Rotational Kinematics

  • Thread starter PhysicsDud
  • Start date
  • #1
24
0
Just wanted to check to ensure I have calculated the correct answers for the following question:

A rotating wheel accelerates uniformly at
6.2 rad/s2, and completes 25 revolutions
during a 5.0 s interval. (a) What is the
angular velocity of the wheel at the start of
the 5.0 s interval? (b) What is the angular
velocity of the wheel at the end of the 5.0
s interval?

A) 31.4 rad/s

B) 157 rad
 

Answers and Replies

  • #2
Tom Mattson
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How about showing your work? As it stands right now, someone has to sit down and completely solve the problem to answer you. If you show your work we can tell very quickly if you've made an error, and we can tell you where it is.
 
  • #3
24
0
My work

1 revolution = 360 degrees = 2 x pie rad = 6.28 rad
25 revolutions = (25 rev) (2 x pie rad/rev) = 50 x pie rad = 157 rad

omega = 157 rad/ 5.0 s = 31.4 rad/s

Therefore, the angular velocity of the wheel at the start of the 5.0 s interval was 31.4 rad/s

And angular velocity at the end of 5.0 s interval was 157 rad.
 
  • #4
Hootenanny
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Looks ok to me.
 
  • #5
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it doesn't look ok to me.

you've just found the average angular velocity, while the question is asking for the initial (and final) angular velocity

in addition, angular velocity is in radians per second. you can't have an angular velocity of 157 radians.
 
  • #6
177
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You have got to use the calculus. But you can solve it graphically by plotting your angular velocity versus time, it is linear sincs ang. acc. is constant, and finding the area under the change in w, 31 rad/s, in the 5 seconds.

This area equals the radians rotated. If this amount is less than the amount rotated, then there needs to be some initial ang. vel., which will be multiplied by the 5s for the difference.

I got 15.9 rad/s initial, add 31 for 46.9 final.
 
  • #7
24
0
Confused

Just wondering how you came up with 15.9rad/s??
 
  • #8
217
0
[tex] \theta = \omega_o(t) + 0.5 \alpha t^2 [/tex]
 

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