Circular Motion of a car around a curve

AI Thread Summary
A car navigating a banked curve must maintain a specific speed range to prevent slipping, determined by the radius of curvature, banking angle, and coefficient of static friction. The discussion emphasizes applying Newton's second law to analyze forces acting on the car, including normal force and friction, to derive the correct speed equations. Participants clarify the importance of distinguishing between horizontal and vertical force components to solve for maximum speed without slipping. The correct formula for maximum speed includes a factor accounting for friction and banking angle, which participants initially miscalculated. Ultimately, the conversation leads to a clearer understanding of the forces involved and the correct method for determining the speed range.
haha1234
Messages
109
Reaction score
0

Homework Statement



A car rounds a banked curve as discussed in Example 6.4 and shown in Figure 6.5.The radius of curvature of the road is R,the banking angle is θ,and the coefficient of static friction is μs.
Determine the range of speeds the car can have without slipping up or down the road.

Homework Equations





The Attempt at a Solution


I've just drawn two free-body diagram representing the situation that the centripetal acceleration is large that the car tends to sliding up and the opposite situation.
But I don't know how to find the range.
 

Attachments

  • DSC_0612-1[1].jpg
    DSC_0612-1[1].jpg
    15.4 KB · Views: 562
  • DSC_0611.jpg
    DSC_0611.jpg
    12.2 KB · Views: 531
Last edited:
Physics news on Phys.org
haha1234 said:
I've just drawn two free-body diagram representing the situation that the centripetal acceleration is large that the car tends to sliding up and the opposite situation.
Good.

But I don't know how to find the range.
Apply Newton's 2nd law to each extreme. That will allow you to solve for the speeds.
 
What forces are there?

Observe that the net force on the car must be strictly horizontal.
 
Doc Al said:
Good.


Apply Newton's 2nd law to each extreme. That will allow you to solve for the speeds.

Is the friction accounts for the centripetal acceleration?
 
haha1234 said:
Is the friction accounts for the centripetal acceleration?
What accounts for the centripetal acceleration is the net force on the car. That includes the friction.
 
Doc Al said:
What accounts for the centripetal acceleration is the net force on the car. That includes the friction.

There is my attempted solution:
tanθ=F/mg
F=mgtanθ
mgtanθ+fcosθ=mvmax2/R
gtanθ+μsg=vmax2/R
vmax=[Rg(tanθ+μs)]1/2
But it's not the correct answer.
Could you tell me the mistakes?
 
haha1234 said:
There is my attempted solution:
tanθ=F/mg
F=mgtanθ
What is F?

Do this: List all forces acting on the car. There are three. Then write Newton's 2nd law for vertical and horizontal components.
 
Doc Al said:
What is F?

Do this: List all forces acting on the car. There are three. Then write Newton's 2nd law for vertical and horizontal components.

Like this?
 

Attachments

  • DSC_0609[1].jpg
    DSC_0609[1].jpg
    12.6 KB · Views: 507
Where is the force of friction?

ehild
 
  • #10
haha1234 said:
Like this?
In your diagram, I'd like to see three vectors only--representing the three forces acting on the car.

Then I'd like you to list the horizontal and vertical components of each force. You'll use those components to write Newton's 2nd law.
 
  • #11
Doc Al said:
In your diagram, I'd like to see three vectors only--representing the three forces acting on the car.

Then I'd like you to list the horizontal and vertical components of each force. You'll use those components to write Newton's 2nd law.

How to find the horizontal component of the weight?
I think Wb in the picture is not parallel to the centripetal acceleration,so it may not be a component of the centripetal force.
 

Attachments

  • DSC_0611-1[1].jpg
    DSC_0611-1[1].jpg
    16.6 KB · Views: 470
Last edited:
  • #12
W is vertical, it does not have horizontal component. ehild
 
  • #13
ehild said:
W is vertical, it does not have horizontal component. ehild

So is the weight giving the centripetal acceleration?
 
Last edited:
  • #14
Why do you think so?

ehild
 
  • #15
haha1234 said:
How to find the horizontal component of the weight?
As ehild explained, since the weight acts vertically, the horizontal component will be zero.

I think Wb in the picture is not parallel to the centripetal acceleration,so it may not be a component of the centripetal force.
For some reason, you found components of the weight parallel (Wb) and perpendiculur (Wa) to the incline surface. This was not necessary--stick to horizontal and vertical components.

(FYI: If you took the horizontal components of both Wa and Wb, you'll find that they cancel out.)
 
  • #16
Doc Al said:
As ehild explained, since the weight acts vertically, the horizontal component will be zero.For some reason, you found components of the weight parallel (Wb) and perpendiculur (Wa) to the incline surface. This was not necessary--stick to horizontal and vertical components.

(FYI: If you took the horizontal components of both Wa and Wb, you'll find that they cancel out.)

Finally I've found that the vmax=[Rg(tanθ+μs)]
But the right answer is [Rg(tanθ+μs)/(1-μstanθ)].
How (1-μstanθ) comes out?
 
