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Your boss at the Cut-Rate 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 one-third as great to save material.

a) By what factor would the frequency change?
b) Would the frequency increase or decrease by the factor found in part (a)?
c) By what factor would the torsion constant need to be changed to make the smaller balance wheel oscillate at the original frequency?
d) Would the torsion constant need to be increased or decreased by the factor found in part (c)?

2007-03-13 06:48:23 · 7 answers · asked by Paul Wall 2 in Science & Mathematics Physics

7 answers

Part A:

As explained in my physics class, frequency f = omega/2pi, where omega = sqrt (k/I), in the case of angular simple harmonic motion, where I is the moment of inertia about the balance wheel’s axis, and k is the torsion constant.

Take equation f=omega/2pi. The 2pi stays the same; the omega affects the f directly – they are directly proportional to each other. So find the change in omega, and you get the change in frequency.

omega = sqrt (k/I). Let’s assume that the torsion constant stays the same, because you’re trying to find the influence a different balance wheel has.

I = CmR^2, with C being a constant dependent on shape, M being mass, and R being radius. Because the C is constant, ignore it.

You know that all the dimensions will be 1/3 of the original. When calculating the area of a solid disc, it is pi R^2 * h. So R is 1/3, and h is 1/3.

I = MR^2 = (pi R^2 h)* (R^2). Pi is a constant, so ignore it.

I = (1/3)^2*(1/3) *(1/3)^2. This means I = 1/(3^5), or I = 1/243

Going back, f is directly proportional to omega = sqrt(k/I) = sqrt (k/(1/243))=sqrt(243k). Ignore the k as it is a constant in this case, and the answer is 15.6

Part B:

By simple logic, if something is smaller, its frequency increases.

Part C:

Now, let I be constant, and let k change. In order to get things to balance,

omega = sqrt(k/l) = 1, so k and I have to balance out to become one, so that the sqrt(1) = 1.

k/I = 1

k = 1*I = 1/243.

In other words, the torsion constant must be 1/243, or change by a factor of 243.

Part D:

Since the torsion constant must now be 1/243, that is a decrease, to counteract the increase brought upon by smaller dimensions.

2007-03-16 11:37:53 · answer #1 · answered by Anonymous 2 · 9 2

Edward, your part a) is not correct. (I have the same problem as Paul).
At first I assumed, as you did, that "one-third as great" could just be applied to the radius, and since I=.5MR^2, then 1/3^2=1/9, and plugging in to (1/2pi)sqrt(k/I) would mean that a 3 would pop out of the sqrt, but it says the answer is incorrect. Which makes sense because you can't just plug in 1/3 for R and call it a day. That would be assuming that the overall mass of the system stays the same but that the radius shrinks. But the mass is reduced as well (at least I assume it is). So following this, I=.5(M/3)(R/3)^2 means we get an extra 1/27 with the I, and plugging in to our "f" equation, we should get a sqrt(27) difference in the frequency. However I tried this answer also, and it was not correct. I tried a few other tricks, so I might as well tell you what all the wrong answers I got so far: (1/9), (1/sqrt3), sqrt3, 3, and sqrt27.

I'm quite confident the fault lies in making assumptions about what "all the balance wheel dimensions were made one-third as great to save material" means for the momentum of the wheel. Perhaps 1/3 the radius and thickness, but the mass is some other fraction? That seems to be escaping the scope of this course, however.

2007-03-15 13:48:02 · answer #2 · answered by nookemol 1 · 0 1

I'm puzzled since I always thought that cuckoo clicks used a typical gravitational pendulum. For a torsional pendulum the frequency can be expressed as (ref1)

(a) f=(1/2pi)sqrt(k/I)
f - frequency
k -torsional constant of the spring/wire
I - moment of inertia of the pendulum.
The problem is in the geometry of the wheel since its geometry will determine I (see ref #2)
for the simples case assume a disk
I=.5 m r^2
then

f1/f2=(1/2pi)sqrt(k/ .5 m r1^2)/(1/2pi)sqrt(k/ .5 m r2^2)=
f1/f2=sqrt(r2^2/r1^2)=
f1=f2=r2/r1
since we shrunk the dimensions by 13 r2=1/3r1
so f1/f2=(1/3 r1)/r1=1/3 r=then
f2=3f1

(b) the frequency would increase 3 times

(c) we have to use
f=(1/2pi)sqrt(k/I)
and we will find that by reducing the k by 1/9 may just solve this problem.
(d) see (c)

2007-03-14 02:44:43 · answer #3 · answered by Edward 7 · 0 0

no longer the 1st time i've got heard it. however the element a pair of somewhat good shaggy dog tale is that it retains on giving. The classics will constantly grant you a good laugh, no count what share circumstances you hear them. thank you for fresh my reminiscence in this one.

2016-10-18 07:04:59 · answer #4 · answered by ? 4 · 0 0

The Cut Rate

2016-10-13 11:01:52 · answer #5 · answered by Anonymous · 0 0

I too have the same question

2016-08-23 21:03:21 · answer #6 · answered by ? 4 · 0 0

thanks everyone for the answers.

2016-09-20 04:26:22 · answer #7 · answered by ? 4 · 0 0

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