- #1
Calculating Kinetic Energy of a Segment: Where Did I Go Wrong?
- Thread starter Saitama
- Start date
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- Tags
- Energy Kinetic Kinetic energy
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Homework Help Overview
The discussion revolves around calculating the kinetic energy of a segment, with participants exploring the setup and evaluation of integrals related to angular motion.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the correctness of initial assumptions regarding the angle measurement and the evaluation of integrals. There are suggestions to consider alternative approaches involving angular speed and distance from a point of contact.
Discussion Status
Participants are actively engaging with each other's attempts, providing corrections and alternative methods. Some have acknowledged errors in their calculations and are seeking clarification on integration limits, indicating a productive exchange of ideas.
Contextual Notes
There is a mention of an attachment with the original problem statement, but its content is not detailed in the discussion. The conversation also reflects a focus on specific mathematical details and assumptions about angle measurement.
- #2
- 42,893
- 10,527
Your first line is wrong. How did you get 1+sin(θ)? You're measuring it down fron horizontal, so should be minus.
And in the last step, you've evaluated the integral of the sin(θ) part to zero. It won't be.
Another approach is to say the angular speed is v/R and work out the distance from the element to the point of contact.
And in the last step, you've evaluated the integral of the sin(θ) part to zero. It won't be.
Another approach is to say the angular speed is v/R and work out the distance from the element to the point of contact.
- #3
Saitama
- 4,244
- 93
Oh yes, the integral sin(θ) won't be zero. I evaluated it correctly during the examination but did it wrong while posting up my attempt.haruspex said:
I did the necessary corrections as you said and got the right answer. One more question, would the limits for integration change from 0 to -pi?
This approach is much easier. I am not much familiar with the instantaneous axis of rotation but this worked out pretty well. Thanks for the alternate method.haruspex said:
- #4
- 42,893
- 10,527
No, you're measuring theta anticlockwise in the diagram from the "9 o'clock" position, so it will be 0 to pi.Pranav-Arora said:
- #5
Saitama
- 4,244
- 93
haruspex said:
Got it, thanks haruspex!