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Solve for question 3   The θ integration fails if α is non-constant.  There is also a type of situation where the step before the integration fails, involving ω. Question 1: derive ω = αt + ω0. Start with the derivatives I began with, and derive this equation. Question 2: derive θ = 1/2αt2 + ω0t + θ0. You may use prior derived results. Question 3: derive θ = θ0 + (ω + ω0) t/2.

Solve for question 3   The θ integration fails if α is non-constant.  There is also a type of situation where the step before the integration fails, involving ω. Question 1: derive ω = αt + ω0. Start with the derivatives I began with, and derive this equation. Question 2: derive θ = 1/2αt2 + ω0t + θ0. You may use prior derived results. Question 3: derive θ = θ0 + (ω + ω0) t/2.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Solve for question 3

 

The θ integration fails if α is non-constant.  There is also a type of situation where the step before the integration fails, involving ω.

Question 1: derive ω = αt + ω0.

Start with the derivatives I began with, and derive this equation.

Question 2: derive θ = 1/2αt2 + ω0t + θ0.

You may use prior derived results.

Question 3: derive θ = θ0 + (ω + ω0) t/2.

You probably need to use prior derived results.

The Notion of Deriving Deriving an equation (or a statement) is more than just asserting it. It's showing how it must be true (and anything contrary must be false). One doesn't understand the material as long as one is limited to memorizing statements or equations, without being able to connect them. As a further, practical thing, understanding the connection makes it less likely that one would use an equation when it doesn't apply. In particular, one would avoid the "equations of constant acceleration" anytime one had non-constant acceleration. In this assignment, I assign you to derive the "equations of constant acceleration" -- all but one that I will derive. We will use 0, w, and a, satisfying these defınitions: dw and OP α W = dt dt and one special case: a is constant. (You should spot the false step when a is non-constant.) w and 0 are functions of time: w(t) and O(t). wo = w(0), and 00 = 0(0). My Example: derive w2 - wo² = 2a(0-0). dw dw do dw a = dt de dt de Multiply by dt and integrate: OP a0 + C = dw = w2 = 2a0 + 2C Substitute the initial (t=0) quantities, and subtract. The constant of integration vanishes. w3 = 2a6, + 2C w? - wz = 2a(0 – 0o) The 0 integration fails if a is non-constant. There is also a type of situation where the step before the integration fails, involving w. Question 1: derive w = at + wo- Start with the derivatives I began with, and derive this equation. Question 2: derive 0 = 1/2at2 + wot + 0o- %D You may use prior derived results. Question 3: derive 0 Oo + (w + wo) t/2. You probably need to use prior derived results.
Transcribed Image Text:The Notion of Deriving Deriving an equation (or a statement) is more than just asserting it. It's showing how it must be true (and anything contrary must be false). One doesn't understand the material as long as one is limited to memorizing statements or equations, without being able to connect them. As a further, practical thing, understanding the connection makes it less likely that one would use an equation when it doesn't apply. In particular, one would avoid the "equations of constant acceleration" anytime one had non-constant acceleration. In this assignment, I assign you to derive the "equations of constant acceleration" -- all but one that I will derive. We will use 0, w, and a, satisfying these defınitions: dw and OP α W = dt dt and one special case: a is constant. (You should spot the false step when a is non-constant.) w and 0 are functions of time: w(t) and O(t). wo = w(0), and 00 = 0(0). My Example: derive w2 - wo² = 2a(0-0). dw dw do dw a = dt de dt de Multiply by dt and integrate: OP a0 + C = dw = w2 = 2a0 + 2C Substitute the initial (t=0) quantities, and subtract. The constant of integration vanishes. w3 = 2a6, + 2C w? - wz = 2a(0 – 0o) The 0 integration fails if a is non-constant. There is also a type of situation where the step before the integration fails, involving w. Question 1: derive w = at + wo- Start with the derivatives I began with, and derive this equation. Question 2: derive 0 = 1/2at2 + wot + 0o- %D You may use prior derived results. Question 3: derive 0 Oo + (w + wo) t/2. You probably need to use prior derived results.
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