Dynamics of Two Falling Rods

In summary, the problem involves two rigid bars of equal length and masses, initially upright and separated by a distance d. The bar on the right is given an infinitesimal push and impacts the bar on the left, with no rebound and sliding without friction as they fall. The equations of motion can be found using the conservation of angular momentum and kinematic equations of rigid bodies, but further analysis is needed to determine the reactions at pins. The problem can be divided into three parts: before impact, during impact, and after impact. The equations for before impact involve the conservation of energy and the moment of inertia for the bar on the right. However, more information is needed to find the reactions at pins and the equations of motion for the entire system
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Homework Statement



Two rigid bars of equal length, masses m1 and m2, and inertias I1 and I2 are initially upright and separated by a distance d. The rods can be assumed to be pinned to the ground such that they are free to have angular rotation. The bar on the right (2) is given an infinitesimal push to the left so that it is set into motion and impacts the bar on the left (1). Upon colliding assume that there is no rebound of bar 2 and it slides without friction along bar 1 as the two fall.

Determine all the equations of motion. Find the reactions at pins.

Homework Equations



ma = Ʃ Forces

conservation of angular momentum

d/dt (I ω) = ƩMoments

Kinematic equations of rigid bodies

Vg = Va + ω x r(ag) , etc.

The Attempt at a Solution



One has to analyze the system in 3 parts, before impact, during impact, and after impact. All I know so far is that for before:

ΔPE + ΔKE = 0
mg(L/2)sin(θ) = 1/2 I(2)ω^2
I = m(2)L^2 / 3

Thus,

ω(2)- = √3gsinθ/L in the k direction
and ω(1)- = 0 k because there is no contact yet.

where the "-" indicated before impact.
 

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I'm not sure how to proceed from this point. How do I find the reaction at pins and equations of motion?
 

1. What is the "Dynamics of Two Falling Rods" experiment?

The "Dynamics of Two Falling Rods" experiment is a physics experiment that involves two rods of different lengths and masses being dropped from the same height at the same time. The experiment explores the concept of rotational motion, energy transfer, and the effects of air resistance on falling objects.

2. How does the length and mass of the rods affect the experiment?

The length and mass of the rods affect the experiment by impacting the rotation and speed of the falling rods. The longer and heavier the rod, the slower it will rotate and the faster it will fall due to its higher inertia. This can result in different patterns of motion and energy transfer between the two rods.

3. What is the significance of air resistance in this experiment?

Air resistance plays a crucial role in this experiment as it affects the speed and trajectory of the falling rods. As the rods fall through the air, they experience air resistance which slows them down and causes them to rotate. This adds an additional factor to the dynamics of the experiment and makes it more complex.

4. How does energy transfer occur in this experiment?

Energy transfer occurs in this experiment through various forms such as kinetic energy, potential energy, and rotational energy. As the rods fall, potential energy is converted into kinetic energy and rotational energy as they gain speed and rotate. Energy is also transferred between the two rods through collisions and air resistance.

5. What are some real-life applications of the "Dynamics of Two Falling Rods" experiment?

The "Dynamics of Two Falling Rods" experiment has real-life applications in fields such as engineering, sports, and physics. It can be used to study the impact of air resistance on objects in motion, the principles of rotational motion, and the transfer of energy between objects. This experiment also has practical applications in designing structures and equipment that can withstand forces and resist rotation, such as buildings and bridges.

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