Q Factor & Run Time of Pendulum

[oddjobmj]

Homework Statement

Consider a small cuckoo clock in which the length of the pendulum is L, suspending a mass m with a period of oscillation T. The clock is powered by a falling weight (mass = M) which falls 2 m between the daily windings. The amplitude of the swing is 0.2 radians.

(A) Determine the quality factor Q of the clock.

Data: m = 55 g; T = 1.0 s; M = 1066 g.

(B) How many days would the clock run if it were powered by a 1.5 volt battery with charge capacity equal to 1800 mAh?

Homework Equations

Q=2*$\pi$*$\frac{stored}{lost/cycle}$

Time=$\frac{(mAh/1000)*(voltage)}{device power}$

Power=J/s

The Attempt at a Solution

For part A my understanding is that the energy stored is the potential energy of M:

m*g*h=(1.066 kg)*9.81 (m/s²)*(2 meters)=20.915 Joules

Also, because that energy is dissipated each day, the energy lost per cycle (which is 1 second) is the stored energy (20.915 Joules) divided by the number of seconds in a day (24*60*60) or:

2.42*10^-4 Joules

If I plug these values into my equation for Q I get a huge, nonsense number. What am I doing wrong?

For part B:
I have mAh and volts but I need power and what I'm getting for J/s above must be wrong.

Thank you!

 

[gneill]

The energy stored per cycle of the clock is related to the swing of the pendulum. I'd look at the potential energy of the pendulum bob when its KE is zero...

Your energy per cycle of 2.42 x 10^-4 J represents the energy lost each cycle that's replenished by the falling weight.

For part B, with the amp-hour and voltage rating of the battery you have the total energy stored in the battery. How many day's weight-falls does that energy represent?

 

[oddjobmj]

Ahh, thank you!

Well, the potential when K=0 is dictated by the length of the pendulum and the bob's mass.

T=2π$\sqrt{L/g}$

So, L=0.24849 meters

Using that in the eq for U:

U=mgL(1-cos(.2))=.002673

Plugging that in as the numerator in my eq for Q:

Q=69.4 which is correct!

For part B:

Energy=V*Amps*3600 seconds
Time=Energy/(2*1.066*9.81)=465 days

Thanks again!

 

[gneill]

Ahh, thank you!

Well, the potential when K=0 is dictated by the length of the pendulum and the bob's mass.

T=2π$\sqrt{L/g}$

So, L=0.24849 meters

Using that in the eq for U:

U=mgL(1-cos(.2))=.002673

Plugging that in as the numerator in my eq for Q:

Q=69.4 which is correct!

For part B:

Energy=V*Amps*3600 seconds
Time=Energy/(2*1.066*9.81)=465 days

Thanks again!

Glad I could help!

 

[Nickie]

Hello,

For part A, your approach is correct, but there are a few things that need to be adjusted. First, the energy stored in the clock is not just the potential energy of the falling weight, but also the kinetic energy of the pendulum. So the total energy stored is:

E = mgh + 1/2mv^2

Where m is the mass of the falling weight, g is the acceleration due to gravity, h is the distance it falls, and v is the velocity of the pendulum at the bottom of its swing. You can calculate v using the formula v = 2πL/T, where L is the length of the pendulum and T is the period.

So the total energy stored would be:

E = (0.055 kg)(9.81 m/s^2)(2 m) + 1/2(0.055 kg)(2π(0.2 m)/1 s)^2 = 0.215 J

Next, the energy lost per cycle would actually be the total energy stored divided by the number of cycles in a day, which is 1/T. So it would be:

Energy lost per cycle = 0.215 J/1 s = 0.215 J/s

Plugging this into the formula for Q, we get:

Q = 2π(0.215 J)/(0.215 J/s) = 2πs ≈ 6.28

For part B, you are correct that you need to convert the mAh to J/s, but you need to do it in two steps. First, convert the mAh to mAs (milliamp-seconds) by multiplying by 3600 (since there are 3600 seconds in an hour):

1800 mAh * 3600 s/h = 6,480,000 mAs

Then, convert the mAs to J by multiplying by the voltage:

6,480,000 mAs * (1.5 V) = 9,720,000 J

This is the total energy that the battery can provide. To find the time it can run the clock, we need to divide this by the power of the clock. The power of the clock can be calculated using the formula:

P = Energy lost per cycle * frequency = (0.215 J/s) * (1 cycle/1 s) = 0.215 W

So the time the battery can