Velocity of a Pendulum

[Dazz4C]

Okay, so I am a student in my first year of physics and I'm doing an EEI on collision; however I am stuck up to a part where I need to figure out the velocity of my pendulum without doing the experiments again (Long story). At the moment, the length of the chain/string is 55cm and the angle is at 90 degrees.

I have also done some googling, and some have ended leading me to here. Is this formula suitable to find the velocity?

v = √{2gL[1-cos(a)]}


I applied my known data into the equation which ended up giving me an answer of about 3.8m/s. I personally suspect that it is incorrect, so I need some experts to help me. (Sorry if i sound like a dag lol.)

[tiny-tim]

v = √{2gL[1-cos(a)]}

Hi Dazz4C! Welcome to PF! :smile:

Is your pendulum a string with a point mass on the end, or does it have a large mass and/or something heavy instead of a string?

Your equation (with a being the angle from the vertical) comes from conservation of energy: KE + PE = constant.

For a point mass, KE/m = 1/2 v2, and PE/m = gL(1 - cosa), and your equation is correct.

But for a large mass, or for a bar instead of a string, KE/m and PE/m will be different. :smile:

[Dazz4C]

Hi Dazz4C! Welcome to PF! :smile:

Is your pendulum a string with a point mass on the end, or does it have a large mass and/or something heavy instead of a string?

Your equation (with a being the angle from the vertical) comes from conservation of energy: KE + PE = constant.

For a point mass, KE/m = 1/2 v2, and PE/m = gL(1 - cosa), and your equation is correct.

But for a large mass, or for a bar instead of a string, KE/m and PE/m will be different. :smile:

It has a rectangular piece of wood (10cm x 4cm) @ 70g with a cushion, that weighs anywhere from 2g-10g. (It's simulating the crumple zone :P)

*EDIT: Sorry, so which formula would I use?

[tiny-tim]

It has a rectangular piece of wood (10cm x 4cm) @ 70g with a cushion, that weighs anywhere from 2g-10g. (It's simulating the crumple zone :P)

is the wood instead of a string (and if so, where is the pivot?), or is it on the end of a string?

in any case, you'll need to use the moment of inertia of a rectangle.

Before I go any further, do you know what moment of inertia is, and how to use it to calculate energy? :smile:

[Dazz4C]

is the wood instead of a string (and if so, where is the pivot?), or is it on the end of a string?

in any case, you'll need to use the moment of inertia of a rectangle.

Before I go any further, do you know what moment of inertia is, and how to use it to calculate energy? :smile:

No idea :(

I don't really understand what you mean by where is the wood.

So here an illustration.
http://i40.tinypic.com/vzvd3t.jpg

[tiny-tim]

oh i see …

ok, if that long straight line is string, then your original equation should work,

with L being the length of the string plus half the height of the block.

(you originally mentioned a chain)

[Dazz4C]

oh i see …

ok, if that long straight line is string, then your original equation should work,

with L being the length of the string plus half the height of the block.

(you originally mentioned a chain)

Ah, yes thankyou. When I meant chain; I kinda meant the weightless support for the bob. Didn't know how to express it.

Thankyou again

[kbaumen]

There is also another way of doing it.

Since


T=2 \pi \sqrt{\frac{l}{g}}


where, T-the period, l-length of the string, we can write a displacement equation (I'm not sure that's the correct term in English, but I hope you'll understand what I mean from the math).


x=A \cos ( \omega t) (1)


where

\omega = \frac{2\pi}{T}

and A - amplitude. Now differentiate (1) and you'll have a velocity equation.

[Dazz4C]

There is also another way of doing it.

Since


T=2 \pi \sqrt{\frac{l}{g}}


where, T-the period, l-length of the string, we can write a displacement equation (I'm not sure that's the correct term in English, but I hope you'll understand what I mean from the math).


x=A \cos ( \omega t) (1)


where

\omega = \frac{2\pi}{T}

and A - amplitude. Now differentiate it and you'll have a velocity equation.
I'll stick to the original equation; it's probably a bit easier to understand. But thankyou for helping aswell.