# Does linear motion have angular momentum?

I'm wondering whether an object that is moving in a linear fashion can be considered as having angular momentum. I am thinking that all that would be needed is to pick a reference point that is not on the axis that the object is traveling on.

So for example, if a bullet is flying in a straight line from west to east at 5 feet above the floor. If a pick a point in the middle of the floor, I can say that at the starting point the bullet is at a 170 degree angle, for example, from that reference point. A moment later, it could be directly above--90 degrees. And so on.

Although this makes sense to me, I am working on a problem in which I am trying to use this and it isn't working. I would like to understand why.

The specific problem can be found here, and is :

A small 4.50kg brick is released from rest 2.00m above a horizontal seesaw and 1.6m to the right of the fulcrum at the middle of the seesaw.
A) Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released.
B) Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.

So I would find the original angle to be about 1.001 radians ($tan^{-1} \frac {2.5}{1.6}$).

The time of the fall: $$\frac{2d}{g}=t^2=\frac{2*2m}{9.8 \frac{m}{s^2}}$$

$$t=0.6388s$$

$$\omega=\frac{1.001}{0.6388s}=1.566s^{-1}$$

And using $L=I\omega$ and treating the square as a point mass, we get $$L=4.5kg*1.6^2m^2*1.566s^{-1}=149kgm^2s^{-1}$$

• Isn't linear motion impossible on Earth unless it's strictly vertical (directly up or directly down)? Just saying. Commented Aug 5, 2016 at 5:26
• Your problem is that you're assuming your angular velocity is constant. The angular momentum will be constant around any reference point that you choose, but the angular velocity and radial velocity will both be changing. You could try finding the equations of motion first in Cartesian coordinates, then transform to polar coordinates. Commented Aug 5, 2016 at 5:47
• @Jo.P, how would knowing the angular momentum at one point help you? If you are considering the bullet to be falling, then gravity will be pulling it in a direction that is not in-line with your chosen point, which means it can be considered a torque, changing the angular momentum. Commented Aug 5, 2016 at 5:53
• @Ricky we ignore such effects here. OT: You can calculate the angular momentum in two ways, either take the component of radius perpendicular to the velocity or take the component of velocity being perpendicular to the line joining the bullet to the reference point. Doing it the first way, it is obvious that the perpendicular distance between the point and the reference point (distance of a point from a line) remains constant and you the bullet moves in uniform motion, hence the angular momentum remains constant. Commented Aug 5, 2016 at 6:17
• Don't worry about the angle. Let the radius be the distance of closest approach. Commented Sep 8, 2016 at 17:20

The (orbital) angular momentum of a body is defined by the cross product of the position vector and the linear momentum L = r x p. Thus a body in linear motion has an angular momentum which depends on the reference point of the position vector. The absolute value of the angular momentum is L = r.p. sin(phi), where phi is the angle between the position and the linear momentum vector.

Thus the answers to the two questions are easy:

A) The angular momentum is L = 0 because at t=0 v=0.

B) The velocity of the brick after a free fall of d = 2.0m is (with v = g·t, g = 9.81m/s^2, t = sqrt(2d/g) = 0.6386s) v = g·t = sqrt(2gd) = 6.26m/s, the linear momentum is p = m·v = 28.19 kg·m/s

Because phi = 0, sin(pi/2) = 1, r = 1.60m just before impact, the absolute value of the angular momentum of the brick just before impact on the seesaw is:

L = 1.60m·28.19kg·m/s = 45.10kg·m^2/s

its direction is perpendicular to the plane of the figure pointing to the back of the figure.

The calculation of the angular momentum with the formula L = I·omega = m·d^2·omega with I = 11.52 kg·m^2 and omega = v/r = 3.915/s gives the same result. You have to take the angular velocity at the impact point. At the starting point the angular velocity and angular momentum is 0.