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kero · Oct 19, 2009, 03:47 PM

a pendulum of mass m and length L is pulled back an angle θ and released. After the pendulum swings through its lowest point, it encounters a peg α degrees out and r meters from the top of the string. The mass swings up about the peg until the string becomes slack with the mass falling inward and hitting the peg. Show that cosθ=r/Lcosα-√3/2(1-r/L)
image: http://s3.amazonaws.com/answer-board...d56a3b27d7.jpg

please guys share any thoughts you have on this even if you don't know the answer.
Thanks

Unknown008 · Oct 20, 2009, 04:40 AM

cosθ=r/Lcosα-√3/2(1-r/L)

Does that mean:

$cos\theta = \frac{r}{Lcos\alpha - \frac{\sqrt3}{2} (\frac{1-r}{L})}$

?

ebaines · Oct 20, 2009, 05:43 AM

Originally Posted by Unknown008

cosθ=r/Lcosα-√3/2(1-r/L)

Does that mean:

$cos\theta = \frac{r}{Lcos\alpha - \frac{\sqrt3}{2} (\frac{1-r}{L})}$

?

I don't think so, as the units don't work out. Since $\theta$ must be dimensionless, I think he means:

$cos\theta = \frac{r}{Lcos\alpha} - \frac{\sqrt3}{2} (1-\frac r L)$

Kero?

kero · Oct 20, 2009, 06:55 AM

$cos\theta = \frac{r}{Lcos\alpha} - \frac{\sqrt3}{2} (1-\frac r L)$

Kero?[/QUOTE]

that's exactly what i mean
i been working on this problem for a long time
hope you guys can help
Thanks

Unknown008 · Oct 20, 2009, 09:32 AM

Sorry, I had just passed through a rather 'traumatising physics paper' just then and didn't think much before posting. :o

ebaines · Oct 20, 2009, 10:13 AM

Kero: are you sure the answer is: $cos\theta = \frac{r}{Lcos\alpha} - \frac{\sqrt3}{2} (1-\frac r L)$ ?

Reason I ask is I just came up with a solution of:
$cos\theta = \frac r L cos\alpha - \frac {\sqrt3} 2 (1-\frac r L)$ .

Note that in my solution the $cos\alpha$ term is in the numerator, which means that if you increase $\alpha$ you have to increase $\theta$ a comparable amount so as to get the velocity up. Your answer would indicate that if you increase $\alpha$ you would use a smaller value of $\theta$ , which doesn't seem right.

Post back, and if my answer is correct I'll give you some pointers on how I got it!

Unknown008 · Oct 20, 2009, 10:18 AM

That makes sense... I'd like to know too. Ok, waiting for kero's response now :rolleyes: Be quick kero ;)

ebaines · Oct 20, 2009, 11:17 AM

OK, here's what I did - refer to the attached drawing:

First, recognize that there are three phases to this problem. Phase 1: you have a mass on a string of length L which swings down in an arc of radius L. Phase 2: then the string hits the peg and you have the mass now moving in an arc of radius L-R. Phase 3: as the mass rises at some point the tension in the string goes to zero, and the mass is essentially lobbed back at the pin. For this phase it moves like a projectile under gravity.

1. In order for the mass to be lobbed back toward the pin the tension in the string must be zero, so that the mass moves in an arc under gravity. The tension in the string is:

$ T = \frac {m V_o ^2} {(L-R)} - mg \cdot sin \beta $

T is zero at the pont that the mass is no longer constraijed by the string. This gives you an expression for $sin \beta$ in terms of $V_0$ :

$ sin \beta = \frac {V_0 ^2} {g (L-R)} $

2. Given an initial launch velocity of $V_0$ , the time for the mass to move horizontally back to the peg is:

$ T = \frac {L-R} {V_0 tan \beta} $

3. In order to precisely hit the peg, the mass has to fall vertically a distance $(L-R) sin \beta$ in that time T:

$ -(L-R)sin \beta = V_0 cos\beta T - \frac 1 2 g T^2 $

4. Combine the equations from 2 and 3 to get $V_0$ in terms of $\beta$ :

$ V_0 ^2 = \frac {g(L-R)} 2 \frac {cos^2 \beta} {sin \beta} $

5. Put the equation from step 1 into 4. This gets you a value for $V_0$ purely in terms of g, L and R:

$ V_0 ^2 = \frac {g(L-R)} {\sqrt 3} $

6. The height of the launch point is:
$ H = L-Rcos \alpha +(L-R) sin \beta $

from step 1 and 5 we know that:
$ (L-R) sin \beta = \frac {V_0 ^2} g = \frac {(L-R)} {\sqrt 3} $

7. You know that when the mass is first released it must reach velocity $V_0$ at this height as it swings downward. So use energy principles to calculate the height from which it must be released for this to occur:

$ A = \frac {V_0 ^ 2} {2g} + H $

Sub in H from step 6, and $V_0 ^2$ from step 5. Plug and chug, and you get:

