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Normal reaction in banked road circular motion

  • Thread starter physea
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  • #1
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Homework Statement


upload_2016-5-11_11-28-0.png


Homework Equations


ΣF=0

The Attempt at a Solution


upload_2016-5-11_11-30-2.png


Basically I don't agree with this solution, I think it is R1 + R2 - mgcos5 + mv^2sin5/r = 0. Can you help me please?
thanks!
 

Answers and Replies

  • #2
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Basically, I don't like that for ΣF it takes the fictional inertial force opposite to centripetal force, instead of taking the actual centripetal force. Isn't this wrong?
Second thing I don't know is the direction of the centripetal force in such situation: isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!
 
  • #3
BvU
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isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!
Fom the centre of gravity of the vehicle to the axis of the circular motion. And when banking a road, they usually have a verical axis (otherwise you would just have a sloping plane).
for ΣF it takes the fictional inertial force opposite to centripetal force
You should see is as: ##R_1+R_2-mg\cos 5^\circ## contribute a fraction ##\sin 5^\circ## to the actual centripetal force ##mv^2/r##.

The presentation is somewhat suggestive: ##\sum F = 0## would mean uniform linear motion, not a circular trajectory. Here ##\sum F \ne 0##. In fact, ##\sum F= {mv^2\over r}##.
 
  • #4
CWatters
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Second thing I don't know is the direction of the centripetal force in such situation: isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!
The drawing is correct. Note how the radius is specified.
 
  • #5
haruspex
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Basically, I don't like that for ΣF it takes the fictional inertial force opposite to centripetal force, instead of taking the actual centripetal force. Isn't this wrong?
Centripetal force is a resultant force, not an applied force. It is the component of the resultant that is normal to the direction of travel. Where the speed is constant, that makes it the entire resultant: ΣFother=Fcentripetal.
Centrifugal force is a fictional force, but is an applied force, not a resultant: ΣFother+Fcentrifugal=0.
Since centripetal and centrifugal are equal and opposite, the equations are the same.
Second thing I don't know is the direction of the centripetal force in such situation: isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!
The vehicle is describing a circle in a horizontal plane. The centre of that motion is therefore in that plane, not down the bottom of the slope.
 

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