Angular Acceleration without Angular Speed.

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Homework Help Overview

The problem involves determining the angular acceleration of a wheel with a given angular position function, specifically theta = 3.0 - 2.0t^3. The context is within rotational motion and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use a kinematic equation to find angular acceleration but questions how to derive angular speed from the given position function. Some participants suggest using the definition of angular acceleration as the second derivative of angular position.

Discussion Status

Participants are exploring different methods to find angular acceleration, with some indicating that the kinematic equations may not apply due to non-constant acceleration. There is a recognition of a simpler method to find the answer, although there is no explicit consensus on the final value.

Contextual Notes

There is a discussion about the applicability of kinematic equations, noting that they are derived under the assumption of constant acceleration, which is questioned in this scenario.

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Homework Statement


A wheel rotating about a fixed axis has an angular position given by theta= 3.0 - 2.0t^3, where theta is measured in radians and t in seconds. What is the angular acceleration of the wheel at t = 2.0s?

Homework Equations


wf = w + at

The Attempt at a Solution


I realize I most likely should use that kinematic equation to find the angular acceleration, but how would I find the angular speed? dtheta/dt? If so, is it like -6t^2/2?
Assuming w is just -6t^2, then I then got -6t^2=at. Which should be -12=a but I don't see that as an answer. Closest would be -24 rad/s^2.
 
Last edited:
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Just look up the definition of angular acceleration:

[tex]\alpha = \frac{d^2\theta}{dt^2}[/tex]
 
zachzach said:
Just look up the definition of angular acceleration:

[tex]\alpha = \frac{d^2\theta}{dt^2}[/tex]

Hmm so it would be -24 rad/s2 wouldn't it?
 
That's what I got.
 
zachzach said:
That's what I got.

Alright thankyou. I was set on using the kinematic equation and completely overlooked the simpler method.
 
Also, the kinematic equations only work when there is constant acceleration (that is how they are derived). As you can see in this case the acceleration is NOT constant.
 
Thankyou, I'll keep that in mind. I appreciate it.
 

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