Why Does a Car Accelerate When Rolling Down a Hill?

In summary, the conversation was about a question regarding a car rolling down a hill and the calculations involved. The author was unsure of whether they should post their working and solution in the thread or not, and asked for suggestions. The author then proceeded to share their working, which involved using conservation of energy and equations of motion to find the velocity and acceleration of the car. The other participant in the conversation gave positive feedback and also provided some advice on the approach to answering questions in the forum.
  • #1
gelfand
40
3
[Mod note: Thread moved from General Discussion forum, so no homework template is shown]

So - I've no idea where to write this but...

I was looking at this post : https://www.physicsforums.com/threads/car-rolling-down-a-hill.302812/

And started to answer it as it was similar to some stuff that I had been doing, and was curious whether I was understanding it.

Also - the thread *kinda* got left for dead, there are no real conclusions for someone to learn from there even though it comes up in searches etc.

I'm not sure what's suggested with this kinda thing?

This is what I was *going* to reply with before I realized the thread was closed >.<

So my working was :I'm not sure if this is poor etiquette but I found this post and did some
working that I would like to check (and perhaps it's best if it's contained
within this thread for future learners)

For the first part about the speed at the bottom -

I note that I have an inclined plane (as said), it's angle and the length that
the car rolls down.

##l = ## length, and this is the hypotenuse for this triangle.

Using this I have the initial height of the car as ##h = l \times \sin(\theta)##.

From here I can use conservation of energy as

$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$

Where ##W##, ##\text{Energy(Lost)}##, ##KE_0## and ##PE_f## are all zero. ##PE_f## is zero
as at the bottom of the hill I'm considering the potential energy (due to
gravity) to be zero.

This gives me

$$
PE_0 = KE_f
$$

Which I can write as

$$
mgl \sin(\theta) = \frac{1}{2}mv^2
$$

Dividing by ##m## and rearranging gives

$$
v = \sqrt{
2 l g \sin(\theta)
}
$$

The acceleration down the hill I'm not so sure about, but I think that I have
the force of gravity pulling the car down, so this should be related to the
acceleration.

in the direction of the hill the force of gravity is

$$
F_{grav} = mg \sin(\theta)
$$

As there is no resistance I would answer with this as the acceleration.

The time that it takes for the car to reach would be dependent on my answer for
acceleration. If as I've suggested it's constant then I'm able to use the
equations of motion$$
x(t) = x_0 + v_0 t + \frac{1}{2}at^2
$$

$$
v(t) = v_0 + at
$$

$$
v^2 = v_0^2 + 2a(x - x_0)
$$

Here I would use the second one as it's purely in ##v, a## and I have this
information. Using the found expression for velocity I have

$$
\sqrt{
2 l g \sin(\theta)
} = 0 + at
$$

Dividing through by the expression found for ##a## gives

$$
\frac{\sqrt{2 l g \sin(\theta)}}{a} = t
$$

Hope this makes sense
 
Last edited by a moderator:
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  • #2
Looks good to me.

(FYI: Old threads are routinely closed after a while.)
 
  • #3
Did you figure out what to use for the acceleration a? You had F = m*g*sin(theta). Your workings out all look solid and well thought out. If you had gone and answered like you just did, it is likely that somebody may have removed it for "too much information, too soon". I found this out when I first started participating, here. The theme here (at least in this homework help section) is to give the person enough information to get them thinking on the right track and see if they can figure it out for themself. If they can't figure it out, then they could post additional questions which can be answered or a little more information can be imparted to help them along the path. Hopefully they will learn and remember it better if they try several solutions with some help, rather than just posting and then seeing the full solution. They may say to themself "oh now I understand". But often it may not sink in and stay with that approach.
 
