How can I find the velocity of the pendulum at the center of the system?

Question

See picture:

http://i15.tinypic.com/2qxny1t.jpg

you guys have given good answers, i do not know the height, i have to remeasure when i get back into the lab (my group forgot to record that info). what im not sure of, however is if i use the PE from the distance it is above the ground or the PEinitial? if i use the height it is above the ground in the middle, wouldnt that be neglecting the energy it already has from being dropped?

okay, so taking into consideration all of your answers, would this be right? ( i know i dont have enough info)

PEmiddle=(0.066kg)(9.81m/s^2)(height)
=unknown value
1/2(0.066 kg)v^2=unknown value

i can plug in the values later, but is that the right setup?

Answer 1

Use mgh for the pendulum at rest at the highest point, ( in other words at one end)

m = mass of the "bobble" at the end of the pendulum

g = force due to gravity

h = height of bobble above where it is at the centre

the mgh above is the potential energy of the bobble due to its height (it is at rest for a moment as it changes direction.

At the centre it no longer has any potential energy, as it is at its lowest point and potential energy is due to height.

All its energy at the centre is kinetic energy

this is 1/2 m v^2

since energy is neither created nor destroyed ( at least for the purposes of questions like this)

mgh = 1/2mv^2

Divide each side by m (that gets rid of the bobble).

The question will have told you the height (or enough about the circle for you to find it)

It's not a very clear diagram. It looks to me as if the height is 50cm = 1/2 metre

good luck!

Answer 2

Energy
Use the physics equation for energy where mgh+.5mv^2=E total
Since no energy is gained or lost by the system then all of the potential energy at the top of arc is represented by mgh and the velocity at the low point in the arc is represented by the kinetic energy. Hence mgh=.5mv^2 where mass cancels because it is on both sides this yields gh=.5v^2 and v^2=gh/.5 and further v=the square root of gh/.5 but assuming g is on earth and it equals 9.8 meters/sec^2 then actually v=the square root of 19.6h where h is the difference in height of the pendulum at the top of the arc and to the bottom of the arc. The height is only need from one side because when the pendulum passes the bottom point in the arc it again starts to convert the kinetic energy back into potential.

OK now using the max height of .5 meters at the top of the arc we need to calculate the angle theta to determine the lowest point in the arc but it does not look like the right information is there to calculate the delta h. If the distance from the center line to the pendulum at max height where known then the angle theta can be found and then the subsequent delta h can be found and the height could be solved. (Unless i am misinterpreting something in the diagram!) and if you could clarify it i will be happy to finish the solution.

Answer 3

they have given long solution go by my method

at the centre whole of p.e(mgh) get converted 2 k.e(1/2mv^2)
hence eqate both

mgh=1/2mv*v
giving v=(2gh)^1/2
apply values in this and it comes = 2.3m/s
ur detail is not very clear
but how can u neglect energy it was given by u to the pendulum which is converted to k.E