Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
8
Rotational Motion
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8-1: Angular Quantities
8-2 and 8-3: Constant Angular Acceleration; Rolling
8-4: Torque
8-5 and 8-6: Rotational Dynamics
8-7: Rotational Kinetic Energy
8-8: Angular Momentum
8-9: Angular Quantities as Vectors

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 18
Q

A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?

A
3.0×104 rev3.0 \times 10^4 \textrm{ rev}
Giancoli 7th Edition, Chapter 8, Problem 18 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The angular displacement of this centrifuge is the average angular velocity multiplied by time. So we have final angular velocity plus initial angular velocity over 2 multiplied by time and it starts from rest so ω i is 0. Well, we have to do some work with units here because we are given revolutions per minute for the final angular velocity of 15000 revolutions per minute. So we are taking 15000, dividing it by 2— since ω i is 0, it's just 15000 for ω f— divide by 2 and then multiply that by 4 minutes and I converted this 240 seconds into minutes by multiplying by 1 minute for every 60 seconds so that the revolutions per minute multiplied by minutes leaves us with revolutions and so we have 3.0 times 10 to the 4 revolutions.

COMMENTS
By carolsilber on Thu, 5/5/2016 - 12:55 AM

wouldnt it be 3.0x10^5 because the answer is 30,000

By Mr. Dychko on Tue, 5/10/2016 - 5:25 AM

Hi carolsilber, thanks for asking. 30,00030,000 is written as 3.0×1043.0 \times 10^4, so it doesn't look like there's any problem here. Let me know if you suspect any other errors though.

All the best,
Mr. Dychko

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