Calculating Angular Speed of a Shaft with Given Angular Acceleration

AI Thread Summary
To calculate the angular speed of a shaft at t = 2.58 seconds, the initial angular speed is 73.8 rad/s, and the angular acceleration is defined as a = -10.1 rad/s² - 4.44t rad/s³. The correct approach involves integrating the angular acceleration to find the change in angular velocity over the given time interval. The integration limits should be set from the initial angular speed to the final angular speed, and from 0 to 2.58 seconds for time. The discussion highlights the importance of proper dimensional analysis and integration techniques in solving the problem.
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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