Angular Velocity Homework: Calculating Higher Derivatives

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Homework Help Overview

The problem involves calculating angular velocity and acceleration based on angular displacement data of the knee during cycling, specifically at an angle of 119.1 degrees. The data is sampled at a frequency of 50 Hz, and participants are exploring the implications of calculating higher derivatives from this sampled data.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between frequency and discrete time variables, questioning how to apply the given frequency to the calculations. There is an exploration of how to derive time intervals from the sampling frequency and how to use these intervals to compute changes in angular displacement.

Discussion Status

Some participants have provided insights on how to approach the calculations, particularly regarding the time intervals between measurements. There is ongoing exploration of the implications of these calculations, with no explicit consensus reached on the final approach or results.

Contextual Notes

Participants note the challenge of calculating higher derivatives from sampled data, which may involve issues related to the accuracy and resolution of the measurements. The original poster expresses uncertainty about the discrete time variable, which is crucial for the calculations.

clokey34
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Homework Statement



The following data are for the angular displacement of the knee in cycling
measured in degrees. Using radian measure, compute the angular velocity
and acceleration when the knee angle is 119.1 degrees. The sampling
frequency was 50 Hz. What is a problem you may encounter when calculating
higher derivatives from sampled displacement data?
98.0, 103.5, 109.0, 114.2, 119.1, 123.4, 126.9


Homework Equations



Angular velocity is degrees per unit time.


The Attempt at a Solution



I only have the angle. And a frequency but not a discrete time variable? Thanks!
 
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clokey34 said:
And a frequency but not a discrete time variable?
What is the definition of frequency? :wink:
 
Yeah I agree it is 50 cycles a second but then I'm not sure how to apply that?
 
clokey34 said:
Yeah I agree it is 50 cycles a second but then I'm not sure how to apply that?
So, you know that every second there are fifty measurements taken. Equally, you know that each measurement is separated by 1/50 = 0.02 seconds.

Can you go from here?
 
Oh right you make it so simple! Thanks.

So change in velocity over change in time. Change in time is 0.02sec

change before
119.1 -114.2 = 4.9deg
4.9/0.02=245deg

change after
123.4 - 119.1 = 4.3deg
4.3/0.02=215deg

(245+215)/2= 230deg.sec-1

Accel= change of rate of velocity is (230-245)/2
=-1500ms-2
 

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