of terminology. A much better term is "torque". This abuse of terminology (using "moment" to mean a torque) is a source of confusion, as evidenced by your question.
The generic concept of a "moment" in this context is (from dictionary.com) "the product of a physical quantity and its directed distance from an axis". Other uses beyond the moment of a force include the first moment of mass (mass times center of mass), the second moment of mass (typically called the moment of inertia), and the moment of (linear) momentum (typically called angular momentum). The concept of moments is very generic, and very useful. Statisticians have a similar concept of moments.
The term "moment of momentum" is now used very rarely. From looking at google ngrams and google scholar, "moment of momentum" briefly held sway over "angular momentum" for a short period around 1900.
Even a point particle can have angular momentum with respect to some other point; it is $\vec L = \vec r \times \vec p = \vec r \times (m\,\vec v)$ where $m$ is the mass of the point particle, $r$ is the vector from the point in question to the point particle and $v$ is the time derivative of $r$. That angular momentum is indeed the moment of linear momentum jumps right out in this context.
It's not so clear in the context of a rotating solid body, where the angular momentum with respect to the center of mass is $L = \mathrm{I}\,\vec \omega$. This still is the moment of momentum: $\mathrm I \vec \omega$ is indeed $\int_V \vec x \times (\rho(\vec x) \dot{\vec x})\, d\vec x$ for a rigid body rotating in three dimensional space. There's a lot of stuff hiding in the moment of inertia tensor and in the assumption that the body is rigid that makes this the case.
A: $$\text{Angular moment}(L) = \text{moment of inertia}(I) \cdot\text{Angular velocity}(o)$$
$$O \cdot \text{radius}=\text{tangential velocity}(V)$$
$$L= MR^2\cdot \frac VR=MRV$$
$MV$ IS linear momentum
$MV$ times $r$ mean moment of liner momentum
