In Homework Problem No. 3/177 (see problem figure in post below), the system is released from rest with y > 0 (block A above the horizontal path of B).
The mathematics of your work-energy equation should show that the speed of B reaches a (local) maximum for some position of A for which y > 0. Can you provide a physical argument as to why this local maximum for vB must exist for y > 0? [HINT: Consider the initial speed of B and the speed of B when AB is horizontal.]
In order to give everyone a chance to see this question, do not submit any responses prior to 5PM on Wednesday, October 15.
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2 comments:
The velocity of block B has a local maximum in this interval because it begins and ends the interval at rest and in different positions.
Please note:
The signs { and } are used for greater than or less than.
Since the system is initially at rest, the velocity of block B is zero to begin with.
We can see that as block A lowers until AB becomes horizontal, block B will move to the left.
At the moment that AB is horizontal, block B will come to rest, since its motion is going to change directions to the right once block A breaks the horizontal axis (y less than 0).
Since block B will START FROM REST, gain some velocity to the left, and then slow to a STOP when AB is horizontal, there must be a local maximum velocity in the interval 0{y{y(o).
In summary:
y=y(o) v(b)=0 (system at rest)
0{y{y(o) |v(b)|}0 to the left
y=0 v(b)=0
y{0 |v(b)|}0 to the right
A change in direction with a system initially at rest always implies a local maximum.
Excellent explanation. Good job.
Thanks for your response.
[BTW, I forgot to state that this blog challenge is worth 20 blog points. This will be the points awarded for ALL blog challenges, including the first.]
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