When given the angular velocity, why not multiply by time to find the # of revs?

  • #1

Main Question or Discussion Point

In an example question the revolutions per minute where 500rev/min. The question asked how many revolutions the disk makes in 5.5s. The solution started with the formula [tex]w=w_o+\alpha t[/tex] I'm wondering why one can't just convert rev/min into raidans/sec and multiply that number by 5.5s? kind of like x=vt

oh ya and it says the disk starts from rest
 
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Answers and Replies

  • #2
1,033
1
you definitely can.
 
  • #3
1,033
1
that formula is equivalent to [tex] v = v_o + at[/tex]
 
  • #4
Nabeshin
Science Advisor
2,205
16
Let me get this straight.. the question is as follows:
A disk starts from rest and accelerates uniformly to 500rev/min in 5.5s. How many revolutions does it make in this time?

If I'm understanding that correctly, then you cannot use your naive solution formula because angular velocity is not constant. Surely you know that x=vt only applies for constant velocity, and the situation is completely analogous for angular motion. Now, knowing what I just told you, how would you solve the problem if it were posed in terms of translation, and then you should be able to easily transfer to rotation (or just think purely in rotation, if you can do that).
 
  • #5
got it now...I miss understood the question and thought that there was no acceleration:rolleyes:
so let me get this straight:
whenever something is rotating there's going to be angular acceleration right? so that means I will NEVER use a formula like x=wt where w is the angular velocity and x is the distance travelled?
 
  • #6
Doc Al
Mentor
44,882
1,129
so let me get this straight:
whenever something is rotating there's going to be angular acceleration right? so that means I will NEVER use a formula like x=wt where w is the angular velocity and x is the distance travelled?
No. Of course you can have something rotating with constant angular speed.

Just like with translational motion, you can have constant velocity or accelerated motion. It depends on the problem.
 

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