Rotational Kinematics - A gymnast is performing a

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Homework Help Overview

The problem involves a gymnast performing a floor routine, where she increases her angular velocity while rotating through half a revolution. The task is to determine the time taken for this maneuver, given the initial and final angular velocities.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the appropriate use of angular displacement and the correct units for angular velocity, questioning the interpretation of revolutions versus radians.

Discussion Status

Some participants have offered guidance on unit consistency, emphasizing the importance of not mixing units. There is an ongoing exploration of the correct formulation of the equation based on the clarified understanding of angular displacement.

Contextual Notes

Participants are navigating the differences between angular displacement units (revolutions, degrees, and radians) and how these affect the calculations. There is a noted confusion regarding the definitions and conversions between these units.

crono_
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NOTE: In playing with the symbols it appears that I've made pi to be an exponent...um...so anytime you see that, please disregard it as pi is not meant to be an exponent anywhere here. Thanks!

Homework Statement



A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 2.60 to 5.40 rev/s while rotating through one half of a revolution. How much time does this maneuver take?

[tex]\omega[/tex]o = 2.60 rev/s

[tex]\omega[/tex]f = 5.40 rev/s

t = ?

[tex]\vartheta[/tex] = 2[tex]\pi[/tex] rev ---> But she only goes through 1/2 revolution. So this would be [tex]\vartheta[/tex] = 2[tex]\pi[/tex] rev / 2 . Or so I thought...

Homework Equations



I figured this equation would be appropriate since all variables, except t, are known.

[tex]\vartheta[/tex] = 1/2 ([tex]\omega[/tex]o + [tex]\omega[/tex]f) t

The Attempt at a Solution



[tex]\vartheta[/tex] = 1/2 ([tex]\omega[/tex]o + [tex]\omega[/tex]f) t

Solve for t

t = 2[tex]\vartheta[/tex] / ([tex]\omega[/tex]o + [tex]\omega[/tex]f

t = 2 (2[tex]\pi[/tex] rev / 2) / (2.60 + 5.40)

t = 6.2831 [STRIKE]rev [/STRIKE]/ 8 [STRIKE]rev[/STRIKE]/s

t = 0.785s

Buuuut...this is wrong. Any thoughts as I seem to be missing something?

Thanks!
 
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[itex]\omega_0[/itex] and [itex]\omega_f[/itex] are given in rev/s, not rad/s
 
I didn't know there was a difference...

My textbook says the following:

Angular displacement is often expressed in one of three units. The first is the familiar degree, and it is well known that there are 360 degrees in a circle. The second unit is the revolution (rev), one revolution representing one complete turn of 360°. The most useful unit from a scientific viewpoint, however, is the SI unit called the radian (rad).

I interpreted that as 360 degrees, 1 rev, and 1 rad are all different ways of saying the same thing. 360 degrees = 1 rev = 1 rad

Apparently not?
 
crono_ said:
I interpreted that as 360 degrees, 1 rev, and 1 rad are all different ways of saying the same thing. 360 degrees = 1 rev = 1 rad

Apparently not?
Nope. Note that the quoted paragraph from your text doesn't define the radian. (I suspect a later paragraph does.)

360° = 1 rev = 2pi radians.

Since your problem used revs, just stick with that. Don't mix units. The angle is 1/2 rev.
 
Hrmmm...

Okay, so then rather than putting [tex]\vartheta[/tex] = 2pi, it should just be [tex]\vartheta[/tex] = 1/2 rev.

So the equation should look like:

t = 2 (1/2 rev) / 2.60 rev/s + 5.40 rev/s

t = 0.125 s

Correct?
 
Correct!
 

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