# Circular Motion RevisionWatch

#1
I'm getting confused with my revision. I've got a question which is simple..but I just need to clarify something:

An ice skater completes 4 revolutions in 1.2s. What is her average angular velocity and her rotation frequency?

Do you have to work out the time for 1 revolution first?
So, it would 1.2/4revs-1
Or is it the other way round, how many revolutions per second (because of the units revs-1)?
So, it would be 4/12revs-1

I'm confused in which way it should be.. I understand what to do from there. Any help would be extremely appreciated - thanks!
0
8 years ago
#2
rotational frequency is the number of rotations per second right? so it = number of rotations/time in s
0
#3
I've got another question:

What is the angular velocity of a person standing on the surface of the earth? What is their linear velocity?

So would it be 1 revolution takes 24 hours. Therefore to obtain rev/s...
24x60x60=86400seconds 1/86400= 1.157x10-5 rev/s
to get that in rad/s x2[pi] which gives 7.27x10-5rad/s which is angular velocity

to obtain linear velocity just use the equation v=wr which means
=7.27x10-5x6.37x106=463.09 so the linear velocity is 463m/s
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#4
0
8 years ago
#5
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#6
(Original post by roar558)
AAH. It's fine, its just I'm trying to revise this chapter by tonight and its so fustrating me. Thanks anyway.
0
8 years ago
#7
What you have done with the circular motion Q about the person on earth's surface is completely right.
0
8 years ago
#8
(Original post by Aiboz)
I've got another question:

What is the angular velocity of a person standing on the surface of the earth? What is their linear velocity?

So would it be 1 revolution takes 24 hours. Therefore to obtain rev/s...
24x60x60=86400seconds 1/86400= 1.157x10-5 rev/s
to get that in rad/s x2[pi] which gives 7.27x10-5rad/s which is angular velocity

to obtain linear velocity just use the equation v=wr which means
=7.27x10-5x6.37x106=463.09 so the linear velocity is 463m/s
That's correct. Standing at the equator (which is what you've worked out the speed for) you are moving with a linear velocity of about 1000mph.
#9
(Original post by Stonebridge)
That's correct. Standing at the equator (which is what you've worked out the speed for) you are moving with a linear velocity of about 1000mph.
Thankyou so much!
I have one other question,

The earth has a mass of 6x1024kg and it orbits the sun at a mean distance of 1.5x1011m once a year.
What is the earth's angular velocity about the sun?
What is the earth's centripetal acceleration?

So is it the same concept of using 1 revolution in 365 days and continue from there? Is the mass just mentioned to confuse you?

Thanks again.
0
8 years ago
#10
(Original post by Aiboz)

Thankyou so much!
I have one other question,

The earth has a mass of 6x1024kg and it orbits the sun at a mean distance of 1.5x1011m once a year.
What is the earth's angular velocity about the sun?
What is the earth's centripetal acceleration?

So is it the same concept of using 1 revolution in 365 days and continue from there? Is the mass just mentioned to confuse you?

Thanks again.
Angular velocity is measured in radians per second, so, yes basically the same thing; just 2Pi radians in a revolution.
Yes, for the centripetal acceleration you don't need the mass of the Earth; but you would need it for the centripetal force, if asked for that.
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