Conservation of Energy,pendulum question

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A pendulum with a 50 cm string is released from a 30° angle, and the problem involves calculating its speed at the lowest point using energy conservation principles. The initial height is determined using trigonometry, where the vertical component of the string is calculated, resulting in a height difference of 0.0670 m. By applying the conservation of energy, the potential energy at the top equals the kinetic energy at the bottom, leading to the equation mgh = 1/2mv^2. After simplifying and solving, the calculated speed at the bottom is approximately 1.15 m/s. The solution confirms that mass cancels out in this scenario, validating the approach taken.
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Homework Statement



A pendulum consists of a mass on a 50 cm long string. It is being held stationary at a 30°angle with the vertical. After release, how fast will it be going at the bottom? (Hint: Use trigonometry to find how high the ball is initally)


Homework Equations



Ek = 1/2mv^2 for sure to find the velocity
Eg = mgh


The Attempt at a Solution


I'm basically brain-dead on this question, I don't know where to start.
 
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Draw a diagram first, that always helps.

Hint: Trigonometry, you're going to have to consider cos of the angle in this situation. It's like a force, you need to consider the vertical component of the string. In forces that would be the vertical force, but here it's the length.
 
Last edited:
Yes, draw a diagram with both the initial position and the final postion(bottom-most).
Then find the height difference between the two positions. It will be a function of the angle.
And then apply energy conservation.
 
So basically I've had a go on the question with the help of the previous posts:
[PLAIN]http://img532.imageshack.us/img532/6052/physicsi.png

So to find the left side of the triangle (lets say n):

50 x cos30 = n
n = 43.3

The string is 50cm
50 - 43.3 = 6.70 cm
x = 6.70cm
=0.0670m

Since Ep @ top = Ek @ bottom
m x 9.80 x 0.0670 = 1/2mv^2, m's cancel out I think?

0.6566 = 1/2v^2
sqrt 1.3132 = sqrt v^2
1.15 m/s = v

I'm not sure about the significant digits, but this is how I tried to solve the question, can anyone confirm the answer?
 
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Assuming your arithmetic is all correct that looks to be correct, and yes for this relationship your mass is negligible.

Joe
 
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