How long does it take for rotating platform to stop

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SUMMARY

The discussion centers on calculating the time it takes for a rotating platform to stop, given an initial angular velocity of \(\omega_0 = 8\pi\) rad/s and a deceleration of \(\alpha = -1.96\) rad/s². The equation used is \(0 = \omega_0 + \alpha t\), where the final angular velocity (\(\omega\)) is zero when the platform stops. The user initially misinterpreted the equation by not recognizing that the final angular velocity is indeed zero, leading to confusion about the time variable.

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Alem2000
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Hi my problem is I have a platform that is rotating and it [tex]\omega_0=8\pi[/tex] and the question is how long does it take it to stop. All my work up to here is correct and I have [tex]\sum\tau=I\alpha[/tex] where my [tex]\sum\tau=3.75Nm[/tex] and [tex]I=1.91kgm^2[/tex] and [tex]\alpha=1.96rad/s^2[/tex] so I wanted to use the [tex]\omega= \omega_0+\alpha t[/tex] equation but i don't know final angular velocity and time is my target variable. The solution manual used [tex]\omega=\alpha t[/tex] why is this justified? The [tex]\omega_0[/tex] does not equal zero so why did the solution manual just take it out?
 
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Alem2000 said:
Hi my problem is I have a platform that is rotating and it [tex]\omega_0=8\pi[/tex] and the question is how long does it take it to stop. ...
I wanted to use the [tex]\omega= \omega_0+\alpha t[/tex] equation but i don't know final angular velocity .

The final state is when the platform stops, it means does not rotate, is not the final angular velocity zero then? :smile:

ehild
 
The final [itex]\omega[/itex] is zero. So you have
[tex]0=\omega_0+\alpha t[/tex]
[itex]\alpha[/itex] is actually negative (angular speed decreasing), and you know the initial angular speed, so solve the above equation for t and you're done.
 

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