please tell me how to solve these!!

**fahd** · Oct4-05, 02:50 AM

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!!!

**balakrishnan_v** · Oct4-05, 04:09 AM

For problem 2,the position vector is given by
$LaTeX Code: p(t)=a \\left(Cos(\\frac{kt^2}{2a}) \\vec{i}+Sin(\\frac{kt^2}{2a}) \\vec{j}\\right)$

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

Now

$LaTeX Code: \\frac{dp(t)}{dt}=kt \\left(-Sin(\\frac{kt^2}{2a}) \\vec{i}+Cos(\\frac{kt^2}{2a}) \\vec{j}\\right)=kt \\vec{u}$
where
$LaTeX Code: \\vec{u}=\\left(-Sin(\\frac{kt^2}{2a}) \\vec{i}+Cos(\\frac{kt^2}{2a}) \\vec{j}\\right)$
is a unit vector
and
$LaTeX Code: \\frac{d^{2}p(t)}{dt^{2}}=-\\frac{k^2 t^2}{a}\\left(Cos(\\frac{kt^2}{2a}) \\vec{i}+Sin(\\frac{kt^2}{2a}) \\vec{j}\\right)+k \\left(-Sin(\\frac{kt^2}{2a}) \\vec{i}+Cos(\\frac{kt^2}{2a}) \\vec{j}\\right) =-\\frac{k^2 t^2}{a} \\vec{v} +k \\vec{u}$

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

ie

$LaTeX Code: \\frac{k^2 t^2}{a}=k$
or
$LaTeX Code: t=\\sqrt{\\frac{a}{k}}$

For problem 3

acceleration of rim along horizontal wrt center is $LaTeX Code: \\alpha R=a$
along vertical it is $LaTeX Code: \\frac{v^2}{R_0}$

hence the net acceleration is the vector sum of both
which is $LaTeX Code: \\sqrt{a^2+\\frac{v^4}{R_0^2}}$

For the next case, the position vector is given by

$LaTeX Code: p(t)=(0.5 a_0 t^2+bt+c)\\vec{i}+R(Sin(\\theta) \\vec{i}+Cos(\\theta) \\vec{j})$

Differentiating twice we get the acceleration which comess to be
$LaTeX Code: (a+a Cos\\theta-\\frac{v^2}{R} Sin\\theta) \\vec{i} -(a Sin\\theta+\\frac{v^2}{R} Cos\\theta) \\vec{j}$

whose magnitude is

$LaTeX Code: a_0 \\sqrt{2+\\frac{v^4}{a_0^2R_0^2}+2Cos(\\theta)-2\\frac{v^2 }{R_0 a_0}Sin(\\theta)}$

For the last problem you put
$LaTeX Code: \\vec{e_u}=Cos(\\theta) \\vec{i} +Sin(\\theta) \\vec{j}$
and
$LaTeX Code: \\vec{e_v}=-Sin(\\theta) \\vec{i} +Cos(\\theta) \\vec{j}$

So we get
$LaTeX Code: \\frac{d\\hat{e_u}}{dt}=\\hat{e_v} \\frac{u\\dot{v}-v\\dot{u}}{2\\sqrt{uv} (u+v)}$
$LaTeX Code: \\frac{d\\hat{e_v}}{dt}=\\hat{e_u} \\frac{v\\dot{u}-u\\dot{v}}{2\\sqrt{uv} (u+v)}$

**fahd** · Oct4-05, 09:21 AM

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