Turn on thread page Beta
    • Thread Starter
    Offline

    6
    ReputationRep:
    A particle of mass 0.3kg is suspended by two identical elastic springs of natural length 1.5m and modulus lamba N. The other ends of the strings are fixed to two points A and B on a horizontal ceiling where AB=3m. P is released from rest at the mid-point of AB and falsl vertically until it is instantaneously at rest at a point 1m below the leve lof the ceiling.

    a) calculate the value of lamba

    b) calculate the speed of P when it is 0.5m below the level of the ceiling
    Offline

    16
    ReputationRep:
    what have you tried so far?
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by didgeridoo12uk)
    what have you tried so far?
    Well I did the first part and I obtained 49N however in the book the answer is given as 48.1N?
    Offline

    16
    ReputationRep:
    (Original post by sulexk)
    Well I did the first part and I obtained 49N however in the book the answer is given as 48.1N?
    cant be bothered to work though it at the moment, but i imagine you did it the right way. just used a different value of g or rounded early or something.
    • Thread Starter
    Offline

    6
    ReputationRep:
    This is in the M3 book exercise 2c question 9

    I did write mg=2(lambda)x^2 / 2l

    and then worked from there.
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by didgeridoo12uk)
    cant be bothered to work though it at the moment, but i imagine you did it the right way. just used a different value of g or rounded early or something.


    Thank you very much!

    I did round early somewhere.

    I have just one more question, can you help me with this please, then I will have completed the exercise.

    Here I go:

    A light elastic string has natural length L and modulus (lamba). One end is fixed to a point A on a ceiling and a particle P of mass m is attached to the other end. P is held vertically below A so that AP=2L and then released. P has speed v when the extension of the string is x. Show that, while the string remains taut,

    1/2mv^2 = L/2 ( (lamba) - 2mg ) + mgx - (lambda)x^2 / 2L

    I have completed that part

    I am stuck with this part:

    By considering the speed of P when x=0 show that the string will never become slack provided 2mg > (lamba)

    Thank you
    Offline

    16
    ReputationRep:
    (Original post by sulexk)

    1/2mv^2 = L/2 ( (lamba) - 2mg ) + mgx - (lambda)x^2 / 2L

    I have completed that part

    I am stuck with this part:

    By considering the speed of P when x=0 show that the string will never become slack provided 2mg > (lamba)

    Thank you
    put x = 0 into that equation and see what happens..
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by didgeridoo12uk)
    put x = 0 into that equation and see what happens..
    I am still confused.

    I have put x=0 and obtained the equation mv^2=L((lambda) - 2mg)
    Offline

    16
    ReputationRep:
    (Original post by sulexk)
    I am still confused.

    I have put x=0 and obtained the equation mv^2=L((lambda) - 2mg)
    re-arrange for v (This equation is correct while the string is taught)

    look what happens if 2mg < lambda...


    i assume this is the root you're meant to be going down. without thinking about it more deeply i'm not sure whether i'm convinced its actually a proof or not
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by didgeridoo12uk)
    re-arrange for v (This equation is correct while the string is taught)

    look what happens if 2mg < lambda...


    i assume this is the root you're meant to be going down. without thinking about it more deeply i'm not sure whether i'm convinced its actually a proof or not
    well if 2mg>lambda then mv^2 would be negative. But v^2 cannot be negative. so then must 2mg be less than lambda?
    Offline

    16
    ReputationRep:
    (Original post by sulexk)
    well if 2mg>lambda then mv^2 would be negative. But v^2 cannot be negative. so then must 2mg be less than lambda?
    exactly
    • Thread Starter
    Offline

    6
    ReputationRep:
    (Original post by didgeridoo12uk)
    exactly
    However does what I stated before answer the question?

    since it says " show that the string will never become slack provided 2mg>(lambda)"

    Where is the connection?


    Thank you
    Offline

    16
    ReputationRep:
    (Original post by sulexk)
    However does what I stated before answer the question?

    since it says " show that the string will never become slack provided 2mg>(lambda)"

    Where is the connection?


    Thank you
    it shows that at x=0 (where the rope is on the verge of becoming slack) provided that 2mg>lambda, then given that the rope isn't slack then the velocity equation holds...


    i guess if lambda = 2mg then the v would be zero and so you could argue that the rope never becomes slack, it just gets to the point then reverses.

    basically i'm not really sure if it you can prove it or if the question is correct :/
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 5, 2011
Poll
Cats or dogs?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.