It is known that for a mass (m) that moves with velocity (v) the linear momentum is (m)x(v). Now, assume we have a rod of mass (m) and radius (R) that

rotates with constant angular velocity (w). Also, isolate only half of the cycle (rotation for 180 degrees). What is the linear momentum of the rod on y-axis? Is it a function of sin(th) where (th) is the angle of rotation? (assume x,y axes and the rod begins rotation when it is on y axis and finishes on the other end of axis y). Thank you.

To find the linear momentum of the rod on the y-axis, we need to consider the rotational motion and relate it to linear motion.

Firstly, let's consider the rotational motion of the rod. The linear velocity (v) of any point on the rod can be given by multiplying the angular velocity (w) by the perpendicular distance of that point from the axis of rotation. In this case, since the rod rotates around its center, this distance is simply the radius (R) of the rod:

v = w * R

Next, let's consider the half-cycle of rotation, from the y-axis to the other end of the y-axis. During this half cycle, the angle of rotation (θ) varies from 0 to π radians (180 degrees).

Now, the linear momentum (p) of any point on the rod is given by the product of its mass (m) and linear velocity (v). Since we are interested in the linear momentum on the y-axis, we need to find the linear momentum of the point which is at the furthest distance from the axis of rotation. This point can be found using trigonometry as the point on the other end of the y-axis:

y = R * sin(θ)

The linear momentum (p) of this point is given by:
p = m * v = m * (w * R)

Substituting the expression for v, we get:
p = m * (w * R) = m * (w * R * sin(θ))

So, the linear momentum on the y-axis is indeed a function of sin(θ), where θ is the angle of rotation during the half-cycle.

Please note that the above explanation assumes that the rod has a uniform mass distribution along its length and that the axis of rotation is at the center of mass of the rod.