Dynamics involving dependent motion

[kc2974]

If cylinder B has a downward velocity in feet per second given by
vB = t2/2 + t3/6 , where t is in seconds. Calculate the acceleration of A
when t = 2 seconds.

How does the velocities of A & B relate...Seems to be at C but I am not clear how to state it.

 

[jamesrc]

If you just look at A and C, you should see that v_A=-v_C. In other words, A and C move at the same speed and when A moves down, C moves up (and vice versa).

You might be able to see the relationship between B and see, but let's try it a little more analytically: Let's call y_C the position of C and y_B the position of B. Now let's write y_B as a function of y_C:

$$y_B=y_c - x$$

where x is a variable I just made up to represent the length of rope between the pulley and weight B. But the length of the rope (let's call it L) is constant and equal to L=y_C-x

Plug this into the equation for y_B:

$$y_B=y_C - (L - y_C)$$
$$y_B = 2y_C - L$$

Take the derivative to find the relationship between velocities (remember L is constant):

$$v_B=2v_C$$

with

$$v_C=-v_A$$

so

$$v_B=-2v_A$$

 

[kc2974]

Thanks, I finally arrived at this answer Thursday.