Physics Help Forum Rotational Motion Problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Nov 28th 2011, 04:01 PM #1 Junior Member   Join Date: Apr 2010 Posts: 12 Rotational Motion Problem A race car starts from rest at point A on a flat, circular track with a radius of 800 feet. The driver presses on the gas pedal such that the speed of the car increases at a constant rate of 7 MPH per second. The coefficient of static friction is 0.70 . What is the speed at which the car begins to slide? I am pretty clueless on this one guys, any help is appreciated.
 Nov 28th 2011, 10:22 PM #2 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Find the point where the force of friction is equal to the required centripetal force to keep the car going in circular motion. Fc = Ff mv^2/r = (mu)mg m cancels out, you need to get v (linear velocity of the car at any instant), you have r, you have g and you have mu (the coefficient of friction). This gives the fastest speed at which the car can go without sliding. Be sure you have consistent units! Have everything in feet or in miles, whatever you're most comfortable with. Once you have the velocity v, just use the kinematic formula with final velocity v, initial velocity 0, acceleration 7 mph and time t which you are being asked for. So, for any time greater than this time, the car will slide. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Nov 28th 2011, 10:33 PM   #3
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 Originally Posted by Unknown008 Find the point where the force of friction is equal to the required centripetal force to keep the car going in circular motion. Fc = Ff mv^2/r = (mu)mg m cancels out, you need to get v (linear velocity of the car at any instant), you have r, you have g and you have mu (the coefficient of friction). This gives the fastest speed at which the car can go without sliding. Be sure you have consistent units! Have everything in feet or in miles, whatever you're most comfortable with. Once you have the velocity v, just use the kinematic formula with final velocity v, initial velocity 0, acceleration 7 mph and time t which you are being asked for. So, for any time greater than this time, the car will slide.
The equation you posted makes sense, but apparently my answer is wrong. When I worked it out, I got 91.6 mph. Did I make an error in my calculations?

 Nov 28th 2011, 10:48 PM #4 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Is mph miles per hour? (or is it metres?) In the case it's miles per hour, I'll have everything in feet. g = 32.2 ft/s^2 a = 10.3 ft/s^2 (7 mph, 7 miles = 36960 ft, then divide by 3600 s to get hours to seconds, to get 10.3 ft) v^2/r = (mu)g v^2/(800) = (0.7)(32.2) v = sqrt(0.7*32.2*800) = 134 ft/s Then using the kinematic equation: v = u + at 134 = 0 + 10.3t t = 13.0097 s = 13 s __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Nov 28th 2011, 10:58 PM   #5
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 Originally Posted by Unknown008 Is mph miles per hour? (or is it metres?) In the case it's miles per hour, I'll have everything in feet. g = 32.2 ft/s^2 a = 10.3 ft/s^2 (7 mph, 7 miles = 36960 ft, then divide by 3600 s to get hours to seconds, to get 10.3 ft) v^2/r = (mu)g v^2/(800) = (0.7)(32.2) v = sqrt(0.7*32.2*800) = 134 ft/s Then using the kinematic equation: v = u + at 134 = 0 + 10.3t t = 13.0097 s = 13 s
This is the same math I had, and yet 134 ft/s (or 91 mph) isn't correct.

I'm not sure why, it looks right.

 Nov 28th 2011, 11:06 PM #6 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Maybe it's the acceleration due to gravity? I know that for m/s^2, some books go by 10 m/s^2 while others use 9.8 m/s^2 and yet others use 9.81 m/s^2. I'm assuming that it's a computer entry that you're having? Well, one other thing to observe is that the there is at least one value given to 1 significant figure, which might mean that you need to put 90 mph instead. __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Nov 29th 2011, 06:16 AM   #7
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 Originally Posted by Unknown008 Maybe it's the acceleration due to gravity? I know that for m/s^2, some books go by 10 m/s^2 while others use 9.8 m/s^2 and yet others use 9.81 m/s^2. I'm assuming that it's a computer entry that you're having? Well, one other thing to observe is that the there is at least one value given to 1 significant figure, which might mean that you need to put 90 mph instead.
You're correct, it is a computer entry. Three significant figures are required and 9.8 m/s is the g that we've been using pretty consistently.

 Nov 29th 2011, 07:43 AM #8 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Hmm I really find it strange that you use feet, miles, metres altogether Try putting g = 10 m/s^2 = 32.8 ft/s^2 ? Otherwise, I can't see what's wrong with the answer __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?
Nov 29th 2011, 12:49 PM   #9
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 Originally Posted by Unknown008 Hmm I really find it strange that you use feet, miles, metres altogether Try putting g = 10 m/s^2 = 32.8 ft/s^2 ? Otherwise, I can't see what's wrong with the answer
Haha, I'm sticking with feet and miles for this problem, I was simply saying we use 9.8 in direct response to the previous post.

We use 32.2 ft/s^2.

The answer is still wrong apparently. We're missing something somewhere.

 Tags motion, problem, rotational

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