Conservation of energy of a pendulum

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SUMMARY

The discussion centers on the conservation of energy in a pendulum system, specifically an 80cm long pendulum with a 0.60kg bob. The initial angle Theta is calculated using the equation Ui + Ki = Uf + Kf, leading to a height (h) of approximately 0.4m. This results in Theta being 60 degrees when the bob reaches a speed of 2.8m/s. For part b, the same energy conservation principles apply to find the angle when the bob's speed is 1.4m/s.

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maniacp08
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An 80cm long pendulum with a .60kg bob is released from rest at an initial angle of Theta with the vertical. At the bottom of the swing, the speed of the bob is 2.8m/s.

a)what is Theta?
b)What angle does the pendulum makes with the vertical when the speed of the bob is 1.4m/s?Revelant equation:
Ui + Ki = Uf + Kf

I converted 80cm to .8m.
mgh + 0 = 0 + 1/2 m v^2
.60kg * 9.81 m/s * h = 1/2 .60kg * (2.8m/s)^2
I need to solve for h which is .3995922528 approx .4m

Length is .8m
h = L - Lcos Theta
.4m = .8m - Lcos Theta
.5 = Cos Theta
Cos^-1 = 60Is this correct?
and Part b is the same procedure correct?
 
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maniacp08 said:
An 80cm long pendulum with a .60kg bob is released from rest at an initial angle of Theta with the vertical. At the bottom of the swing, the speed of the bob is 2.8m/s.

a)what is Theta?
b)What angle does the pendulum makes with the vertical when the speed of the bob is 1.4m/s?Revelant equation:
Ui + Ki = Uf + Kf

I converted 80cm to .8m.
mgh + 0 = 0 + 1/2 m v^2
.60kg * 9.81 m/s * h = 1/2 .60kg * (2.8m/s)^2
I need to solve for h which is .3995922528 approx .4m

Length is .8m
h = L - Lcos Theta
.4m = .8m - Lcos Theta
.5 = Cos Theta
Cos^-1 = 60Is this correct?
and Part b is the same procedure correct?

Looks OK so far.
And you can solve part b in a similar way.
 

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