Calculating Angular Speed of a Shaft with Given Angular Acceleration

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Homework Help Overview

The problem involves calculating the angular speed of a shaft given its initial angular speed and a time-dependent angular acceleration. The context is within the subject area of rotational motion in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to integrate the angular acceleration to find the angular speed. There are questions about the correct setup for integration and the interpretation of the equations involved.

Discussion Status

The discussion is ongoing, with some participants providing guidance on the integration process while others express confusion about the dimensional correctness of certain expressions. There is a mix of interpretations regarding the approach to solving the problem.

Contextual Notes

There is an indication of uncertainty regarding the proper application of integration limits and the relationship between angular acceleration and angular speed. Participants are navigating through the problem without a clear consensus on the method.

Chuck 86
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Homework Statement


A shaft is turning at 73.8 rad/s at time zero. Thereafter, its angular acceleration is given by
a = -10.1 rad/s2 - 4.44 t rad/s3.
where t is the elapsed time. Calculate its angular speed at t = 2.58 s

Homework Equations


1/2(Angular accel formula)+ 73.8t+0= theta final


The Attempt at a Solution


do i take the integral of the accel formula then the velocity the plug in 2.58?
 
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You need to integrate both sides of

dω = αdt

The limits of integration should be

73.8 to ω (left side) and 0 to 2.58 (right side)

After you integrate, solve for ω.
 
so i set 1/2(angular accel)= 73.8+ theta final? then take the integral?
 
so i set 1/2(angular accel)= 73.8+ theta final? then take the integral?
I never said that. I have no idea where you get that 1/2(angular accel) = 73.8 + theta final means something when it is not even dimensionally correct. Do the following integral and solve for ω. That's what I suggested in the first place.

\int_{73.8}^{\omega}d \omega=\int_{0}^{2.58}(-10.1-4.44 t) dt
 
whoa dude calm down. i appericiate the help but remember i was just asking a question, i already feel stupid enough as it is
 

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