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physea
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Homework Statement
Homework Equations
ΣF=0
The Attempt at a Solution
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!
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).physea said: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!
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##.physea said:for ΣF it takes the fictional inertial force opposite to centripetal force
physea said: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!
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.physea said: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?
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.physea said: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 normal reaction in banked road circular motion is the force exerted by a surface on an object in contact with it, perpendicular to the surface. In the case of a banked road, the normal reaction is split into two components: the vertical component, which balances the weight of the vehicle, and the horizontal component, which provides the centripetal force needed for circular motion.
The normal reaction increases with the angle of banking. As the angle increases, the vertical component of the normal reaction also increases, providing more support for the vehicle. This allows for a higher speed to be maintained without the vehicle slipping off the road.
The factors that affect the normal reaction in banked road circular motion include the angle of banking, the speed of the vehicle, the weight of the vehicle, and the coefficient of friction between the tires and the road surface. These factors all work together to determine the amount of normal reaction needed for the vehicle to maintain a stable circular motion.
Yes, in some cases the normal reaction can be greater than the weight of the vehicle. This can happen when the angle of banking is steep and the speed of the vehicle is high, resulting in a larger vertical component of the normal reaction. However, the horizontal component of the normal reaction will still be equal to the centripetal force needed for circular motion.
The normal reaction plays a crucial role in determining the turning radius of a vehicle on a banked road. As the normal reaction increases, the centripetal force needed for circular motion also increases, resulting in a smaller turning radius. This allows for sharper turns to be made at higher speeds, as long as the angle of banking is appropriate for the speed of the vehicle.