1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newton's Laws of motion!

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of Mass M (=2m) as shown in the given figure. At a instant at the string between the ring and the pulley makes an angle [tex]\theta[/tex] with the rod, (a)Show that, if the ring slides with a speed v, the block descends with speed v cos[tex]\theta[/tex]. (b) With what acceleration will the ring start moving if the system is released from rest with [tex]\theta=~30^\circ[/tex]?

    2. Relevant equations

    Newtons Equations, free body diagram

    3. The attempt at a solution

    Well, this question happens to be from a textbook. And its a solved one too.. The solution which they have given is quite complicated! I thought I'd get a brief explanation of "why and how" over it, and yes an alternative method would be highly appreciated!

    The solution, is like this (as given in the book),

    Suppose in a small time interval [tex]\delta t[/tex] the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacements A'B~ PB (I didnt get this!), since the lenght is constant (?????), we have,
    AB+BC= A'B+BC'
    AP+PB+BC=A'B+BC'
    AP=BC'-BC=CC' (as A'B=PB)

    AA'[tex]cos\theta[/tex]= CC'

    or [tex]\frac{AA'cos\theta}{\delta t}=~\frac{CC'}{\delta t}[/tex]

    Therefor, (velocity of the ring)[tex]cos\theta[/tex]= (velocity of the block)


    Please help :cry:
     

    Attached Files:

    Last edited: Jan 23, 2009
  2. jcsd
  3. Jan 23, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    When the ring moves with a velocity v horizontally, its component along the string is v*cos(theta). Since the length of string between the ring and block is costant, the velocity of the block is equal to v*cos(theta).
     
  4. Jan 23, 2009 #3
    Well, did you have a look at the diagram? It hasnt been approved yet..
     
  5. Jan 23, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper


    Suppose in a small time interval LaTeX Code: \\delta t the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacements A'B~ PB (I didnt get this!), since the lenght is constant (?????), we have,

    When A and A(dash) are very close, AB and A(dash)B are nearly equal. If you take BA(dash) equal to BP, AP becomes A*A(dash)cos(theta) and A*A(dash)/t = velocity. The length of the string is ABC = A(dash)BC(dash)
     
  6. Jan 24, 2009 #5
    okay, thats fine with me, I've several more problems over the free body diagrams! I'll post it in mean time! Thank You!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?