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Please tell me how to solve these

  1. Oct 4, 2005 #1
    please tell me how to solve these!!

    hi there.i really want someone to help me with questions 2,3 and 5
    in the attached document.

    i am really stuck.I tried solving question 2 using planar polar coordinates..and tried to get the given time value for the angle set at 45..however i am getiing the time to be 1/K^0.5

    Also i seriously dunno how to go about 3(b)...and 5.can u please guide me.this is a practise sheet for the exams.And i wanna get good grades!PLEASE HELP!!!

    Attached Files:

  2. jcsd
  3. Oct 4, 2005 #2
    For problem 2,the position vector is given by
    [tex]p(t)=a \left(Cos(\frac{kt^2}{2a}) \vec{i}+Sin(\frac{kt^2}{2a}) \vec{j}\right)[/tex]

    if the center is the origin and the line joining the center and the starting point is the x-axis


    [tex]\frac{dp(t)}{dt}=kt \left(-Sin(\frac{kt^2}{2a}) \vec{i}+Cos(\frac{kt^2}{2a}) \vec{j}\right)=kt \vec{u}[/tex]
    [tex]\vec{u}=\left(-Sin(\frac{kt^2}{2a}) \vec{i}+Cos(\frac{kt^2}{2a}) \vec{j}\right)[/tex]
    is a unit vector
    [tex]\frac{d^{2}p(t)}{dt^{2}}=-\frac{k^2 t^2}{a}\left(Cos(\frac{kt^2}{2a}) \vec{i}+Sin(\frac{kt^2}{2a}) \vec{j}\right)+k \left(-Sin(\frac{kt^2}{2a}) \vec{i}+Cos(\frac{kt^2}{2a}) \vec{j}\right) =-\frac{k^2 t^2}{a} \vec{v} +k \vec{u}[/tex]

    It can be seen that u and v are othogonal and velocity vector is along u
    Hence for 45 we need the components along u and v to be equal in magnitude


    [tex]\frac{k^2 t^2}{a}=k[/tex]

    For problem 3

    acceleration of rim along horizontal wrt center is [tex]\alpha R=a[/tex]
    along vertical it is [tex]\frac{v^2}{R_0}[/tex]

    hence the net acceleration is the vector sum of both
    which is [tex]\sqrt{a^2+\frac{v^4}{R_0^2}}[/tex]

    For the next case, the position vector is given by

    [tex]p(t)=(0.5 a_0 t^2+bt+c)\vec{i}+R(Sin(\theta) \vec{i}+Cos(\theta) \vec{j})[/tex]

    Differentiating twice we get the acceleration which comess to be
    [tex](a+a Cos\theta-\frac{v^2}{R} Sin\theta) \vec{i} -(a Sin\theta+\frac{v^2}{R} Cos\theta) \vec{j} [/tex]

    whose magnitude is

    [tex]a_0 \sqrt{2+\frac{v^4}{a_0^2R_0^2}+2Cos(\theta)-2\frac{v^2 }{R_0 a_0}Sin(\theta)}[/tex]

    For the last problem you put
    [tex]\vec{e_u}=Cos(\theta) \vec{i} +Sin(\theta) \vec{j}[/tex]
    [tex]\vec{e_v}=-Sin(\theta) \vec{i} +Cos(\theta) \vec{j}[/tex]

    So we get
    [tex]\frac{d\hat{e_u}}{dt}=\hat{e_v} \frac{u\dot{v}-v\dot{u}}{2\sqrt{uv} (u+v)}[/tex]
    [tex]\frac{d\hat{e_v}}{dt}=\hat{e_u} \frac{v\dot{u}-u\dot{v}}{2\sqrt{uv} (u+v)}[/tex]
    Last edited: Oct 4, 2005
  4. Oct 4, 2005 #3

    thanks a bunch balakrishnan
    ill definitely compare and see where i went wrong!
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