# Why does angular momentum have sine in its formula?

Angular momentum is defined as the cross product of the radius vector and the (linear)momentum vector. Its magnitude is given by the formula: r * m * v sinθ .

Angular Momentum comes into picture only in rotational motion. So, if we have an object of mass 'm' rotating in a circle of radius 'r' and moving with tangential velocity 'v', then the angular momentum would simply be m * v * r [since sin90(degrees) = 1].

I have the following doubts:

1) Since angular momentum comes into play only during rotational motion, and since the tangential velocity vector is always perpendicular to the radius vector, there cannot be any situation where θ can take values other than 90(degrees), right? If it does so, the motion cannot be rotational, correct?

2) Then, why is angular momentum not defined as simply "m * v(tangential) * r" but defined as the cross product of 2 vectors (since even the direction of angular momentum has no physical significance) ?

• And the angular momentum of your motion about an axis in the plane of motion would be? That is why you need the sine(). – Jon Custer Mar 2 '18 at 19:17
• Putting the sine in means that you can find the angular momentum about a given point of a body with any velocity. It's $\vec{L} = m \vec{r} \times \vec{v}$ No need for any pre-determined circle or radius. Useful, for example, when studying planets' elliptical orbits. And the direction of $\vec{L}$ $does$ have physical significance: it's at right angles to the plane of motion. – Philip Wood Mar 2 '18 at 19:31
• But can you please tell me what good is saying that it's at right angles? – Gokulakrishnan Shankar Mar 2 '18 at 19:40