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Thread: Rotational Speed and Force

  1. #1 Rotational Speed and Force 
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    Im having trouble with a physics question to do with speed, and force, around a radius. The question i was given was:

    "A car goes around a curved stretch of flat roadway of radius R = 100.5 m. The magnitudes of the horizontal and vertical components of force the car exerts on a securely seated passenger are, respectively, X = 210.0 N and Y = 515.0 N"

    And then to work out the speed of the car. Any ideas? What formulae would i use?


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  3. #2  
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    I wll try to give a few hints without actually doing your homework for you. Obviously this is a problem involving centripetal force, so look up the formula for that. You will notice that the formula for centripetal force needs the mass, which is not given directly. So you will have to determine that from the clues given.


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  4. #3  
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    Magic. Thanks for that i got it. I was stuck because the answer is in m/s and it was supposed to be in km/h. Thanks for your reply
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  5. #4  
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    sorry to be a bother, but ive got another related question now. How do you work out speed or the coefficent of friction when there is an angle invloved in the corner? for example. What is the maximum speed this car can round the corner with no friction if the angle of the road is 23.5 above horizontal. And then with a coefficent of friction of 0.2 Any idea how that works?
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  6. #5  
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    Quote Originally Posted by chris083210
    sorry to be a bother, but ive got another related question now. How do you work out speed or the coefficent of friction when there is an angle invloved in the corner? for example. What is the maximum speed this car can round the corner with no friction if the angle of the road is 23.5 above horizontal. And then with a coefficent of friction of 0.2 Any idea how that works?
    what you are thinking about is called a banked curve. if there is not much friction between the wheels and the street, it is dangerous and the car may skid because it doesnt have enough centripetal force to round the corner. a banked curve tries to negate this by allowing the normal force vector to provide sufficient centripetal force to round the curve even with frictionless wheels. it would be easier to show this with a diagram, but basically, since the road is tilted up at an angle, and normal force is perpendicular to the surface, the normal force vector would point diagonally upwards as opposed to straight upwards on a flat road. The horizontal vector of this normal force that points inward provides the centripetal force, instead of friction, which would normally be used to provide centripetal force
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  7. #6  
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    Quote Originally Posted by Harold14370
    I wll try to give a few hints without actually doing your homework for you. Obviously this is a problem involving centripetal force, so look up the formula for that. You will notice that the formula for centripetal force needs the mass, which is not given directly. So you will have to determine that from the clues given.
    I think we can find the mass by using the formula F=M*A
    The force is given and the acceleration can found out.

    This is the explanation till my extent of knowledge.
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    I think we can find the mass by using the formula F=M*A
    The force is given and the acceleration can found out.

    This is the explanation till my extent of knowledge.
    I'm afraid that won't do much good in this case. Since Chris already solved it I'll explain. The centripetal force is F=m*V^2/r. Then V = square root of mrF where F is 210 Newtons, the horizontal force. The vertical force is the weight, 515 Newtons and is equal to the mass times the acceleration due to gravity, so m = 515 (kg m·s−2)./9.8 (m·s−2) = 52.55 kg. You can solve from there.
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