LHC@home

Hello, I am really stuck on my physics coursework and was wondering if you where able to help me.
I have to calculate gravity using a compound pendulum, but i have 2 values for h because h1+h2=L
i worked out h by doing h(squared)- Lh +k(squared)=0

i am using the formula T= 2(pie)(Square root) (k (squared)+(h squares)/ gh)

Im stuck please help

Jabeen

Hello, I am really stuck on my physics coursework and was wondering if you where able to help me.
I have to calculate gravity using a compound pendulum, but i have 2 values for h because h1+h2=L
i worked out h by doing h(squared)- Lh +k(squared)=0

i am using the formula T= 2(pie)(Square root) (k (squared)+(h squares)/ gh)

Im stuck please help

Jabeen

If a=b
then does a squared = b squared ?

I think squaring both sides will help get rid of that scary square root sign.

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I'm not the LHC Alex. Just a number cruncher like everyone else here.

yeah i done that and made g the subject of the formula but i dont understand wat h1 and h2 are

Excellent place to ask such a question.

A simple pendulum has its mass concentrated at a point called the bob. Although the definition of a point is a position without dimension, a pendulum is considered to be simple if the mass of the supporting device (string) is negligible in comparison to that of the bob, and the supporting device is long in comparison to the diameter of the bob.

When displacement of a simple pendulum is less than 5 degrees, the period of simple harmonic motion can be describe by:

T=2*Pi*(L/g)^(1/2)

Where T is the period and L is the length of the pendulum.

A rigid body with distributed mass able to freely pivot about a horizontal axis 'O' which does not coincide with the centre of gravity (C) is called a compound pendulum.

For rotary motion of any rigid body about a fixed axis, the angular acceleration is equal to the torque (T) about O divided by the moment of inertia (I) about O. The moment of inertia (I) of a compound pendulum, about O, is given by:

I_O = m*(k_O)^2, where k_O is the length of the pendulum.

From the parallel axis theorem, the moment of inertia about C is:

I_C = m * ((k_C)^2 + h^2), where k_C is the radius of gyration about C, and h is the length of OC

if m*k_O = -mgh sin theta, then

theta" = -(gh/k_O^2) sin theta

and for small theta, sin theta ~ theta, so

theta" + (gh/k_O) theta = 0

if Wn^2 = gh/k_O and 2Pi/Wn = t, then

t = 2Pi * SQRT(k_O^2/gh) = 2Pi * SQRT ((k_C^2 + h^2)/gh)
The center of oscilation (P), also lies on the line OC, and is some distance from C, where L>h is the length of OP and can be determined by I/mh. P is reciprocal to O, and if the pendulum was suspended from P, then O would be the center of oscillation.

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