solution

- ParthKohli

solution

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- chestercat

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- ParthKohli

|dw:1447233261702:dw|

- Willie579

I don't get it. ;-;

- Willie579

Lol.

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## More answers

- ganeshie8

.

- ParthKohli

Suppose at this instant that the mass is going with a velocity \(v\) and the radius of its motion is \(r\).
In a very small time interval \(dt\), so small that we can approximate the motion as a straight line motion, the length of the string that wraps around the pole wouldn't be \(v dt\) right?

- ParthKohli

|dw:1447233656278:dw|

- Astrophysics

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- ParthKohli

I'm confused... working with so many variables ugh!

- ParthKohli

Ohhh, lol, I don't know why I separately introduced \(r\) when I actually meant\[d \ell = - a d \theta \]and\[d\theta = \frac{v~dt}{\ell }\]Therefore,\[\frac{v~dt }{\ell } = - \frac{d \ell }{a}\]\[\Rightarrow d t =- \frac{\ell d \ell }{v a}\]But\[\frac{mv^2}{\ell} = k \ell \]So\[v=\ell \sqrt{\frac{k}{m}}\]Now\[dt = -\sqrt{\frac{m}{ka^2}} d \ell \]

- ParthKohli

\[t = -\sqrt{\frac{m}{ka^2}}(\ell_f -\ell_0)\]And \(\ell_f = 0,~ \ell_i = L\).The answer should be\[t = \sqrt{\frac{mL^2}{ka^2}} \]

- ParthKohli

anything wrong with this solution?

- ganeshie8

@IrishBoy123

- ParthKohli

http://prntscr.com/91gqmv

- IrishBoy123

my goodness! that's not easy, is it?
my instinct is that the spring would extend out as the radius of the thing tried to diminish. both on interial grounds and on energy conservation.

- ParthKohli

here's one thing: what is providing the tangential force to the mass?

- ParthKohli

which means that the speed must remain constant throughout the motion!

- ParthKohli

or wait a minute, no. the angular momentum must.

- ParthKohli

\[v =\frac{v_i r_i}{r}\]

- ParthKohli

\[= \frac{v_iL}{\ell }\]we know \(v_i\) because\[\frac{mv_i^2}{L}=kL\Rightarrow v_i = \sqrt{\frac{k}{m}}L = 200 ~ms^{-1}\]

- ParthKohli

so we had\[dt =- \frac{\ell d \ell}{av}=-\frac{\ell^2 d \ell}{200a L }\]

- ParthKohli

no, not even close

- ParthKohli

let me post an easy problem to cheer y'all up
I hang a small, heavy bob using a thread of length \(a\).
Now I hit a nail into the wall \(b\) distance to the right of the point that supports the thread.
What is the minimum velocity that I must project the bob with, so that it completes the whole circle?

- lochana

is b smaller than a?

- lochana

I am little bit confused now, let me draw what I think the path of bob
|dw:1447241039037:dw|

- lochana

|dw:1447241173555:dw|

- lochana

is that right? bob will winds around b.

- lochana

actually , it winds around nail. not b

- lochana

V0 = initial velocity
V1 = velocity at nail and thread level
V2 = velocity at top of the circle
the minimum velocity required at the top of circle is zero. So it will completes the circle
so V2 = 0
considering top of the level and nail and thread level, from energy conservation law
\[\frac{1}{2}mV1^2 = mg(a-b)\]\[V1^2 = 2g(a-b)\] considering initial level and nail, thread level, from energy conservation\[\frac{1}{2}mV0^2 = mga + \frac{1}{2}mV1^2\]\[V0^2 = 2ga+ V1^2\] finally we get \[V0^2 = 2ga + 2g(a-b)\]\[V0^2 = 2g(2a-b)\]\[V0 = \sqrt{2g(2a-b)}\]

- ParthKohli

yes, b < a.

- ParthKohli

that's actually not the correct answer because we do need a velocity at the top of the circle.

- lochana

mm.. I thought It will complete the circle if the bob somehow reach the top.
yes , you are correct. only vertical velocity becomes zero

- ParthKohli

the condition is that at the top, there is some tension in the thread.
now we have\[T + mg = \frac{mv^2}{a-b}\]\[\Rightarrow T = \frac{mv^2}{a-b} - mg \ge 0\]The limiting case is the velocity being\[v = \sqrt{g(a-b)}\]Now the bob is at the height \(a + a-b = 2a - b\) so we'll apply conservation of energy accordingly.\[\frac{1}{2}mv^2_{bottom} = \frac{1}{2}m(g(a-b)) + mg(2a-b)\]\[\Rightarrow v^2_{bottom} = g(a-b+4a-2b)\]\[\boxed{v_{bottom}=\sqrt{g(5a-3b)}}\]

- lochana

so the idea is that the minimum velocity is given when tension at the top is zero. right?

- ParthKohli

exactly

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