Pendulum velocity derivation

1. Nov 20, 2008

cideraid

Been stuck on this problem for a bit. Anyways here it is:
A pendulum bob on an ideal string of length (L) is released from an angle (theta) with respect to the direction of gravity.
Determine the velocity of the bob at the bottom of its path in terms of (theta) using only Newton's Laws of motion.
The solution to the problem should reaffirm that the v=Sqrt(2gL(1-cos(theta))) which one gets using the equations for conservation of mechanical energy.

*Currently I'm taking AP analytical physics, and we've covered dynamics, kinematics, energy, and uniform circular motion, but nothing like torque or angular velocity.

So far, I figured out that the tangential force on the pendulum is equal to: -mgsin(theta).
since ma=-mgsin(theta), the tangential acceleration is equal to: -gsin(theta).
So I tried to find the velocity by taking the integral of that, but it's nothing close to what the problem says I'm supposed to get.
I think I'm missing something really big here, but i don't know what...

2. Nov 20, 2008

No need to drag forces into this. It says right in the problem to "use conservation of mechanical E"

Write out the formula for that first and see what you have.

3. Nov 20, 2008

cideraid

Yea that's actually the second part of the problem which I already solved doing:
1/2(mv)^2=mgh
1/2(v)^2=gh
v=sqrt(2gh)

so to get h, I found the vertical displacement of the pendulum using trig:
so so a right triangle which has sides (L, L-h, and some hypotenuse)
cos(theta)=L/(L-h)
solving for h, h=-Lcos(theta)+L
factoring out L, h=L(1-cos(theta))
plugging that back in v=sqrt(2gL(1-cos(theta)))

Anyways, that was the second part of the problem, but the first part of the problem states that one needs to derive the same formula using only the Newton's laws of motion, which is the part I'm stuck on...

Last edited: Nov 20, 2008
4. Nov 20, 2008

Oops. Sorry. My fault. Let me have another look.

5. Nov 20, 2008

I don't know why I cannot remember how to do this! You are correct in saying there IS an integral involved. But I am coming up blank here... perhaps someone else can chime in here.

6. Nov 20, 2008

cideraid

Yea, I was thinking about approaching this using non-uniform circular motion; finding the tangential and radial acceleration at the bottom of the pendulum's path. Then using pythagorean theorem to find the acceleration. I would then take the integral of that in order to get the velocity.
The trouble is i'm having trouble finding the radial acceleration.
tangential acceleration should be -gsin(theta)?

7. Nov 20, 2008

naresh

Your variable of motion should be theta, just like you would have had x for linear motion. I find it better to just use the angular position, angular velocity and angular acceleration for these problems. You can also work by converting theta into a tangential distance variable.

That said, there is nothing wrong with your force equation. How did you do the integration?

8. Nov 20, 2008

Yes; however, I chose downward to be positive to simplify the calculations. Thus I have a=gsin(theta)

9. Nov 20, 2008

cideraid

Well i took the integral of just the tangential acceleration which i then get -gcos(x)+C, but now thinking about it, I need to also take into account the radial acceleration (V^2)/r?

Last edited: Nov 20, 2008
10. Nov 20, 2008

$$a=\frac{dV}{dt}=-gsin\theta$$

thus

$$V=\int-gsin\theta\, dt$$

however in order to integrate this, you need sin(theta) as a function of t
or
a as a function of theta

11. Nov 20, 2008

naresh

Deleted. Salad said all the relevant stuff.

12. Nov 20, 2008

cideraid

hmmm that clarifies somethings for me

13. Nov 20, 2008

I wish I could be of more help, but I am all tuckered out. Hopefully Dick will chime in, I notice he is online.

14. Nov 20, 2008

cideraid

thanks for everything so far though

Last edited: Nov 20, 2008
15. Nov 21, 2008

naresh

Try relating V to \theta - that way you can obtain a differential equation for \theta and hopefully be able to solve it.