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Homework Help: Rod sliding along a block

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Rod AB is placed against a block which is moving towards right with a speed of 1m/s. If at an instant when the rod makes an angle ##60^{\circ}## with the horizontal and end A is sliding towards left with a speed of 1m/s. Then the speed of the point of contact B of the rod is?

    (Ans: ##\sqrt{3}/2\,\,\text{m/s}##)

    2. Relevant equations

    3. The attempt at a solution
    Let the horizontal distance from the dashed line of point A be ##x_A## and that of B be ##x_B##. Also, let the distance of B from ground be ##y##. Clearly, ##dx_A/dt=-1\,\,\text{m/s}## and ##dx_B/dt=1\,\,\text{m/s}##. From Pythagoras theorem,

    $$(x_B-x_A)^2+y^2=l^2 \Rightarrow 2(x_B-x_A)\left(\frac{dx_B}{dt}-\frac{dx_A}{dt}\right)+2y\frac{dy}{dt}=0$$
    where ##l## is the length of rod.
    $$\Rightarrow -y\frac{dy}{dt}=2(x_B-x_A) \Rightarrow \frac{dy}{dt}=-\frac{2}{\sqrt{3}}\,\,\text{m/s}$$
    where I have used ##\tan(60^{\circ})=y/(x_B-x_A)=\sqrt{3}##.

    Hence, the net velocity is

    Where did I go wrong? :confused:

    Any help is appreciated. Thanks!

    Attached Files:

  2. jcsd
  3. Jan 12, 2014 #2
    Hi Pranav...

    Your work looks alright to me .
  4. Jan 12, 2014 #3

    I like Serena

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    Hey Pranav!

    Can it be they wanted to know the speed of point B relative to the block?
    And furthermore that in the given answer the numerator and denominator are swapped around?
  5. Jan 12, 2014 #4
    Hi Tanya! :)

    I too feel that there is nothing wrong with my work but then I don't see what's wrong with the given solution to this problem. I have attached the solution.

    Hi ILS! :smile:

    I copied the problem word by word. I have attached the solution given and I can't follow what's going on in the solution. It looks as if they are equating the component of velocity along the rod. Please have a look. :)

    Attached Files:

  6. Jan 12, 2014 #5
    The solution is simply wrong for the reason that 1/2 m/s may be the component of velocity of tip B along the rod ,but it is surely not the velocity of the tip with respect to the block .

    vB = vB,Block + vBlock

    Does this makes sense ?
  7. Jan 12, 2014 #6
    Umm...why do we talk about the velocity with respect to block? :confused:
  8. Jan 12, 2014 #7

    I like Serena

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    Interesting. :wink:

    With their speed along the bar, point B would become detached from the block since its horizontal speed would exceed 1 m/s.
    But if that is supposed to be the case, we do not have enough information, and the problem does refer to "the point of contact B".

    Anyway, I don't understand either why they would take the component of the velocity at A along the rod.
    And I certainly do not understand how they added the two velocities at point B - they appeared to have applied the cosine rule the wrong way.

    I'd stick with your solution.
    As an alternative and a verification, you can look at a frame that moves with point A.
    In that frame the horizontal speed at point B is 2 m/s and point A is fixed.
    Then it follows immediately that the vertical speed at point B has to be 2/√3.
  9. Jan 12, 2014 #8
    Yes, I tried that frame when Tanya talked about velocity with respect to block and I get the same answer. :)

    But why do you say it exceeds 1m/s? I think the horizontal velocity would become less than 1 m/s that's why it detaches, right?

    And the sad part is, I lost 4 marks. :(
  10. Jan 12, 2014 #9
    Decompose the velocity vector into any two perpendicular directions x and y, say v=vx+vy .

    Again decompose the velocity vector into another set of perpendicular direction m and n ,say v=vm+vn .

    Do you think v=vx+vm ?

    The problem with the solution is that they are combining components of velocity in different sets of perpendicular directions .If they wanted to use velocity of tip B along the rod then they should have combined that with the component perpendicular to the rod .
    Last edited: Jan 12, 2014
  11. Jan 12, 2014 #10

    I like Serena

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    You're quite right. It's less than 1 m/s.
    Perhaps you can get your points back if you ask for them.
  12. Jan 12, 2014 #11
    Agreed, thanks a lot Tanya! :smile:

    If the results are not declared by tomorrow, I can ask them to do the corrections. :)
  13. Jan 12, 2014 #12


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    Maybe it comes to the same thing, but I would characterise the blunder a little differently. Looks to me like they took the 0.5 m/s along the rod as a component velocity to be added to the horizontal velocity. Instead, it is a component of the resultant velocity (of the horizontal and vertical movements). If the vertical velocity is u and the overall velocity v we have v2 = u2 + 1, and resolving along the rod: u sin 60 - 1 cos 60 = 0.5.
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