please tell me how to solve these!!

[fahd]

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]

For problem 2,the position vector is given by
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

\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
\vec{u}=\left(-Sin(\frac{kt^2}{2a}) \vec{i}+Cos(\frac{kt^2}{2a}) \vec{j}\right)
is a unit vector
and
\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

\frac{k^2 t^2}{a}=k
or
t=\sqrt{\frac{a}{k}}


For problem 3

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

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

For the next case, the position vector is given by

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
(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

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
\vec{e_u}=Cos(\theta) \vec{i} +Sin(\theta) \vec{j}
and
\vec{e_v}=-Sin(\theta) \vec{i} +Cos(\theta) \vec{j}

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

[fahd]

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