Need confirmation with my way of solution.

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Homework Help Overview

The discussion revolves around a problem related to rotational motion, specifically involving angular velocity, angular acceleration, and the calculation of revolutions before coming to a stop. Participants are examining the original poster's approach to the problem and the calculations provided in attachments.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessity of converting units to radians per second versus keeping them in revolutions per minute. There are attempts to clarify the calculations related to angular acceleration and the time taken to stop. Questions arise regarding the correctness of the calculations and the interpretation of the results.

Discussion Status

Some participants have provided feedback on the calculations, indicating errors and suggesting a more straightforward approach using revolutions. There is acknowledgment of mistakes in the original poster's calculations, and some guidance has been offered regarding the interpretation of the problem.

Contextual Notes

Participants are working with specific units of revolutions per minute and are discussing the implications of these units on the calculations. There is mention of attachments containing calculations, which are referenced but not detailed in the discussion.

Sanosuke Sagara
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I am facing problem on question regarding rotational motion.I have my solution and my question in the attachment that followed.
 

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Firstly, let me comment that you don't need to do all that extra work converting everything into rad/s. You can simply leave things in rev/min, as long as you're consistent. The 1.3 answer is the time in minutes for it to come to a rest, not the number of revolutions. Assuming constant angular acceleration:

[tex]\alpha = \frac{\Delta \omega}{\Delta t} = \frac{120\textrm{rev/min} - 150\textrm{rev/min}}{1/3\textrm{min}} = -90 \frac{\textrm{rev}}{\textrm{min}^2}[/tex]

where the negative rate makes sense as it's slowing down. Now, to compute the time it takes to come to a stop, assume it's angular position and time is 0 after then end of that first 20 seconds. Then,

[tex]\omega = \omega_0 + \alpha t \implies 0 = 120\frac{\textrm{rev}}{\textrm{min}} + -90\frac{\textrm{rev}}{\textrm{min}^2}t \implies t=\frac{120\frac{\textrm{rev}}{\textrm{min}}}{-90\frac{\textrm{rev}}{\textrm{min}^2}} \implies t=1.\overline{3} \textrm{min}[/tex]

Now, if it takes [itex]1.\overline{3}[/itex]minutes to stop from an initial angular velocity of 120 rev/min at a constant acceleration of -90 rev/min/min, then how many revoutions will it take to stop?
 
Confirm with the number of revolutions made

I have my calculation in the attachment that followed.Do comment if I have any calculation error.
 

Attachments

Sanosuke Sagara said:
I have my calculation in the attachment that followed.Do comment if I have any calculation error.

Nope,it's wrong.The last 2n3 lines of your document are wrong.Using these special units (revolutions/minute) will defintely enable u to get to the answer right away.That's because the result for [itex]\theta[/itex] is the final answer,as it already put in terms of revolutions.Atually the angle is 80revolutions*2\pi radians/revolution approx 500 radians.
So the final answer to yor question is 80 revolutions.This is the number the problem is waiting from u.

Daniel.
 
Thanks for your confirmation and mark with my mistake.Now I have understand what the question want.Thanks again.
 

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