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Some question from the oxford entrance test

  1. Oct 26, 2008 #1
    14. You should already be familiar with the mathematical treatment of an
    ideal pendulum, in which the pendulum bob is modelled as a point
    mass on the end of a rigid rod of negligible mass. In this problem you
    will consider the behaviour of more complex types of pendulum. You
    will be given all the information you need in the sections below.
    For a general pendulum of any shape and size the period P is given by
    p= 2*pie*root(I/gM(LCM))
    where g is the acceleration due to gravity, M is the total mass of the
    pendulum, LCM is the effective length of the pendulum, defined as the
    distance from the pivot to the centre of mass, and I is the moment of
    inertia around the pivot point. For a point mass m fixed at a distance
    r from the pivot I = mr^2, while for a uniform rod of mass m and length
    r attached to the pivot at one end I = 1/3mr^2
    . For more complex objects
    the total moment of inertia can be calculated by adding together values
    for the component parts.

    (c) Now consider the case of a real pendulum, with a bob of mass
    Mb (which you may treat as a point mass) attached to the pivot
    using a uniform rod of mass Mr and length L, and find the period
    in this case. Show that the result for a real pendulum reduces to
    the results for an ideal pendulum and a rod pendulum by taking
    appropriate limits.
     
  2. jcsd
  3. Oct 26, 2008 #2

    gabbagabbahey

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    What have you tried? You are given the equation for the period of the real pendulum...all you need to do is determine LCM and I. What happens to p as Mr approaches zero?
     
  4. Oct 26, 2008 #3
    okokok thank you dat s the direction i was thinking but i dont kno whether it s the right one

    (d) Most substances expand with increasing temperature, and so a
    metal rod will expand in length by a fraction ®dT if the tem-
    perature is changed by dT, where ® is called the coe±cient of
    linear thermal expansion. Consider the effect of this expansion
    on a pendulum clock with a pendulum made from brass, with
    ® = 19 £ 10¡6 K¡1. What temperature change can this clock
    tolerate if it is to remain accurate to 1 second in 24 hours? [5]

    do u just try to seperate p and dp and stuff??
     
  5. Oct 26, 2008 #4
    wat is the question saying wen it is asking u to take appriopriate limit??
     
  6. Oct 26, 2008 #5

    gabbagabbahey

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    For the ideal pendulum, Mr=0 so the "appropriate limit is the limit as Mr approaches zero.
     
  7. Oct 26, 2008 #6

    gabbagabbahey

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    the period p will be a function of the length of the pendulum and the length of the pendulum will be a function of temperature....therefor p is a function of temperature.... if the temperature change is [itex]\Delta T=T_2-T_1[/itex] what is the change in period [itex]\Delta p = p(T_2)-p(T_1)[/itex]? If the pendulum is accurate to within 1 second in 24 hours, then [itex]\Delta p \leq \frac{1s}{24h}[/itex] and you can solve for [itex]\Delta T[/itex].
     
  8. Oct 27, 2008 #7
    thabk you
     
  9. Nov 4, 2008 #8
    24. A black box has three electrical connections labelled A, B, and C,
    arranged in a triangle as shown below.

    A
    b c
    The box contains three components: a resistor, a small capacitor and a
    diode. You know that one component is connected between each pair
    of terminals, but you cannot see exactly how they are arranged. You
    make the following observations with a 9V battery connected in series
    with an ammeter
    ² When the battery is connected with + to A and ¡ to B a current
    of 3mA °ows
    ² When the battery is connected with + to B and ¡ to C a very
    large current °ows
    ² When the battery is connected with + to C and ¡ to A no current
    is measured
    ² When the battery is connected with ¡ to B and + to C no current
    is measured
     
  10. Nov 4, 2008 #9
    Will you be tested on "ure" grammar?
     
  11. Nov 4, 2008 #10
    OK so I got to √L2 - √L1 ≤ 5.825 x 10^-6

    How do I get ΔL from there?
     
  12. Nov 4, 2008 #11
    You got Δ√L haven't you?
     
  13. Nov 4, 2008 #12
    Yeah, but how can I get ∂T from ∆√L ?
     
  14. Nov 4, 2008 #13
    You are looking for the maximum change in length allowed aren't you? So if you have the square root of the change in length allowed..
     
  15. Nov 4, 2008 #14
    I'm sorry, I still don't understand, if I had √∆L then I could do it, but that's not equal to ∆√L is it?
     
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