Question

The Earth’s rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 2006 the Earth took about 0.840 s longer to complete 365 revolu- tions than it did in the year 1906. What was the average angular acceleration, in rad>s2 , of the Earth during this time?

Answer 1

Answer:

Approximately $\rm 6.1 \times 10^{-22}\; rad \cdot s^{-2}$.

Explanation:

Angular acceleration is equal to $\displaystyle \frac{\text{Change in angular speed}}{\text{Time taken}}$.

Apparently, for this question, the time taken is $100\; \text{years} \approx 100 \times 365.24\times 24 \times 3600 \; \text{seconds}$. The challenge is to find the change in angular speed over that period of time.

Let the time (in seconds) it took to complete $365$ revolutions be $t$ in the year 1906. In 2006 that number would be $(t + 0.840)$.

Each revolution is $2\pi$ radians. $365$ revolutions will be an angular displacement of $365 \times 2\pi$ in radians. Angular speed is equal to $\displaystyle \frac{\text{Angular Displacement}}{\text{Time Taken}}$.

The average angular speed in 1906 could thus be written as $\displaystyle \frac{365\times 2\pi}{t}$.

Similarly, the average angular speed in 2006 could be written as $\displaystyle \frac{365\times 2\pi}{t + 0.840}$.

The difference between the two will be equal to:

$\begin{aligned} & \; \Delta \omega \cr = &\; \frac{365\times 2\pi}{t} - \frac{365\times 2\pi}{t + 0.840}\cr =& \; 365 \times 2\pi \times \left.\frac{(t + 0.840) - t}{t(t + 0.840)}\right. \cr =& \;365 \times 2\pi \times \left.\frac{0.840}{t(t + 0.840)}\end{aligned}$.

Since the value of $t$ (about the same as the number of seconds in 365 days) is much bigger than $0.840\; \rm s$, apply the approximation $t + 0.840 \approx t$.

$\begin{aligned} &\;365 \times 2\pi \times \left.\frac{0.840}{t(t + 0.840)} \cr \approx &\; 365 \times 2\pi \times \left.\frac{0.840}{t^2} \cr \approx & \; 365 \times 2\pi \times\frac{0.840}{(365 \times 24 \times 3600)^2} \cr \approx &\; 1.9370\times 10^{-12}\; \rm rad \cdot s^{-1}\end{aligned}$.

And that's approximately the average change in angular velocity over a period of 100 years. Apply the formula for average angular acceleration:

$\displaystyle \dfrac{\Delta \omega}{\Delta t} = \rm \dfrac{1.9370\times 10^{-12}\; rad \cdot s^{-1}}{100 \times 365.24 \times 24 \times 3600\; s} \approx 6.1\times 10^{-22}\; rad \cdot s^{-2}$.