Determine the height to which the center of the baton travels

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SUMMARY

The discussion focuses on calculating the height to which the center of a baton travels when thrown straight up by a baton twirler completing 4 revolutions at an angular speed of 1.80 revolutions per second. The correct approach involves using angular kinematics, specifically the equation θ = ω₀t + 1/2(α)(t²), where the initial angular velocity (ω₀) is derived from the angular speed. The final calculated height is approximately 6.05 meters, with minor discrepancies attributed to rounding errors.

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Homework Statement


a baton twirler throws a baton straight up in the air, it goes up and comes straight back down. it completes 4 rev. ignoring air resistance and assuming the angular speed is 1.80 rev/s , determine the height to which the center of the baton travels above the point of release.


Homework Equations



i tried to use a kinematic eq. x=1/2(vnot+v)t

3. The Attempt at a Solution [/b
i tried that equation and solved for x, i put zero for vnot and 1.8 for v. i fount t to be 2.22 seconds i got my answer to be 2m but its wrong ...thanks!
 
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1.8 is the angular velocity. Time is correct. Again initial velocity is not zero. You are using the equation in linear form so you will have use linear velocity and displacement only.
 
ok so i used the formula to find angular acceleration then used that answer and put it into this...theta=(wnot)t + 1/2 (alpha)(t^2) i got 6.08 the answer in the book is 6.05m so must just be some rounding errors. thanks!
 

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