# The advisory speed for a car on a sloped bend

• Grizzly_1
In summary, the conversation is about a student struggling with a circular motion question and seeking help to understand it. The student has attached their working and the mark-scheme answer, noting that their approach and the mark-scheme's approach are different. They ask for assistance in understanding the correct method to solve the problem. The conversation then delves into discussing the correct equation for Newton's second law in the vertical direction. The expert suggests using LaTeX to write equations for better readability and quotes the correct equation for the motion.

#### Grizzly_1

Homework Statement
At bends on motorways the road is sloped so that a car is less likely to slide out of its
lane when travelling at a high speed.

Figure 7 shows a car of mass 1200 kg travelling around a curve of radius 200 m.
The motorway is sloped at an angle of 5.0°.

Figure 8 shows the weight W and reaction force N acting on the car. The advisory
speed for the bend is chosen so that the friction force down the slope is zero.

Suggest an appropriate advisory speed for this section of the motorway
Relevant Equations
F=mv^2/r
Hello all, I am sadly stuck on the last part of a circular motion question sheet I was given for homework. I have a mark-scheme with me, but it has actually given me more questions than answers. I have attached my working, and how I arrived at my answer, and the differences it has with the mark-scheme answer. To note, the mark-scheme answer rounds to my answer, but the trigonometry they used is entirely opposite to mine. I would really appreciate some help in understanding:

Thanks to anyone who can lend me a hand on this. I assume it is just poor physics on my end, so getting through this will help keep me up to scratch on circular motion.

Grizzly_1 said:
Homework Statement: At bends on motorways the road is sloped so that a car is less likely to slide out of its
lane when travelling at a high speed.

Figure 7 shows a car of mass 1200 kg travelling around a curve of radius 200 m.
The motorway is sloped at an angle of 5.0°.

Figure 8 shows the weight W and reaction force N acting on the car. The advisory
speed for the bend is chosen so that the friction force down the slope is zero.

Suggest an appropriate advisory speed for this section of the motorway
Relevant Equations: F=mv^2/r

Hello all, I am sadly stuck on the last part of a circular motion question sheet I was given for homework. I have a mark-scheme with me, but it has actually given me more questions than answers. I have attached my working, and how I arrived at my answer, and the differences it has with the mark-scheme answer. To note, the mark-scheme answer rounds to my answer, but the trigonometry they used is entirely opposite to mine. I would really appreciate some help in understanding:
View attachment 325957View attachment 325958View attachment 325959
Thanks to anyone who can lend me a hand on this. I assume it is just poor physics on my end, so getting through this will help keep me up to scratch on circular motion.

I get ##13.10 \frac{m}{s}##, is that the correct answer?

Grizzly_1
erobz said:
I get ##13.10 \frac{m}{s}##, is that the correct answer?
Well yes, 13.1 is the answer from the mark-scheme (see the second attachment), my problem is how you arrive at this result. You must have resolved the forces differently to me, and if you don't mind I would like to know how/why you did it this way. Thank you for your reply!

Grizzly_1 said:
Well yes, 13.1 is the answer from the mark-scheme (see the second attachment), my problem is how you arrive at this result. You must have resolved the forces differently to me, and if you don't mind I would like to know how/why you did it this way. Thank you for your reply!
I put my coordinates parallel and perpendicular to the slope. Then in the non-inertial frame of the car did a force balance parallel to the slope. Care to try that, and see what you get?

EDIT: Oh, and please use Latex to write your equations (see the LaTeX Guide in the lower left corner of your reply). Pictures are supposed to be for diagrams. It's hard to read handwritten equations, and it makes it easier to quote any parts of your working that may contain an error.

If you don't want to go that route, then in your approach ask yourself "What direction is the radial acceleration" That's where you made the mistake (as far as I can tell).

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erobz said:
I put my coordinates parallel and perpendicular to the slope. Then in the non-inertial frame of the car did a force balance parallel to the slope. Care to try that, and see what you get?

EDIT: Oh, and please use Latex to write your equations (see the LaTeX Guide in the lower left corner of your reply). Pictures are supposed to be for diagrams. It's hard to read handwritten equations, and it makes it easier to quote any parts of your working that may contain an error.
erobz said:
If you don't want to go that route, then in your approach ask yourself "What direction is the radial acceleration" That's where you made the mistake.
Thank you for both of these replies, I would think the centripetal acceleration is horizontally towards the centre. I apologise for not using LaTeX, I will get up to scratch to use it next time. Finally I am unsure of what you mean in your first reply.

berkeman
You have an expression of Newton's second law for the motion in the horizontal direction. That is ##\cancel{N\cos\!\varphi=\dfrac{mv^2}{r}}## and I agree.

How about writing an expression for Newton's second law in the vertical direction?

On edit:
The correct equation is
##N\sin\!\varphi=\dfrac{mv^2}{r}##.

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Grizzly_1 said:
Thank you for both of these replies, I would think the centripetal acceleration is horizontally towards the centre. I apologise for not using LaTeX, I will get up to scratch to use it next time. Finally I am unsure of what you mean in your first reply.
Did you resolve ALL vectors parallel and normal to the slope (the radial acceleration is a vector too)

kuruman said:
You have an expression of Newton's second law for the motion in the horizontal direction. That is ##N\cos\!\varphi=\dfrac{mv^2}{r}## and I agree.

How about writing an expression for Newton's second law in the vertical direction?
Don't you mean ##N \sin \varphi = \frac{mv^2}{r}##?

kuruman
Grizzly_1 said:
I apologise for not using LaTeX, I will get up to scratch to use it next time.
Sending you a PM with LaTeX tips...

Grizzly_1 said:
I assume it is just poor physics on my end,
Your mistake is the line ##W\cos(\theta)=N##. For that, you presumably resolved normal to the slope, but the centripetal acceleration has a component normal to the slope, so those two forces are not in balance.

erobz
erobz said:
Don't you mean ##N \sin \varphi = \frac{mv^2}{r}##?
Yes, that's what meant for the horizontal component. Thanks.

erobz