Understanding the Direction of Beam Tilt After Release

[Parallel]

Homework Statement

figure attached:

Uniform beam with mass M,is placed horizontally,pivoted about it's center
on the beam there are two small masses m1 and m2,connected by a spring,the system is in equilibrium,and at t=0 the spring is cut,and m1 and m2 slide on the beam.(their speed is not given),there is friction between the beam and the masses.
given: m2>m1

the problem is to find the direction to which the beam will tilt

The Attempt at a Solution

I thought that I should look at the speed of the center of mass.

motion of m1:

V1 = V'1 - m1g*Y*t (v'1 is the speed upon release(I assumed) Y is coefficient of friction)

same for m2 so Vcm is:
[-V1*m1 + V2*m2] / (M+m1+m2)

I cannot get anything out of this.

thanks for the help.

 

[Parallel]

can someone please try to help..(sorry for bumping but my post,got to the second page).

thanks

 

[P3X-018]

Maybe it's because my english is a little poor, but I find it really hard to see what the problem even is about, and what is happening. Maybe you could give a better figure? Don't you mean a "string" instead of a "spring"? Why is the spring cut?
If you can give a little better description on the situation, I might be able to help.

 

[Parallel]

hmm o.k i'll try to explain it again.

2 masses are connected by a SPRING,the spring is contracted,you can think of it as someone is holding those masses with his hands,bringing them closer,so the system will be in equilibrium..and another person comes and cuts the spring..
and at the same time the person holding the masses let's go(why did he cut it?..I don't know :)..)

so now you have to analyze,in which direction the beam will "go",it can rotate about the pivot,as in the figure.

is it clear now?

 

[P3X-018]

This sounds wired, cutting a spring would make the springs completely useless in this situation. Are you sure it's not cutting a STRING, so that the system begins to oscillate, because of the spring? So there is both a spring and a string. The string get's cut.

 

[Parallel]

there is no oscillation here...and yes the spring is useless here,the problem is to find in which direction the beam will start rotating.

cutting the spring only gives the masses initial speed,and from now on,you just need to focus on the masses,and friction,to find the direction the beam will start to rotate.

 

[P3X-018]

there is no oscillation here...and yes the spring is useless here,the problem is to find in which direction the beam will start rotating.

cutting the spring only gives the masses initial speed,and from now on,you just need to focus on the masses,and friction,to find the direction the beam will start to rotate.

I have a real hard time imagening how that initial speed can come from cutting the spring. It's like saying that they are connected with a sTring and when you cut the string they start moving...

Anyways, let's just assume that the blocks somehow get some inital velocity. Now originally the beam didn't tilt, therefor the center of mass of the 2 blocks must coinside with the center of the beam. Now since there is friction (external force) the center of mass (for the 2 blocks) won't stay at the same place.
Now you need to know what side CM moves towards. Assume that the initial velocity of m1 is $u$ and the initial velocity of m2 is $v$. What is the connection between these two velocities, assuming that the time $\Delta t$ it takes to give them the impulses (and hence give them the initial velocities) is so short that we can neglect the force from friction? What does this mean for the total momentum of the 2 blocks (not including the beam(!)) for this short amount of time?
Now again, consider only the 2 blocks at time $t > \Delta t$. What is the equation of motion for the center of mass for these blocks? What is the direction of the center of mass velocity? What does this mean?

I hope this'll help you.

 

[Parallel]

I wrote the euqation of motion for the CM,in my first post,it looks o.k to me.

[-u*m1 + v*m2] / (m1+m2)

where u and v are found by using the equation for constant accel': u=u0-at

but from this equation I can't deduce anything about the direction.

I don't know how to get rid of the friction coefficient(or maybe I don't need to use it?)

thank you very much for your help :)

 

[P3X-018]

Your first equation wasn't incorrect, it just included the mass of the beam, which you don't need.
You don't need to get rid of the friction coefficient. Read the question in my last reply, there is a connection between the blocks initial velocities, when you find that use it in your equation, [-u*m1 + v*m2] / (m1+m2) that you have, it may be easier just to consider the total momentum for the the 2 blocks, that is only the term, [-u*m1 + v*m2]. Insert the expressions for u and v in that equation. Consider then the questions I asked earlier.
Just ask if this is unclear

 

[Parallel]

is the relation between the velocities is just: 0 = -m1u + m2v,that is total momentum has to be zero(at least for a short time dt,as you suggested)

but if I do it this way,then i get Vcm=0,which is not true.

 

[P3X-018]

is the relation between the velocities is just: 0 = -m1u + m2v,that is total momentum has to be zero(at least for a short time dt,as you suggested)

Indeed, this should give you the relation between the initial velocities.

but if I do it this way,then i get Vcm=0,which is not true.

Ofcourse Vcm = 0 in that short amount of time, but you have to consider the situation after that, where the velocities is time dependend. You know that $u(t) = u_0-\mu m_1 gt$ and $v(t) = v_0-\mu m_2 gt$. Use these velocities to calculate the total momentum of the 2 blocks, and use the relation you have between the initial velocities, you should get an expression for the total momentum that allows you to read of it's direction, and hence the CM velocity's direction.

 

[denverdoc]

It is initially, but as friction kicks in the masses will slow at different rates. Thats when things will pivot if I understand P3-x post above.

 

[Parallel]

thank you so much P3X-018...I finally got it.

thanks again :)