$ cos \theta = \frac R L cos \alpha - \frac { sqrt 3} 2 (1 - \frac R L ) $

kero · Oct 20, 2009, 01:09 PM

Sorry guys you are right that's what I meant $cos\theta = \frac r L cos\alpha - \frac {\sqrt3} 2 (1-\frac r L)$

Thanks sooooooooooo much for the help, I really appreciate it
You guys saved my life today
God bless

kero · Oct 20, 2009, 08:49 PM

Originally Posted by ebaines

OK, here's what I did - refer to the attached drawing:

First, recognize that there are three phases to this problem. Phase 1: you have a mass on a string of length L which swings down in an arc of radius L. Phase 2: then the string hits the peg and you have the mass now moving in an arc of radius L-R. Phase 3: as the mass rises at some point the tension in the string goes to zero, and the mass is essentially lobbed back at the pin. For this phase it moves like a projectile under gravity.

1. In order for the mass to be lobbed back toward the pin the tension in the string must be zero, so that the mass moves in an arc under gravity. The tension in the string is:

$ T = \frac {m V_o ^2} {(L-R)} - mg \cdot sin \beta $

T is zero at the pont that the mass is no longer constraijed by the string. This gives you an expression for $sin \beta$ in terms of $V_0$ :

$ sin \beta = \frac {V_0 ^2} {g (L-R)} $

2. Given an initial launch velocity of $V_0$ , the time for the mass to move horizontally back to the peg is:

$ T = \frac {L-R} {V_0 tan \beta} $

3. In order to precisely hit the peg, the mass has to fall vertically a distance $(L-R) sin \beta$ in that time T:

$ -(L-R)sin \beta = V_0 cos\beta T - \frac 1 2 g T^2 $

4. Combine the equations from 2 and 3 to get $V_0$ in terms of $\beta$ :

$ V_0 ^2 = \frac {g(L-R)} 2 \frac {cos^2 \beta} {sin \beta} $

5. Put the equation from step 1 into 4. This gets you a value for $V_0$ purely in terms of g, L and R:

$ V_0 ^2 = \frac {g(L-R)} {\sqrt 3} $

6. The height of the launch point is:
$ H = L-Rcos \alpha +(L-R) sin \beta $

from step 1 and 5 we know that:
$ (L-R) sin \beta = \frac {V_0 ^2} g = \frac {(L-R)} {\sqrt 3} $

7. You know that when the mass is first released it must reach velocity $V_0$ at this height as it swings downward. So use energy principles to calculate the height from which it must be released for this to occur:

$ A = \frac {V_0 ^ 2} {2g} + H $

Sub in H from step 6, and $V_0 ^2$ from step 5. Plug and chug, and you get:

$ cos \theta = \frac R L cos \alpha - \frac { sqrt 3} 2 (1 - \frac R L ) $

Thanks sooooooooooo much for the help, I really appreciate it.
I got everything except for the last part where you did the plug in and hope you can help me with that.
Again thanks a lot !

kero · Oct 20, 2009, 09:13 PM

And also if you please could explain how you got # 5, I will really appreciate it

ebaines · Oct 21, 2009, 05:47 AM

Originally Posted by kero

Thanks sooooooooooo much for the help, I really appreciate it.
I got everything except for the last part where you did the plug in and hope you can help me with that.
Again thanks alot !

Kero: I've given you a pretty good outline of how to do this problem. This is your homework problem - so before I give any more details, please show us your work so far.

kero · Oct 21, 2009, 10:20 PM

Originally Posted by ebaines

Kero: I've given you a pretty good outline of how to do this problem. This is your homework problem - so before I give any more details, please show us your work so far.

OK here is what I did following your outline so far

#3

kero · Oct 21, 2009, 10:21 PM

combining 2 &3

x

kero · Oct 21, 2009, 10:23 PM

I get stock on number 5 because when I plug in #1 into # 4 I get

Also when I get to plug in at the last step, I don't know what to plug in for A

Unknown008 · Oct 21, 2009, 11:09 PM

Ok, both of you made me understand better. Here's my small part! :)

$V_o\,^2 = \frac{grcos^2\beta}{2sin\beta}$

$sin\beta = \frac{V_o\, ^2}{gr}$

I'm letting r = L-R to reduce confusion between the variables.

$V_o\, ^2 = \frac{gr$ 1-\frac{V_o\, ^4}{g^2r^2}$}{2$\frac{V_o\, ^2}{gr}$}$

From:
$cos^2\beta = 1 - sin^2 \beta$

$V_o\, ^2 = \frac{gr - \frac{V_o\,^4}{gr}}{\frac{2V_o\,^2}{gr}}$

Rationalise by multiplying by gr/gr:

$V_o\, ^2 = \frac{g^2r^2 -V_o\,^4}{2V_o\,^2}$

$2V_o\, ^4 = g^2r^2 -V_o\,^4$

$3V_o\, ^4 = g^2r^2$

$V_o\, ^4 = \frac{g^2r^2}{3}$

$V_o\, ^2 = \sqrt{\frac{g^2r^2}{3}} = \frac{gr}{\sqrt{3}} = \frac{g(L-R)}{\sqrt{3}}$

Here! :)