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Edit - Consider an homogeneous bar of length $L$ and mass $M$. This bar can rotate on a horizontal plane with no friction around a point $A$. The distance between $A$ and the center $O$ of the bar is $a$. Since we are on a plane, the gravity is not working.

Suppose that this bar is rotating with constant angular velocity $\omega_0$.

Moreover, suppose that this bar hits a mass $m$ in a point $B$. The distance between $B$ and the center $O$ is $b$. Regardless the nature of the collision (elastic or inelastic), I was told that linear momentum is not conserved while angular momentum is.

The explanation I received is the following: during the collisions, an impulsive force arises on the fulcrum in $A$; this force is external and hence the linear momentum is not conserved, while the angular one is conserved since this impulsive force does not produce torque in $A$.

This explanation does not convince me totally.

I post a picture.
enter image description here

I would like to figure out which are the forces that arises during the collisions. Moreover, I would like to know in which points they act.

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  • $\begingroup$ "F1+F2=0" is not a sensible statement as each acts on a different body. It looks like Newton's 3rd law but one cannot add forces acting on different objects to predict anything about the system. Also, you need to include the force on the rod due to the axis at A. As for the torque and axis force. Why do you expect that the force acting on A is in the x-direction? This could be part normal force (contact between touching surfaces) and friction. If the hinge is frictionless the force will point along a line through the center point (cross product = 0). $\endgroup$ – ggcg Jun 5 '18 at 21:45
  • $\begingroup$ is this homework? $\endgroup$ – ggcg Jun 5 '18 at 21:46
  • $\begingroup$ @ggcg no, it isn't. It is just a setup that I don't understand after the mechanics course. $\endgroup$ – the_candyman Jun 5 '18 at 21:56
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    $\begingroup$ If I can find time I may post a complete solution. But in the mean time think of whether you can determine the force at the axis (that the axis exerts on the rod) after the collision and when the rod is rotating. That process may illuminate how to analyze the impulse case. $\endgroup$ – ggcg Jun 6 '18 at 23:33
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    $\begingroup$ Also, one of my comments is not correct. When I say that the impulse at the axis is from below that isn't true (always). It depends on where the small mass hits the rod. For example, if it hits at the CoM then the rod will move down and the axis impulse will be up. Etc. $\endgroup$ – ggcg Jun 6 '18 at 23:41
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I try to solve this problem exactly \begin{align*} &\text{I) The EOM's bevor impact:} \\\\ M\,L\,\ddot{\varphi}&=\cos(\varphi)\left(F_2+M\,g\right)\\ m\,\ddot{y}&=F_1-m\,g \end{align*} \begin{align*} &\text{II) The EOM's after impact:} \\\\ M\,L\,\ddot{\varphi}&=\cos(\varphi)\left(F_2+M\,g-F_c\right)\,,&(1)\\ m\,\ddot{y}&=F_1-m\,g+F_c\,,\quad& (2)\\ \tan(\varphi)&=\frac{y}{b}\,,\quad \text{The impact condition} &(3) \end{align*} So we habe three equations for three unknowns

$\ddot{\varphi}\,,\quad \ddot{y}\,\quad$ and the impact force $F_c$

Results:

\begin{align*} \ddot{y}&=\frac{2\,\dot{y}^2\,y}{1+y^2}\,,\quad y(0)=h_0\,,\quad D(y)(0)=0\\ \ddot{\varphi}&=f(y,\dot{y}\,,F_1\,,F_2\,,b)\\ F_c&=M\,g+\frac{M}{1+y^2}\left(2\,\dot{y}^2\,y\right)+F_1 \end{align*} The EOM's don't depent on the "geometry parameter $a$" !!

I hope it is helpful for you ?

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  • $\begingroup$ I think the rod is on a horizontal table, hence gravity is cancelled by Normal force. $\endgroup$ – ggcg Jun 6 '18 at 19:56
  • $\begingroup$ So you can put g=0 on the equations $\endgroup$ – Eli Jun 6 '18 at 20:35
  • $\begingroup$ It seems that there may be errors in your angular equation. How is phi defined and where are your lever arms? $\endgroup$ – ggcg Jun 6 '18 at 23:46

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