Question

1. Pendulum clocks are very important for the history of exploration.   Accurate time was the best way to determine one's longitude, since the time of sunrise or sunset depends sensitively on your longitude. In this regard, the great enemy of accuracy was thermal expansion of the pendulum used for the chronometer. We are considering a large mass attached at the end of a brass rod.   The mass at the end is much greater than the mass of the rod, so we are considering a classic pendulum, just held by a brass rod.

(A) If a brass rod lengthens by 0.002% for a 1°C temperature rise, by how much would the period of a 1 m long brass-rod pendulum change if the temperature was changed 10°C?

(B) How much error would that lead to in a day?

(C) Given that 24 hours corresponds to 360 degrees of longitude, how many degrees of longitude error does your answer to B imply?

(D) Using the radius of the earth from your book, and your answer above, how far away from your expected location would you be,  assuming you are at the equator?   (recall that this is the error in 1 day; the total will accumulate.)

2.  It is said that walking at a comfortable speed involves a pendulum-like motion of your legs.  We want to see how good this claim is quantitatively.   

    (A) First we are going to predict your gait.   The legs are not a SIMPLE pendulum, in that all the mass is not concentrated at your foot, but is distributed.    For such a pendulum the period T is given by the equation  T = 2π √(I/mgL) where I is the moment of inertia.   m is the mass of the pendulum; L is its length.   As you recall, for a uniform rod,   I=1/3 mL ²   This is a tough estimation because you know your leg has more mass at the thigh than the ankles, but for the moment, it's going to be our best guess.     Now m is not your whole mass, but the mass of your leg, which is typically about 5% of your m.   So, with a ruler and a scale, make a PREDICTION of the period of your leg as it swings in a walking motion.

    (B)  CHECK your prediction!    Time yourself (or do it with a friend, more fun) walking 20 steps.    Now there's just one more little trick here.   If you watch carefully, a "step" is only half a cycle, since the leg must then swing back.   So 20 steps = 10 cycles.   Take the time you measure for 20 steps, divide by 10, and see if you are any where near the prediction from part A.  

Note that because of your length of leg and weight, everybody can have a different answer to this question!

Screen Shot 2021-02-02 at 6.36.07 PM.png

Answer 1

One Question Per Post!

Let L be the initial length of the brass rod. \(L=1 \mathrm{~m}\)

For \(1^{0} \mathrm{C}\) temperature rise, the change is \(0.002 \%\)

for \(10^{\circ} \mathrm{C}\) temperature rise, the change will be \(10 \times 0.002 \%=0.02 \%\)

So the new length will be \(L^{\prime}=\) Actual length \(+\) change in length \(=L+0.02 L=1.02 L\)

The formula for time period of pendulum is \(T=2 \pi(L / g)^{1 / 2}=2 \pi(1 / 9.8)^{1 / 2}=2.00\) seconds

New Time period \(\mathrm{T}\) ' will be \(\mathrm{T}^{\prime}=2 \mathrm{~m}\left(\mathrm{~L}^{\prime} / \mathrm{g}\right)^{1 / 2}=2 \mathrm{~m}(1.02 \mathrm{~L} / \mathrm{g})^{1 / 2}=2 \pi(\mathrm{L} / \mathrm{g})^{1 / 2} \times 1.02^{1 / 2}=1.02^{1 / 2} \times \mathrm{T}\)

\(\mathrm{T}^{\prime}=1.02^{1 / 2} \times \mathrm{T}=1.009950494 \mathrm{~T} \sim 1.01 \mathrm{~T}=1.01 \times 2=2.02 \mathrm{~seconds}\)

So the time period increases by .02 seconds.

b.) In a day there are 24 hours \(=24 \times 60\) minutes \(=24 \times 60 \times 60\) seconds \(=86400\) seconds

Actual Time period should be \(\mathrm{T}=2\) seconds

So with in a day, that corresponds to \(86400 / T=86400 / 2=43200\) cycles.

For every cycle, there is an error of 0.02 seconds.

So in a day, for 43200 cycles the error would be \(43200 \times 0.02=864\) seconds \(=14\) minutes 24 seconds

c.) If 24 hours, i.e 86400 seconds correspond to 360 degrees, 864 seconds correspond to \((864 / 86400) \times 360\)

\(=3.6\) degrees

d.) The radius of the earth is \(6371 \mathrm{~km}\). So, the perimeter around will be \(=2 \pi \mathrm{xR}=2 \pi \times 6371=40030.17359 \mathrm{~km}\)

This corresponds to complete angle around the earth, i.e., 360 degrees

So the corresponding length of arc subtended by an angle of 3.6 degrees will be \((3.6 / 360) \times 40030.17359 \mathrm{~km}\)

\(=400.3 \mathrm{~km}\)

This is just the error of one day. Imagine an ancient sailor who was on a voyage of, say, 2 or 3 months!

Answer 2

A) Given, α: 0.002% -0100274 7.8 m/s 2,00 089 s Neo time od .000m 7.3 m/s2 2.00 290s = 2,01 X1 , S

for 0 20o8 9 LC 즈メ。. O 36 Radius 4 km Rocot on em尸echod