My Math Forum Circular motion question

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 April 14th, 2011, 06:30 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 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.
 April 14th, 2011, 07:37 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Circular motion question 1.) Yes, you did it correctly! Good work. 2.) We are given $T=98.0\text{ min}$ and $r=$$6370+640$$\text{ km}=7010\text{ km}$ You see, we need to add the satellite's altitude to the radius of the Earth to get the radius of the orbit. a) $v=r\omega=r\frac{2\pi}{T}=$$7010\text{ km}$$\frac{2\pi}{98.0\text{ min}}\cdot\frac{1\text{ min}}{60\text{ s}}\approx7.49\text{ \frac{km}{s}}$ b) $a_r=\frac{v^2}{r}=\frac{$$r\frac{2\pi}{T}$$^2}{r}= r$$\frac{2\pi}{T}$$^2=$$7010\text{ km}$$$$\frac{2\pi}{98\cdot60}\:\text{ s^{-1}}$$^2\approx8.00\text{ \frac{m}{s^2}}$ 3.) We are given r = 0.15 m and f = 1200 rpm. a) This is simply the circumference of a circle of radius r: $d= C = 2\pi r=2\pi$$0.15\text{ m}$$\approx0.94\text{ m}$ b) It takes the tip 1/1200 of a minute to travel d, and using v = d/t, we have: $v=\frac{2\pi r}{T}=\frac{2\pi$$0.15\text{ m}$$}{\frac{1}{1200}\:\text{min}}\cdot\frac{1\text { min}}{60\text{ s}}\approx18.85\text{ \frac{m}{s}}$ c) $a_r=\frac{v^2}{r}=\frac{$$\frac{2\pi r}{T}$$^2}{r}=r$$\frac{2\pi}{T}$$^2=$$0.15\text{ m}$$$$\frac{2\pi}{\frac{1}{1200}\:\text{min}}\cdot \frac{1\text{ min}}{60\text{ s}}$$^2\approx2368.7\text{ \frac{m}{s^2}}$ d) $T=\frac{1}{f}=\frac{1}{1200\text{ min^{-1}}\frac{1\text{ min}}{60\text{ s}}}=\frac{60}{1200}\:\text{s}=\frac{1}{20}\:\text {s}$ 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.
 April 15th, 2011, 10:24 PM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Circular motion question Thank you!!!!!!!!!!!!!!
 April 15th, 2011, 10:32 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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.
 April 15th, 2011, 10:58 PM #5 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 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.
 April 15th, 2011, 11:14 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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?
 April 16th, 2011, 02:24 AM #7 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 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.
 April 16th, 2011, 02:43 AM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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 $a_r= r\omega^2$. So how do we put this together? b) $a_r=r$$\frac{2\pi}{T}$$^2=g$ solve for T. 2.) We know v, and we know r. You did this correctly! Good job! 3.) We know r, and we know $a_r$. 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: $a_r=\frac{v^2}{r}$ Solve for v. Post your solutions now. I will check them later.
 April 16th, 2011, 07:26 AM #9 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 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.
 April 16th, 2011, 01:33 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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: $a_r=\frac{v^2}{r}$ c) $v^2=r\cdot a_r$ $v=\sqrt{r\cdot a_r}=\sqrt{$$1.00\text{ m}$$$$0.500\text{ \frac{m}{s^2}}$$}\approx0.71\text{ \frac{m}{s}}$

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# what is the magnitude of the acceleration of a sprinter running at 10m/s when rounding a turn of a radius 25m

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