Register to reply 
Circular motion 
Share this thread: 
#1
Jan913, 08:52 AM

P: 5

how can rcosθ in a banked road be equal to mg; since r is equal to normal reaction which is equal to mgcosθ. rcos is even smaller than r.
so mg>mgcosθ mgcosθ=r r>rcosθ so mg>rcosθ then how can mg=rcos when banking in curved road? 


#2
Jan913, 11:27 AM

Emeritus
Sci Advisor
PF Gold
P: 5,196

Welcome to PF regan1,
mg cannot be equal to rcos(theta) simply on dimensional grounds. The former is a force, and the latter is a length. Could you post the equations that you are having trouble with more carefully and in greater detail? For a banked curve, if you look at the road in crosssection, it is essentially like an inclined plane, with the "downhill" direction being the direction towards the centre of the circle. In the absence of friction, the only two forces that act on the car are gravity, and the normal force from contact with the road, which can contribute to the centripetal force. This should be helpful to you as a starting point for working out the force balance equations. 


#3
Jan913, 02:33 PM

P: 999

As I read it, regan1 has used an infelicitous choice of variable names and used "r" to denote the normal reaction force that is supporting the car against gravity.
When a physicist or mathematician sees "r cos θ" the automatic impulse is to read the "r" as radius, however that does not appear to be the intended reading in this case. mg > mg cos θ... yes, that's true. mg cos θ = r... no, that's false. The correct version is mg = r cos θ That seems to be the problem; a multiplication where division was called for. 


#4
Jan913, 03:32 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 7,292

Circular motion
If you are taking a curve on a banked road, the car is accelerating towards the center of the curve. You seen to have forgotten about that in your equations.
This means that the reaction force r can be bigger than the weight of the car. 


#5
Jan1013, 06:55 PM

P: 5

In my opinion the maximum value that R can have is equal to the weight of the car. But since it is a slanted surface it is always less than mg 


#6
Jan1013, 08:20 PM

P: 999

AlephZero has it right. The force of the road on the car must be able to both support the car against gravity and accelerate the car in its circular path  it must be equal to the vector sum of those two components. That means that its magnitude is greater than either one of those components individually. It must be both greater than mg and greater than mv^2/radius. I'm not clear on why you are calling this a "reaction force". 


#7
Jan1113, 06:57 AM

P: 5

while banking the car in a curved road from where does the car get the force to balance it's weight and force to balance it's centrifugal force. As far as i know that force is given by the reaction force of the car's weight itself. and when you divide the reaction force in to it's component forces one force is directed toward the center which provide the centripetal force and the next balances the weight. since the car is in equilibrium the remaining component must balance the weight of the car. so
R cos θ=mg but my point if R is smaller than mg itself how can this value be equal in magnitude. i am not clear how you said R > mg can you please mathematically demonstrate it i could not get the point how road give so much force on the car greater than the car gives on the road ( max force car can give on the road is equal to mg, isn't it? ) . can you explain it further more? 


#8
Jan1113, 07:16 AM

Mentor
P: 41,579

That force would equal the weight of the car if the car were in equilibrium. But it's not. Acceleration changes things. Imagine yourself on an elevator. When the elevator is stationary (or moving with constant speed) the 'reaction' force of the floor on you will equal your weight. But if the elevator accelerates upward, that force will be greater than your weight. The same thing is going on here, but it's a bit obscured by the fact that the motion is circular. 


#9
Jan1113, 12:24 PM

P: 999




#10
Jan1113, 03:58 PM

P: 5




#11
Jan1113, 04:31 PM

Mentor
P: 41,579

The correct Newton's 3rd law pair to the force of the road on the car is the equal and opposite force that the car exerts on the road. That force depends on various things, not just the weight of the car. 


Register to reply 
Related Discussions  
Block moving on a circular trackWork energy Circular Motion problem  Introductory Physics Homework  34  
Circular motion: Car driving along a circular hill  Introductory Physics Homework  2  
Vertical circular motion in uniform circular motion  Introductory Physics Homework  3  
Uniform circular motion  plane in a circular arc  Introductory Physics Homework  1  
Identifying the forces in a difficult circular / noncircular motion hybrid problem.  Introductory Physics Homework  5 