
#1
Mar1607, 09:25 PM

P: 5

1. The problem statement, all variables and given/known data
Your boss at the CutRate Cuckoo Clock Company asks you what would happen to the frequency of the angular SHM of the balance wheel if it had the same density and the same coil spring (thus the same torsion constant), but all the balance wheel dimensions were made onethird as great to save material. By what factor would the frequency change? 2. Relevant equations frequency = 1/2(pi) * square root of (torsion constant/I) 3. The attempt at a solution I tried finding the factor by setting up a second equation where frequency = 1/2(pi) * square root of (torsion constant/ m(r/3)^2 since I=mr^2 and got square root 3 but it didn't work. Is it because I'm using the wrong moment of inertia equation or is it something else? Please help and many thanks in advance. 



#2
Mar1607, 09:38 PM

P: 1,351

I thought for a thin disk the I=1/2Mr^2, if you reduce the radius by a factor of three, the I shoud will be 1/9th the original, hence...




#3
Mar1707, 04:08 PM

P: 8

You also have to take into account the mass that change. Since the balance wheel dimensions were made 1/3 the original, you should have 1/27 the original mass.




#4
Mar1807, 09:12 PM

P: 5

Angular Simple Harmonic Motion
thanks for the help denverdoc and enter260!



Register to reply 
Related Discussions  
Deriving angular frequency for simple harmonic motion  Introductory Physics Homework  2  
Simple Harmonic Motion  Introductory Physics Homework  4  
Motion of object (simple harmonic motion?)  Introductory Physics Homework  3  
Simple Harmonic Motion and Wave Motion  General Physics  2  
Simple Harmonic Motion/Wave Motion  Introductory Physics Homework  2 