Marion and Thornton Dynamics Problems

[MARX]

Homework Statement

PROBLEM 7-10[/B]
Two blocks, each of mass M, are connected by an extensionless, uniform string of length l. One block is placed on a smooth horizontal surface, and the other block hangs over the side, the string passing over a frictionless pulley. Describe the motion of the system when the string has a mass m.

Homework Equations

L = T -U

The Attempt at a Solution

NOT HOMEWORK I AM SELF LEARNING
I did solve this question but I have a sign ambiguity in U
U= -Mgy- ∫ (m/l)dy*g*y
in the solutions second term in Lagrangian is negative
mine positive
where did I go wrong isit because U = -W and my integral represents work (∫ F.dx) so I have to flip sign of U
just want to make sure
THanks

 

[BvU]

Hello again
Note that y upward is positive. So y itself is negative. If they wrote ##U = Mgy - {m\over l}yg{y\over 2} \ ## it might be a bit clearer.
In your case it might have to do with the integration boundaries ?

 

[zwierz]

The integrals are not needed. Express the potential energy in terms of center of mass's height

 

[MARX]

They did write it the way you said
is the 1/2 factor because the center of mass of the y piece below is at y/2
ok let's define y>0, U = -Mgy+(m/l)*(y)*g*(-y/2) correct?
how about my way-I wanted to integrate F to get U (I am in the habit of calculating things from scratch-to an extent-without memorizing)
my bounds are 0 to y as that dy piece of the rope descended y below x-axis no?

 

[MARX]

They did write it the way you said
is the 1/2 factor because the center of mass of the y piece below is at y/2
ok let's define y>0, U = -Mgy+(m/l)*(y)*g*(-y/2) correct?
how about my way-I wanted to integrate F to get U (I am in the habit of calculating things from scratch-to an extent-without memorizing)
my bounds are 0 to y as that dy piece of the rope descended y below x-axis no?

Hello again
Note that y upward is positive. So y itself is negative. If they wrote ##U = Mgy - {m\over l}yg{y\over 2} \ ## it might be a bit clearer.
In your case it might have to do with the integration boundaries ?

Helloooo to you as well
and thanks so much for your help

 

[MARX]

CORRECTION:
They did write it the way you said
is the 1/2 factor because the center of mass of the y piece below is at y/2
ok let's define y>0, U = -Mgy+(m/l)*(y)*g*(-y/2) correct?
how about my way-I wanted to integrate F to get U (I am in the habit of calculating things from scratch-to an extent-without memorizing)
my bounds are 0 to -y as that dy piece of the rope descended y BELOW x-axis no?

 

[vela]

CORRECTION:
They did write it the way you said. Is the 1/2 factor because the center of mass of the y piece below is at y/2?

Yes.

ok, let's define y>0, U = -Mgy+(m/l)*(y)*g*(-y/2) correct?

Yes.

how about my way-I wanted to integrate F to get U (I am in the habit of calculating things from scratch-to an extent-without memorizing)
my bounds are 0 to -y as that dy piece of the rope descended y BELOW x-axis no?

I wouldn't do it that way. Say you define the positive direction to be downward. The top of the string would be at y=0 and the bottom at y=L, where L>0 is the length of the string hanging down. The mass of an infinitesimal piece of the string would be ##dm = (m/l)dy##. If you integrate from y=0 to y=L, then dy>0 and dm>0, which is what you want for a mass. The potential energy of that piece of the string is ##dU = dm\,g (-y)##. The negative sign comes in because you need the potential energy to decrease as y increases. Put it all together and you get
$$U = \int_0^L (-y) g \left(\frac m l \right) dy $$
If you choose the other sign convention so that you integrate from y=0 to y=-L, then dy<0 so you want dm = -(m/l)dy so that dm>0. You no longer need the other minus sign though because the potential energy decreases when y decreases with this sign convention. So you get
$$U = \int_0^{-L} y g \left[-\left(\frac m l \right) dy\right].$$ Note if you change variables from ##y \to u=-y## in this second integral, you get the first integral.

 

[MARX]

Yes.Yes.I wouldn't do it that way. Say you define the positive direction to be downward. The top of the string would be at y=0 and the bottom at y=L, where L>0 is the length of the string hanging down. The mass of an infinitesimal piece of the string would be ##dm = (m/l)dy##. If you integrate from y=0 to y=L, then dy>0 and dm>0, which is what you want for a mass. The potential energy of that piece of the string is ##dU = dm\,g (-y)##. The negative sign comes in because you need the potential energy to decrease as y increases. Put it all together and you get
$$U = \int_0^L (-y) g \left(\frac m l \right) dy $$
If you choose the other sign convention so that you integrate from y=0 to y=-L, then dy<0 so you want dm = -(m/l)dy so that dm>0. You no longer need the other minus sign though because the potential energy decreases when y decreases with this sign convention. So you get
$$U = \int_0^{-L} y g \left[-\left(\frac m l \right) dy\right].$$ Note if you change variables from ##y \to u=-y## in this second integral, you get the first integral.

Amazing great understood thanks so much every time I get a clear cut explanation like this here I know my physics is improving tremendously!
What do you think of this book I am learning mechanics from? good? the problems are tough I like them they enforce understanding!

 

[BvU]

From what I've seen it's a good book !