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April 14th, 2011, 07:30 PM   #1
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Circular motion question

1.What is the magnitude of the acceleration of a sprinter running at 10 ms^-1 when rounding a turn with a radius of 25m?

i use the formula

a= v^2 /r
=(10)^2 / (25)
= 4 ms^-2

i get the correct answer but is this the right way to solve the question?

2.An Earth satellite moves in a circular orbit 640km above Earth's surface with a period of 98.0min.If radius of Earth is 6370km,what are

a)the speed and
b)the magnitude of the centripetal acceleration of the satellite?

i use the formula v=rw
w= 2?/(98 x 60)
=1.09 x 10^3
v=rw=(6370)(1.09 x 10^3)
= 7.0 x 10^ 7 (the correct answer is 7.49kms^-1),i need help and part (b) too

3.A rotating fan completes 1200 revolutions every minutes.Consider the tip of blade,at a radius of 0.15m.

a)through what distance does the tip move in one revolution?What are
b)the tip's speed and
c)the magnitude of its acceleration
d)What is the period of the motion

This question i dunno how to solve bcs my book just give me limited information only.i need yr help,sir.

and one more important thing,can i post 5 question in a post at the same time,because i have so many question to ask.Thank you very much.
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April 14th, 2011, 08:37 PM   #2
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Re: Circular motion question

1.) Yes, you did it correctly! Good work.

2.) We are given and

You see, we need to add the satellite's altitude to the radius of the Earth to get the radius of the orbit.

a)

b)

3.) We are given r = 0.15 m and f = 1200 rpm.

a) This is simply the circumference of a circle of radius r:



b) It takes the tip 1/1200 of a minute to travel d, and using v = d/t, we have:



c)

d)

I would actually limit it to no more than 3 questions per post. Any more than that, and it becomes a lot of work at one time to answer all the questions in one post.
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April 15th, 2011, 11:24 PM   #3
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Re: Circular motion question

Thank you!!!!!!!!!!!!!!
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April 15th, 2011, 11:32 PM   #4
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Re: Circular motion question

I just noticed an error, which I have corrected, in problem 2, part b). I failed to convert km to m, and the answer I gave was 1/1000 what it should be. It is approximately 8.00, not 0.008, m/s².

I trust you have done enough of these problem now that you can make an effort to solve each one, and post your efforts. You will gain MUCH more this way.
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April 15th, 2011, 11:58 PM   #5
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Re: Circular motion question

sir i don understand about the part 3(b) that you said It takes the tip 1/1200 of a minute to travel d, and using v = d/t, we have:

how does 1/1200 come from? is it 1200 revolutions?but how 1200 revolutions become 1/1200.i am confuse.

and the part 3(d)

i don understand how the value of f come from.
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April 16th, 2011, 12:14 AM   #6
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Re: Circular motion question

We are told "A rotating fan completes 1200 revolutions every minute." This means every minute the fan makes 1200 revolutions, thus it takes 1/1200 min. to make one revolution. If this doesn't seem intuitive. we can look at it this way:

Period = 1/frequency

Period is the time it takes to complete one cycle (time/cycle), while frequency is the number of cycles per unit time (cycle/time).

This means:

time/cycle = 1/(cycle/time)

This is how we get the relationship:

T = 1/f

T = period, f = frequency

As you can see period and frequency are tied intimately together; they are the multiplicative inverse (reciprocal) of one another. Do you understand now?
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April 16th, 2011, 03:24 AM   #7
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Re: Circular motion question

yes i understand now. Thank you sir.

and i need another help

1.a)What is the magnitude of the centripetal acceleration of an object on Earth's equator owing to the rotation of Earth if Earth's radius is 6370km?

b)What would the period of rotation of Earth have to be for objects on the equator to have a centripetal acceleration with a magnitude of 9.8ms^-2?

2.Determine the acceleration of a kid on a bike traveling at a constant speed 10.0 ms^-1 around a flat circular track of radius 200m.

3.A small ball rolling at a constant speed in a horizontal circular groove having a diameter of 2.00m experiences a centripetal acceleration of 0.500 ms^-2.

a)What is the radius of the ball's orbit?
b)How does the centripetal acceleration depend on the speed?
c)Determine the speed

My solution

1.a) i use the formula a=v^2/r, a=rw^2 and so on but still can't the problem.i only have the value r=6370km.

2. a=v^2/r
=(10.0)^2 / 200
= 0.5ms^-2

3.i have no idea to solve the question.

Please help me sir.Thank you.
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April 16th, 2011, 03:43 AM   #8
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Re: Circular motion question

1.)

a) You have r = 6370 km = 6.37 x 10^6 m and T = 24 hr. We know ? = 2?/T and we know . So how do we put this together?

b) solve for T.

2.) We know v, and we know r. You did this correctly! Good job!

3.) We know r, and we know .

a) We are given r = (2.00 m)/2 = 1.00 m.

b) What is the centripetal acceleration as a function of v?

c) You are given the centripetal acceleration, and the radius:

Solve for v.

Post your solutions now. I will check them later.
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April 16th, 2011, 08:26 AM   #9
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Re: Circular motion question

1.a) a=rw^2
=(6370 x 10^3)( 2?/(24 x 60 x 60)^2
=0.034 ms^-2

b) a= r(2?/T)^2=g
=(6370 x 10^3)(2?/T)^2=9.81
(6370 x 10^3)(2?)^2 =9.81T^2
T=5063 (the answer given is 84 min,i think i done something wrong but i can't see it)

3.b) a=vw
=(0.500)(2?/60)
= 0.0524 ms^-2

c)a= v^2/r
= (0.500)^2 / 1
= 0.25 ms^-2


Here is my solution.please check for me ,sir.Thank you.
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April 16th, 2011, 02:33 PM   #10
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Re: Circular motion question

1.)

a) Correct.

b) I get approx. 5066 s. which is 84.43 min.

3

b) The centripetal acceleration varies as the square of the speed:

c)

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what is the magnitude of the acceleration of a sprinter running at 10 m/s
,
a rotating fan completes 1200 revolutions
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an earth satellite moves in a circular orbit

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the magnitude of the centripetal acceleration of the satellite?

,

what is the magnitude of the acceleration of a sprinter running

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an earth satellite moving in acircular orbit of 640 km above the earth surface has aperiod of 98 minutes what is its speed in m/s

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An earths satellite moves in a circular orbit with an orbital speed 6280 ms . find the time of revelation

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what is the magnitude of the acceleration of a sprinter running at 9.0 m/s when rounding a turn of a radius 25 m?

,

An earth satellite moves in a circular orbit of 640km above earth's surface. Its takes 98 mins for the satellite to make one complete revolution. Calculate the acceleration of the satellite?

,

a rotating fan completes 1200 revolutions every minute. consider a point on the tip of the blade, at a radius of 0.15m. through what distance does the point move in one revolution

,

What is the period of the motion, the time its takes the ball to complete one revolution?

,

an earth satellite moves in a circular orbit 640 km above earth’s surface with a period of 98.0 min. what are the (a) speed and (b) magnitude of the centripetal acceleration of the satellite?

,

an earth satellite moves in a circular orbit 640 km above

,

what is the magnitude of the acceleration of a sprinter running at 9.0 m/s when rounding a turn of a radius 23 m?

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An Earth satellite moves in a circular orbit 640km above Earths surface with a period of 98mins. What are the Speed of the satellite

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