Angular Acceleration of a Flywheel Lab

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Homework Help Overview

The discussion revolves around the angular acceleration of a flywheel in a lab setting, focusing on the relationships between forces, torques, and accelerations in a system involving a weight and a flywheel. Participants are exploring concepts related to free body diagrams, torque, moment of inertia, and the effects of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the construction of free body diagrams for both the flywheel and the weight, questioning the direction of forces such as tension and gravity.
  • There is an exploration of the relationships between angular and linear accelerations, with attempts to derive equations connecting them.
  • Some participants are unsure about how to relate linear acceleration to angular acceleration and are seeking clarification on the implications of friction in the system.

Discussion Status

The discussion is active, with participants providing feedback on each other's free body diagrams and reasoning. Some guidance has been offered regarding the relationships between forces and accelerations, and there is a collaborative effort to clarify the connections between different physical quantities.

Contextual Notes

Participants are working under the assumption of no friction for initial questions, but later considerations include the effects of friction on the system. There is an emphasis on understanding the relationships between various forces and accelerations without reaching a final conclusion.

Shelby8
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Homework Statement


Assume no friction for 1 - 6
1. Draw a free body diagram of the fly wheel (from above), and a free body diagram of the weight (from side).

2. What force appears in both diagrams?

3. What is the relationship between the torque on the flywheel and the tension in the string?

4. What is the relationship between the angular acceleration of the flywheel and the linear acceleration of the weight?

5. What is the formula for the moment of inertia of a solid disk?

6. Based on 1 - 5, write an equation that predicts angular acceleration of the flywheel, given mass and radius and g.

7. Suppose there is friction between fly wheel and stand that holds the fly wheel, so eventually it stops. As the weight drops, will the angular acceleration of the fly wheel be larger than, smaller than, that predicted by the equation in 6?

Homework Equations


angular acceleration = change in angular velocity / change in time
angular momentum = moment of inertia X angular velocity
lever arm = r X sin(theta) (perpendicular distance between axis of rotation and line of action of force) See image
moment of inertia = m X r^2
torque = r X F X sin(theta)[/B]

The Attempt at a Solution


1. See image

2. Tension

3. Torque = radius X Tension (because tension is the force)

4. distance from rest (d) = 1/2 a t^2 --> a = 2 x d / t^2
Not sure how to relate this to angular acceleration?

5. I = M x R^2 / 2

6. Can't answer without #4..

7. Same as 6
 

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Check your FBD of the flywheel from above. Which way does gravity act?

I don't see your FBD for the weight.
 
For 4, we can take it in stages. First, if the wheel turns through an angke ##\theta##, how far does the weight move?
Next, if the wheel rotates at a rate ##\dot \theta##, how fast does the weight move? Finally, Q4.
 
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The mg should act down while the T acts toward the side, I see how I drew that wrong. Am I missing other forces for the flywheel though?

If the wheel turns with theta, like if it turns one rotation (2 pi) then the weight moves the same distance as the circumference of the flywheel? (2 pi r)?
Would the speed be the relation between angular acceleration and linear acceleration? Because F net of the weight = T - mg = m (linear a), and a = α * r.
Is that how to relate them?

so a = (T - mg) / m = α * r ?
 
CWatters said:
Check your FBD of the flywheel from above. Which way does gravity act?

I don't see your FBD for the weight.
 

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Shelby8 said:
so a = (T - mg) / m = αr
Yes, except which way are you taking as positive for the acceleration? Will T be more or less than mg?
 
haruspex said:
Yes, except which way are you taking as positive for the acceleration? Will T be more or less than mg?
Oh the weight is moving down so the down is positive. So rewrite it like: a = (mg - T) / m = αr
 
Shelby8 said:
Oh the weight is moving down so the down is positive. So rewrite it like: a = (mg - T) / m = αr
Ok. Can you attempt 6 now?
 

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