What is the Speed and Total Mechanical Energy of a Simple Pendulum?

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[SOLVED] Pendulum Speed and total ME

Homework Statement



The length of a simple pendulum is 0.84 m and the mass of the particle (the "bob") at the end of the cable is 0.68 kg. The pendulum is pulled away from its equilibrium position by an angle of 7.8 ° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

Homework Equations



E = .5mv^2+.5Iw^2+mgh+.5kx^2
w = sqrt(g/L)
L-Lcos@ = max height of pendulum bob
I of pendulum = (1/3)ML^2

The Attempt at a Solution



so w = 3.416 rad/s
The total energy:
.5(.68)(v^2) + .5(1/3 ML^2)(w^2) + mgh + .5kx^2
kx^2 is potential energy, which is 0 at the bottom of the swing
.34v^2 + (1/6)(.68*.84^2)(3.416^2) + .68*-9.8*.00777
.34v^2 + .88137
v^2 = 2.5922
v = 1.61 m/s

Plugging this in gives a total ME of 1.7627 J

The v and total E are wrong, but I'm not sure how to remedy this. Ideas, please? Thanks!
 
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Whao, slow down there, your using way too many equations! When the plumb bob is at it's highest point, what type of energy does it have?
 
It has potential energy, mgh. But the question asks for total mechanical energy at the bob's LOWEST point.
 
thatgirlyouknow said:
It has potential energy, mgh. But the question asks for total mechanical energy at the bob's LOWEST point.
Yes, but if you know that at it's highest point it only has potential energy and you know the energy is conserved, then what is the value of the total energy at the lowest point?
 
So I totally overanalyzed that one... oh well.

Thanks for the help!
 
thatgirlyouknow said:
So I totally overanalyzed that one... oh well.

Thanks for the help!
No worries :approve: