How Do You Calculate Initial Angular Velocity with Constant Acceleration?

Click For Summary

Homework Help Overview

The problem involves calculating the initial angular velocity of a flywheel with a constant angular acceleration of 2 rad/sec² over a time period of 19 seconds, during which it rotates through an angle of 15 radians.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the angular motion equation and question the correctness of the original poster's equation setup. There are attempts to derive the initial angular velocity based on the given parameters.

Discussion Status

Some participants have provided feedback on the calculations, with one suggesting a possible miscalculation and another proposing an alternative form of the angular motion equation. Multiple interpretations of the equation are being explored.

Contextual Notes

There is a noted confusion regarding the correct form of the angular motion equation, with participants referencing similarities to linear motion equations. The discussion reflects uncertainty about the initial conditions and the setup of the problem.

blue5t1053
Messages
23
Reaction score
1
Question:
A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1?

Hint: let t1=0 sec, and t2=t

Equations:
[tex]\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}[/tex]

My Work:
[tex](15 radians) - (0 radians) = \omega_{0} (19 sec) + (2 \frac{rad}{sec^{2}})(19 sec )^{2}[/tex]

[tex](15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2} = \omega_{0} (19 sec)[/tex]

[tex]\frac{(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2}}{(19 sec)} = \omega_{0}[/tex]

[tex]\omega_{0} = (-18.2)\frac{rad}{sec} = (18.2)\frac{rad}{sec} \ for \ magnitude; \ at \ t1[/tex]

Did I do everything right?
 
Last edited:
Physics news on Phys.org
blue5t1053 said:
Question:
A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1?

Hint: let t1=0 sec, and t2=t

Equations:
[tex]\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}[/tex]

My Work:
[tex](15 radians) - (0 radians) = \omega_{0} (19 sec) + (2 \frac{rad}{sec^{2}})(19 sec )^{2}[/tex]

[tex](15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2} = \omega_{0} (19 sec)[/tex]

[tex]\frac{(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2}}{(19 sec)} = \omega_{0}[/tex]

You're good up to here.

[tex]\omega_{0} = (-18.2)\frac{rad}{sec} = (18.2)\frac{rad}{sec} \ for \ magnitude; \ at \ t1[/tex]

You've miscalculated. You should get something close to 37 rad/s for the magnitude.
 
I think your equation is wrong

Should it not be

theta(final) - theta(initial) = time * angular velocity(initial) + (1/2) * angular acceleration * time^2.

similar to the equation in linear motion?
 
manjuvenamma said:
I think your equation is wrong
You are right.
blue5t1053 said:
Equations:
[tex]\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}[/tex]
That should be:
[tex]\vartheta - \vartheta_{0} = \omega_{0} t + (1/2)\alpha t ^{2}[/tex]
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
1K
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
2K