  • #17
haha1234 said:
Finally I've found that the vmax=[Rg(tanθ+μs)]
But the right answer is [Rg(tanθ+μs)/(1-μstanθ)].
How (1-μstanθ) comes out?
You still have not clearly explained your working.
If N is the normal force and F the frictional force, what is the net force in the vertical direction, and what is the net force in the horizontal direction? (You can take F as being either up slope or down slope as you wish, so long as you are consistent.)
 
  • #18
haruspex said:
You still have not clearly explained your working.
If N is the normal force and F the frictional force, what is the net force in the vertical direction, and what is the net force in the horizontal direction? (You can take F as being either up slope or down slope as you wish, so long as you are consistent.)

Is the net force in vertical direction 0N?
Is the net force in horizontal direction Nsinθ+fcosθ?
 
  • #19
haha1234 said:
Finally I've found that the vmax=[Rg(tanθ+μs)]
But the right answer is [Rg(tanθ+μs)/(1-μstanθ)].
How (1-μstanθ) comes out?
Again, don't just give answers, show your work step by step. Once again I request that you give the vertical and horizontal components of each force.
 
  • #20
haha1234 said:
Is the net force in vertical direction 0N?
Yes, the net force in the vertical direction will equal 0, but express this as an equation: ∑Fy = 0. What is ∑Fy?

Is the net force in horizontal direction Nsinθ+fcosθ?
Yes, for one of the extremes.
 
  • #21
Doc Al said:
Yes, the net force in the vertical direction will equal 0, but express this as an equation: ∑Fy = 0. What is ∑Fy?Yes, for one of the extremes.

Ny+W=0?

And I would like to know whether the sum of the the horizontal component of each force is the centripetal force.
 
  • #22
haha1234 said:
Ny+W=0?
There are three forces. Please list the horizontal and vertical components of each. Pay attention to sign.

And I would like to know whether the sum of the the horizontal component of each force is the centripetal force.
That happens to be true, but the better way to understand it is to think of the acceleration and then apply Newton's 2nd law. In this case the acceleration is centripetal and it acts horizontally.
 
  • #23
Doc Al said:
There are three forces. Please list the horizontal and vertical components of each. Pay attention to sign.


That happens to be true, but the better way to understand it is to think of the acceleration and then apply Newton's 2nd law. In this case the acceleration is centripetal and it acts horizontally.

Vertical component:
Ny-W-fsin=0?

Horizontal component:
Nx+fcosθ=mv2/R?
 
  • #24
haha1234 said:
Vertical component:
Ny-W-fsin=0?
Express Ny in terms of θ and N.

Horizontal component:
Nx+fcosθ=mv2/R?
Same comment: Express Nx in terms of θ and N.

But you are getting closer! Also: Express the friction in terms of the normal force.
 
  • #25
Doc Al said:
Express Ny in terms of θ and N.


Same comment: Express Nx in terms of θ and N.

But you are getting closer! Also: Express the friction in terms of the normal force.

Nx+fcosθ=mvmax2/R
mgtanθ+μsNcosθ=mvmax2/R
gtanθ+μs(g/cosθ)cosθ=vmax2/R
vmax=[Rg(tanθ+μs)]1/2
But it's not the correct answer.:cry:
 
  • #26
haha1234 said:
Nx+fcosθ=mvmax2/R
mgtanθsNcosθ=mvmax2/R
Why do you think Nx = mgtanθ?

Just call the normal force N; it is one of the unknowns. Do as I suggested: Express Nx and Ny in terms of N and θ. (Don't solve for N in terms of anything else. That comes later.)
 
  • #27
Doc Al said:
Why do you think Nx = mgtanθ?

Just call the normal force N; it is one of the unknowns. Do as I suggested: Express Nx and Ny in terms of N and θ. (Don't solve for N in terms of anything else. That comes later.)

Nsinθ+μsNcosθ=mvmax2/R
N(sinθ+μscosθ)=mvmax2/R
vmax=[RN(sinθ+μscosθ)/m]1/2?

But why Nx not equals to mgtanθ?
I think Ny=mg,so Nx=mgtanθ.
 
Last edited:
  • #28
haha1234 said:
Nsinθ+μsNcosθ=mvmax2/R
Good! That equation comes from applying Newton's 2nd law in the horizontal direction. Now get a second equation by applying Newton's 2nd law in the vertical direction.
 
  • #29
haha1234 said:
But why Nx not equals to mgtanθ?
I think Ny=mg,so Nx=mgtanθ.
Well, you are wrong. Don't guess! Write the equations and solve for the unknowns. (The only one you need to solve for is the velocity.)
 
  • Like
Likes 1 person
  • #30
Doc Al said:
Good! That equation comes from applying Newton's 2nd law in the horizontal direction. Now get a second equation by applying Newton's 2nd law in the vertical direction.

Thank you!
I can find the correct answer now!
 
Last edited:
  • #31
Good!
 
Back
Top