Can angular momentum also be defined for a straight line motion?

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Aadit Singh Uppal | Contributor-Level 10

5 months ago

Yes, an object moving in a straight path can also possess angular momentum with respect to a referenc epoint which doesn't lie on it's path. In such cases. the referenc epoint may lie perpendicular distance from the axis. If an object is moving in a straight line (not through the reference point) th

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The position vector of 1kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) m s^{- 1} .$ The magnitude of is angular momentum is $\sqrt[]{x} N m$ where x is________.

Vishal Baghel

$\vec{L} = \vec{r} \times m \vec{v}$

= $(3 \hat{i} - \hat{j}) \times (3 \hat{j} + \hat{k})$

$= 9 \hat{k} + 3 (- \hat{j}) - \hat{i}$

$= - \hat{i} - 3 \hat{j} + 9 \hat{k}$

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Three rods of equal mass having length L, L and √2L are joined to form a triangular frame as shown in the figure. The system is rotating about an axis passing through the vertex of the triangle and perpendicular to the plane of the triangle with an angular velocity ω. Find the angular momentum of the system.


Vishal Baghel

L = Iω
I = (mL²/3) + (mL²/3) + [m (√2L)²/12] + m (L/√2)² = mL² [2/3 + 1/6 + 1/2] = (4mL²/3); L = (4/3)mL²ω

Do the planets in outer space also possess angular momentum?

Aadit Singh Uppal

Yes, how much time these planets take in completing an orbit around the sun hugely depends on their distance from the sun as well as their mass. Since these planets move in a circular orbit, the term angular momentum is used in such cases for computing relevant information.

Why is angular momentum not a scalar quantity?

Aadit Singh Uppal

Angular momentum also depends on how the mass is spread over the body and not just mass only. If you closely look at the formula of angular momentum:

L=r*p

You will observe that it is a cross product of two vector quantities. Hence, the formula won't be practical without direction being involved.

Raj Pandey

Using conservation of Angular momentum along axis of rotation, we can write
Mr²ω = (Mr² + 2mr²)ω? ⇒ ω? = Mω / (M + 2m)