Conservation of energy of a pendulum

AI Thread Summary
The discussion centers on calculating the angle Theta for a pendulum based on its height and speed. An 80cm pendulum with a 0.60kg bob is analyzed, with the speed at the bottom of the swing being 2.8m/s. The height is calculated to be approximately 0.4m, leading to the conclusion that Theta is about 60 degrees. For part b, the same method is suggested to find the angle when the bob's speed is 1.4m/s. The approach and calculations are confirmed as correct.
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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