Verified answer

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.

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.

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)

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The Cut Rate

I too have the same question

thanks everyone for the answers.