Pendulum conservation of energy question

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Homework Help Overview

The discussion revolves around a pendulum problem involving conservation of energy. A 300g ball is attached to a 1.4m string and is released from a 30-degree angle with the vertical. Participants are tasked with finding the speed of the ball at a 20-degree angle, the maximum speed, and the angle when the speed is one-third of the maximum value.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using potential and kinetic energy equations to analyze the pendulum's motion. Some attempt to calculate distances using trigonometric functions, while others question the assumptions about height and potential energy references.

Discussion Status

Some participants have offered guidance on setting up the problem and calculating energy at different points. There is an ongoing exploration of the relationship between potential and kinetic energy, with some participants noting discrepancies in calculated speeds and discussing potential rounding errors.

Contextual Notes

Participants are working under the assumption that the pendulum's height is 1.4m above the ground, but there is uncertainty about this reference point. The discussion also highlights the importance of maintaining precision in calculations to avoid errors.

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Homework Statement


To make a pendulum, a 300g ball is attached to one end of a string that has a length of 1.4m and negligible mass. (The other end of the string is fixed) The ball is pulled to one side until the string makes a 30 degree angle with the vertical, then (with the string taut) the ball is released from rest. Find a) the speed of the ball when the string makes a 20 degree angle with the vertical and b) the maximum speed of the ball. c) What is the angle between the string and the vertical when the speed of the ball is one-third is maximum value?

m=0.300kg
length=1.4m
angle=30 degrees with vertical
vinitial=0 m/s

Homework Equations


ETinitial=mgh
ETfinal=0.5mv^2 + mgh
ETinitial = ETfinal


The Attempt at a Solution


a) I tried using cos20=adj/hyp and isolated adj, and subtracted adj from 1.4 to figure out the distance from the floor to the bottom of the mass on the string. I then tried substituting that answer into the equation above and tried to isolate final velocity without much success. (The answer should be 1.42m/s)

b) I tried doing the same thing as a, except that ETfinal would not have mgh because it would be at its terminal velocity when h=0. The answer should be 1.92 m/s

c) I tried solving this by multiplying all the terms with speed by 1/3, but it requires knowledge gained from the above two questions to be solved. The answer should be 29.5 degrees

Thanks.
 
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Here is how we start.
We will take the lowest point of the path of pendulum to 0 potential energy (our reference line). (we will call the fixed point O)

Draw the pendulum at 30 degrees. From the pendulum draw a horizontal line to meet a line from O to the lowest point (where the lines meet, we will call A). The distance from O to the lowest point is 1.4m and the distance from O to the pendulum is also 1.4m.
Can you find the length OA?

EDIT: since you are offline and won't probably be online when I am again I will continue the explanation.

When you get the distance OA, you can find the Total energy at the initial position with mgh.

Now re-draw the diagram with 20 degrees and similarly find the perpendicular distance from the line through the lowest point in the path of the pendulum. You can now get the potential energy at this distance (at this angle of 20 degrees).

So the potential energy at 20 degrees + kinetic energy at 20 degrees=Total energy of pendulum.

You should now be able to get out part a)

For part b) conservation of energy says that

Total Energy = potential energy + kinetic energy and the values of potential and kinetic increase/decrease along the the path of the pendulum, at which point will all the energy be entirely kinetic?

Part c) When you find vmax, use a little backwards logic with part a)
 
Last edited:
cos 30 = adj/hyp
cos 30 = adj/1.4
1.21m = adj

1.4 - 1.21 = 0.19m off the ground

cos 20 = adj/hyp
cos 20 = adj/1.4
1.32m = adj

1.4 - 1.32 = 0.08m off the ground

mgh = 0.5mv2 + mgh
gh = 0.5v2 + gh
9.8(0.19) - 9.8(0.08) = 0.5v2
1.47 m/s = v

Although it's close, it's about 0.05 m/s off from the answer. D:
 
Complete Solution Removed

Hootenanny
 
Last edited by a moderator:
What is lcos?
 
Your method is correct. Two points, first off you assume the pendulum is hung 1.4meters above the ground, which we don't know. It does not matter either, because the only thing that matters is the difference in potential energy. If you don't see it just replace 1.4 with h. You will see it drops out.

Point 2. You are making a lot of small steps and every time you calculate you make a rounding error, this is why your answer is not correct. I would suggest you to not round up or down until the very end.
 

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