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?

Answers

Answer 1

Answer:

Answer:

Approximately $\rm 6.1 * 10^(-22)\; rad \cdot s^(-2)$ .

Explanation:

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

Apparently, for this question, the time taken is $100\; \text{years} \approx 100 * 365.24* 24 * 3600 \; \text{seconds}$ . 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 $365 * 2\pi$ 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 $\displaystyle (365* 2\pi)/(t)$ .

Similarly, the average angular speed in 2006 could be written as $\displaystyle (365* 2\pi)/(t + 0.840)$ .

The difference between the two will be equal to:

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

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 $t + 0.840 \approx t$ .

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

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

$\displaystyle (\Delta \omega)/(\Delta t) = \rm (1.9370* 10^(-12)\; rad \cdot s^(-1))/(100 * 365.24 * 24 * 3600\; s) \approx 6.1* 10^(-22)\; rad \cdot s^(-2)$ .

Answer 2

Answer:

The average angular acceleration of the Earth during this time was approximately -7.42 × $10^{-22$ rad/s², indicating that the Earth's rotation is slowing down.

To calculate the average angular acceleration of the Earth during this time, we can use the following formula:

Average angular acceleration = (Δω) / Δt

where Δω is the change in angular velocity and Δt is the elapsed time.

The angular velocity of the Earth can be calculated using the formula:

ω = 2π / T

where T is the period of rotation. In this case, T is equal to 365.25 days, or approximately 31.54 × $10^6$ seconds.

Using the given information, we can calculate the change in angular velocity as follows:

Δω = (2π / 31.54 × $10^6$ seconds) - (2π / (31.54 × $10^6$ + 0.840 seconds))

Δω ≈ -2.34 × $10^{-15$ rad/s

The elapsed time is simply the difference between the two years, or 100 years. Therefore, the average angular acceleration is:

Average angular acceleration = (-2.34 × $10^{-15$ rad/s) / (100 years × 3.154 × $10^7$ seconds/year)

Average angular acceleration ≈ -7.42 × $10^{-22$ rad/s²

Therefore, the average angular acceleration of the Earth during this time was approximately -7.42 × $10^{-22$ rad/s². The negative sign indicates that the angular acceleration is causing the Earth's rotation to slow down.

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