Angular deceleration of the Earth

In summary: T = (2π)/(3600*365) rad/sw_77 =(2π)/T = (2π)/2.43×10^9 rad/sAnd delta t being 1.01 s. Replacing the equation and making alpha sof will give the answer?
  • #1
24
2
Homework Statement
The rotation of the Earth is slowing down. In 1977, the Earth
took 1.01 s longer to complete 365 rotations than in 1900.
What was the average angular deceleration of the Earth in the
time interval from 1900 to 1977?
Relevant Equations
w=w_o+αt
365 rotations - 365 days
365 days - 31536000 s

apart from that I do not know how to continue the question
 
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  • #2
You are given w_o, w and t. Can you recognize what their numerical values are from what is given? Make a list. You may assume that over the period of one year, w does not change significantly.
 
  • #3
How many seconds in 77 years?

What are the angular velocities of the Earth in 1900 and 1977? (Remember that there are ##2\pi## radians in each daily rotation)

And what is the difference in those two angular velocities divided by the total time? (that would be the average angular deceleration over that time period) What units is your answer in?
 
  • #4
Zoubayr said:
Homework Statement:: The rotation of the Earth is slowing down. In 1977, the Earth
took 1.01 s longer to complete 365 rotations than in 1900.
What was the average angular deceleration of the Earth in the
time interval from 1900 to 1977?
Relevant Equations:: w=w_o+αt

365 rotations - 365 days
365 days - 31536000 s

apart from that I do not know how to continue the question
What is the definition of the average acceleration? If you understand that you will see how to get this.
 
  • #5
berkeman said:
How many seconds in 77 years?

What are the angular velocities of the Earth in 1900 and 1977? (Remember that there are ##2\pi## radians in each daily rotation)

And what is the difference in those two angular velocities divided by the total time? (that would be the average angular deceleration over that time period) What units is your answer in?
In 77 yrs there are approx 2.43 x 10^9 s
how to find the angular velocity given that the frequency f is not given?
 
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  • #6
Can you get the frequency from the period?
 
  • #7
This problem is simpler if you stick to the variables given, angular velocities at two different times and the basic definition of average angular acceleration. In fact, you only care about the difference of the two angular velocities over the time period of concern.
 
  • #8
kuruman said:
Can you get the frequency from the period?
w=(2π)/T with T being 2.43 x 10^9 s?
 
  • #9
Zoubayr said:
w=(2π)/T with T being 2.43 x 10^9 s?
What is 2.43×109 s in years?
 
  • #10
kuruman said:
What is 2.43×109 s in years?
77 yrs
 
  • #11
Zoubayr said:
In 77 yrs there are approx 2.43 x 10^9 s
how to find the angular velocity given that the frequency f is not given?
Should a physics student be expected to know how long is a day on Earth? Or, is that data not generally common knowledge?
 
  • #12
Zoubayr said:
77 yrs
That 77 yrs is the time interval from 1900 to 1977. What do the symbols w and w_o that you used in your relevant equation represent and how are they related to the 77 yrs?
 
  • #13
PeroK said:
Should a physics student be expected to know how long is a day on Earth? Or, is that data not generally common knowledge?
a day on earth is approx 24 hrs
 
  • #14
Zoubayr said:
a day on earth is approx 24 hrs
I agree. How is that related to w and w_o?
 
  • #15
One does not actually need to know the exact angular velocities ##\omega## or ##\omega_0## or the equivalent periods to do this problem, only the difference.
 
  • #16
kuruman said:
I agree. How is that related to w and w_o?
w=(2π)/T with T being 24 hrs?
 
  • #17
bob012345 said:
One does not actually need to know the exact angular velocities ##\omega## or ##\omega_0## or the equivalent periods to do this problem, only the difference.
how to find the difference?
 
  • #18
Zoubayr said:
how to find the difference?
If you assume the change in angular velocity over the period of 1 year is negligible (the years which they made their measurements)

You have that:

$$\omega_{77} = \omega_{00} - \alpha \Delta T$$

Put each ## \omega## ( measured over a period of a year) in terms of the angle turned in a year ##\theta## and the times ##t, t+\Delta t ## it took to turn ##\theta##.
 
  • #19
erobz said:
If you assume the change in angular velocity over the period of 1 year is negligible (the years which they made their measurements)

You have that:

$$\omega_{77} = \omega_{00} - \alpha \Delta T$$

Put each ## \omega## ( measured over a period of a year) in terms of the angle turned in a year ##\theta## and the times ##t, t+\Delta t ## it took to turn ##\theta##.
w_00=(2π)/T = (2π)/3600 rad/s
w_77 =(2π)/T = (2π)/2.43×10^9 rad/s
And delta t being 1.01 s. Replacing the equation and making alpha sof will give the answer?
 
  • #20
Zoubayr said:
w_00=(2π)/T = (2π)/3600 rad/s
w_77 =(2π)/T = (2π)/2.43×10^9 rad/s
And delta t being 1.01 s. Replacing the equation and making alpha sof will give the answer?
1) Does the earth make a full revolution in an hour?
2) They measured the revolutions over a year. Does the earth make one revolution in a year?
 
