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Oct 9, 2008

Homework Problem 3/177


Suggestions:
  1. FBD: Draw an FBD of A, B and link AB. Determine which, if any, nonconservative forces that do work on the system.
  2. Work-energy equation: Since there are no nonconservative forces that do work, the WE equation becomes T1 + V1 = T2 + V2 where: T1 = 0 and T2 = (1/2)*m*vA^2 + (1/2)*m*vB^2. Write down the potential energy of the system based on your choice of datum position (recommended datum position: horizontal line passing through B).
  3. Kinematics: Look back at your WE equation in 2. At what position will B have its maximum speed? (Hint:  The system has its maximum kinetic energy when its potential energy is a minimum -- when A is at its bottom-most position. What is the speed of A at this position? Use this in your work-energy equation.) 
  4. Solve: Use the WE and kinematics equations to find vB.

NOTE:  As A falls through the vertical slot, the speed of B reaches a LOCAL maximum value at y = 2/3 of the initial height of A (this comes from differentiating the work energy equation with respect to time and setting d_vB/d_t = 0, as required for a maximum). This is the answer that is reported in the textbook. However, as A continues to fall, the speed of B reaches its absolute maximum when y = -L giving vB_max = 5.49 m/sec. Following the hint in 3. above will give you this absolute maximum. We will accept either answer.

Let us know if you have any questions on this.







7 comments:

Dale Szul said...

I am a slightly confused with the comment that V_b will reach its max. speed when A is y=2/3 of the original height. So far my final equation before plugging anything in is:
(2)*g*(y_inital-y_final)=((x/y)*V_b)^2-V_b^2

V_a = (x/y)*V_b
y_initial = .636 m
and all I need in y_final, but when using the y_final = (2/3)*y_initial rule, I do not get the right answer for V_b(max).

Where does this 2/3 value come from and what could be my error? Thank you.

CMK said...

* The left hand side of your work energy equation looks fine.

* The last term involving vB^2 should have a "+" sign in front. All KE terms should add since KE is always positive.

BoilerBrian said...

where does v_a = (x/y)v_b come from? I assume that it has something to do with the fact that the blocks innitally started equaldidstant from the "origin".

CMK said...

Actually, that is a general result, independent of the initial conditions. In Step 3 where you do kinematics, you will write down:

x^2 + y^2 = L^2 ; L = 0.9 m = constant

Differentiating this w.r.t. time gives:

2*x*x_dot + 2*y*y_dot = 0

which gives:

y_dot = -(x/y)*x_dot

(The minus sign is not significant since you square the speeds in the WE equation.)

Does this help?

Adam Watson said...

In the hint at the bottom it says the absolute maximum velocity is reached when y = -L. What is L supposed to be? (2/3)y or just y or something else?

CMK said...

L is the length of bar AB (0.9 meters).

Byron said...

After working the problem and reading the comments, I'm still a little confused about the derivation about finding the maximum "y" position. I understand differentiating the x^2 + y^2 = L, but I don't understand how to find the "y" value where vb = is a maximum.