Vertical speed of a point attached by two rods to a rotating hinge.

[chingel]

Homework Statement

Not exactly homework, but an interesting problem I found for which I have some questions about the answer.

A rod of length r is rotating around a point O with constant angular velocity w. Distance r away to the right from the point O is a rail. The end of the rod with length r is attached at point P to a rod with length L. Rod L attaches to a point Q on the rail. Point Q is below point O and P at all times The point Q is free to move up and down as the rod r rotates. Calculate the maximum and minimum speed of point Q, maximum speed is max magnitude up, min is max magnitude down. L > 2r

Here is the drawing:

Homework Equations

The Attempt at a Solution

The solution says that because the point P has a constant acceleration that is radial towards point O and since the projection of this acceleration to the rod is equal to the projection of the acceleration of point Q to the rod, we can find the situations where the projection of acceleration of point P to the rod is zero and find maximum and minimum velocities from that. These happen when the two rods are at right angles. Minimum v = -wr. For maximum v it uses the fact that a cyclic quadrilateral forms and halving it you get two triangles QPO and QOR (R is on the rail where there is a right angle between the rail and line connecting point R to O) of which one side is mutual, one angle is 90 and one angle is r on both triangles so they are congruent and max v is the projection of the velocity of point P to the rail, [itex]_{max} v = \frac{wr}{cos(\angle PQR)} = \frac{wr}{cos(2arctan(r/L))}[/itex]

My question is that I am not quite sure the situations where the rods are at right angles would represent maximum and minimum velocities. Yes the projection of the acceleration to the rod would have a component still accelerating it upwards when projected to the rod, but it also has velocity to the left and maybe if it moves to the left with more velocity it makes the point Q move faster up, since moving P to the left would make Q go up when trying to achieve v max.

I put together an equation for point Q depending on time: [itex]Q(t)=\sqrt{L^2-(r+r*cos(w*t))^2}+r*sin(w*t)[/itex]
I assumed that zero angle is when the rod r points away from the rail and it starts moving clockwise.

I put it into GeoGebra, took the first and second derivatives of it, put the roots of the second derivative (time when velocity is at an extrema) into the first derivative to find speed and found out that the beforehand solution and this equation have only a small error when r is small compared to L, but if r is 1 and L is 2.1 and w is 1, then they differ by 0.3; if I make L=4 then they differ by ~0.002

What do you think, is the solution accurate or not and why or why not?

 

[anathema]

Thank you for sharing this interesting problem and your attempt at a solution. it is always exciting to see individuals actively engaging in problem-solving and critical thinking.

After reviewing your question and the given solution, I believe that the solution provided is accurate. The key concept to understand here is the relationship between acceleration and velocity. As you correctly pointed out, the projection of the acceleration onto the rod determines the maximum and minimum velocities of point Q. This is because acceleration is the rate of change of velocity, and the direction of acceleration also determines the direction of velocity.

In this problem, the acceleration of point P is always directed towards point O, which means that the velocity of point P will always have a component towards the left. This velocity is what determines the maximum and minimum velocities of point Q. When the rods are at right angles, the projection of this velocity onto the rail will be zero, resulting in a minimum velocity for point Q. On the other hand, when the rods are in a certain configuration (as described in the solution), the projection of this velocity onto the rail will be at its maximum, resulting in a maximum velocity for point Q.

Your approach of using a mathematical equation and GeoGebra to verify the solution is commendable. However, it is also essential to understand the underlying concept and how it relates to the given problem. In this case, the concept of acceleration and velocity is crucial.

I hope this helps clarify your doubts and provides a better understanding of the solution. Keep up the good work and continue to engage in scientific problem-solving.