  • #21
erobz said:
1) Does the earth make a full revolution in an hour?
2) They measured the revolutions over a year. Does the earth make one revolution in a year?
Opps my bad. So,
w_00=(2π)/T = (2π)/(3600×365) rad/s
w_77 =(2π)/T = (2π)/2.43×10^9 rad/s
And delta t being 1.01 s. Replacing the equation and making alpha sof will give the answer?
 
  • #22
Zoubayr said:
how to find the difference?
Sorry, I was looking at the periods difference not angular velocity. I think you do need the value at each end not just the difference. But they give enough to figure them.
 
  • #23
Zoubayr said:
Opps my bad. So,
w_00=(2π)/T = (2π)/(3600×365) rad/s
w_77 =(2π)/T = (2π)/2.43×10^9 rad/s
And delta t being 1.01 s. Replacing the equation and making alpha sof will give the answer?
You know what, I may be leading you on a wild goose chase. You can't solve for ##\alpha## that way. Sorry.
 
  • #24
bob012345 said:
Sorry, I was looking at the periods difference not angular velocity. I think you do need the value at each end not just the difference. But they give enough to figure them.
I don't know... I can't seem to find ##\Delta \omega## from the information given. I think we stil need ##\omega## at one of the end points?
 
  • #25
erobz said:
You know what, I may be leading you on a wild goose chase. You can't solve for ##\alpha## that way. Sorry.
My fault. Sorry.
 
  • #26
bob012345 said:
My fault. Sorry.
I don't think it's your fault. I can't seem to deduce the answer from the given data?
 
  • #27
Not enough info. Thats my current position.
 
  • #28
erobz said:
Not enough info. Thats my current position.
I assume the angular velocity in 1900 is in units of revolutions or rotations 365 revs/yr and in 1977 it is 365 rev/(1yr -1.01s). Then just figure the average acceleration from 1900 to 1977.
 
  • #29
bob012345 said:
I assume the angular velocity in 1900 is in units of revolutions or rotations 365 revs/yr and in 1977 it is 365 rev/(1yr -1.01s). Then just figure the average acceleration from 1900 to 1977.
But I think we need the time it took to make 365 revolutions the year in 1900 or 1977?
 
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  • #30
erobz said:
Not enough info. Thats my current position.
The goal is to find the angular acceleration using the equation ##\alpha=\dfrac{\Delta\omega}{\Delta t}##.
We know that
  1. The time over which the period changes is ##\Delta t## = 77 years = 2.4×109 s.
  2. The change in period over that time is ##\Delta T=T_{\text{1977}}-T_{\text{1900}}= 1~## s.
We need to find the change in frequency ##\Delta \omega## corresponding to the change in period ##\Delta T.## $$\Delta \omega=\Delta \left(\frac {2\pi}{T}\right)= 2\pi\left(\frac{1}{T+\Delta T}-\frac{1}{T} \right)=~?$$Of course, it can also be done by using differentials.
 
  • #31
kuruman said:
The goal is to find the angular acceleration using the equation ##\alpha=\dfrac{\Delta\omega}{\Delta t}##.
We know that
  1. The time over which the period changes is ##\Delta t## = 77 years = 2.4×109 s.
  2. The change in period over that time is ##\Delta T=T_{\text{1977}}-T_{\text{1900}}= 1~## s.
We need to find the change in frequency ##\Delta \omega## corresponding to the change in period ##\Delta T.## $$\Delta \omega=\Delta \left(\frac {2\pi}{T}\right)= 2\pi\left(\frac{1}{T+\Delta T}-\frac{1}{T} \right)=~?$$Of course, it can also be done by using differentials.
My interpretation is:

$$ \omega_{77} = \omega_{00} - \alpha \Delta T $$

Where ##\Delta T ## is the time in seconds between the year 1977 and 1900.

Let ##t## be the time to make 365 revolutions ##\theta## in 1900. It follows that:

$$ \frac{2 \pi \theta}{ t + \Delta t} = \frac{2 \pi \theta}{ t } - \alpha \Delta T$$

So we need some more info, because I count two variables ##t, \alpha##, and one equation?
 
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  • #32
erobz said:
My interpretation is:

$$ \omega_{77} = \omega_{00} - \alpha \Delta T $$

Where ##\Delta T ## is the time in seconds between the year 1977 and 1900.

Let ##t## be the time to make 365 revolutions ##\theta## in 1900. It follows that:

$$ \frac{2 \pi \theta}{ t + \Delta t} = \frac{2 \pi \theta}{ t } - \alpha \Delta T$$

So we need some more info, because I count two variables ##t, \alpha##, and one equation?
You can safely take, and I think this was presumed, that t=1 year. But a small variation of a few moments will not change the final answer within the accuracy of the 1.01s change in time over 77 years.
 
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  • #33
bob012345 said:
You can safely take, and I think this was presumed, that t=1 year. But a small variation of a few moments will not change the final answer within the accuracy of the 1.01s change in time over 77 years.
Yeah, that's quite reasonable (I still feel like one of the time measurements should just be stated).
 
  • #34
@Zoubayr Sometimes it's hard to tell how close, "close enough" is to the desired outcome in physics problems (this probably wasn't one of those times). Let my ignorance be a lesson to you.
 
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  • #35
how to do the problem then? I am struggling to understand
 

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