  • #4
scottdave said:
Did you figure out what to use for the acceleration a? You had F = m*g*sin(theta). Your workings out all look solid and well thought out. If you had gone and answered like you just did, it is likely that somebody may have removed it for "too much information, too soon". I found this out when I first started participating, here. The theme here (at least in this homework help section) is to give the person enough information to get them thinking on the right track and see if they can figure it out for themself. If they can't figure it out, then they could post additional questions which can be answered or a little more information can be imparted to help them along the path. Hopefully they will learn and remember it better if they try several solutions with some help, rather than just posting and then seeing the full solution. They may say to themself "oh now I understand". But often it may not sink in and stay with that approach.
cheers Scott - I wrote this because the thread was *really* old (if you look the original thread is from 2009), and there's not much information with it (so it's useless as a resource).

Given it's age I don't (personally) see the problem with a more 'fully fleshed' solution. However, I *do* appreciate the 'socratic' method that's taken initially (and appreciate when people take the time to question me in the same way).

Also - I posted this in the general section first as I didn't know where to put it - or whether such thoughts (answering old threads that have been left dead and appear in searches) was of any use (as the threads themselves aren't that useful as it stands)

No worries though, cheers
 
  • #5
Doc Al said:
Looks good to me.

(FYI: Old threads are routinely closed after a while.)

All good, thanks :)
 
  • #6
You may have noticed that we have an "Open Practice Problems" forum which is intended to contain unanswered homework threads from before 2015, and which are now "up for grabs" for anyone to have a go at. I don't know why the thread in question was not moved there, as it's from 2009. Maybe @Greg Bernhardt can provide some insight here.
 
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  • #7
jtbell said:
I don't know why the thread in question was not moved there, as it's from 2009.
It has 4 replies. Only unanswered homework threads are in the practice problem forums.
 
  • #8
jtbell said:
You may have noticed that we have an "Open Practice Problems" forum which is intended to contain unanswered homework threads from before 2015, and which are now "up for grabs" for anyone to have a go at. I don't know why the thread in question was not moved there, as it's from 2009. Maybe @Greg Bernhardt can provide some insight here.
Ah no I didn't realize - cheers.

I just posted this in the general and it's been bounced over here :S

no worries.
 
  • #9
Greg Bernhardt said:
It has 4 replies. Only unanswered homework threads are in the practice problem forums.
Right fair. I wish I'd never unearthed it now ha .
 
  • #10
IMO it's OK. The question was effectively "unanswered" because the original poster never finished it, at least not here, despite the responses. I bet Greg used an automated query to pick out threads containing exactly one post, for moving to the practice problems forum. Identifying threads like the one in question would have required a visual inspection of all threads, which would have been very tedious!
 
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  • #11
Yes I see the confusion, when I say unanswered I mean without a reply.
 
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  • #12
jtbell said:
IMO it's OK. The question was effectively "unanswered" because the original poster never finished it, at least not here, despite the responses. I bet Greg used an automated query to pick out threads containing exactly one post, for moving to the practice problems forum. Identifying threads like the one in question would have required a visual inspection of all threads, which would have been very tedious!
Yeah I wouldn't fancy it :P

I only wrote it because it popped up whilst I was asking a related question. Cheers
 
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Related to Why Does a Car Accelerate When Rolling Down a Hill?

1. How does gravity affect a car rolling down a hill?

Gravity is the force that pulls objects towards the center of the Earth. When a car is on a hill, gravity causes it to roll down the slope as it is pulled towards the ground.

2. What factors can affect the speed of a car rolling down a hill?

The speed of a car rolling down a hill can be affected by several factors, including the incline of the hill, the weight and size of the car, the force of gravity, and the presence or absence of friction or air resistance.

3. How does the shape of the hill affect the motion of a rolling car?

The shape of the hill can impact the motion of a rolling car in several ways. A steeper incline will result in a faster acceleration due to the increased force of gravity. A curved hill can also affect the trajectory of the car, causing it to turn or curve as it rolls down.

4. What happens to the energy of a car rolling down a hill?

As a car rolls down a hill, it gains kinetic energy due to its increased speed. This kinetic energy comes from the potential energy of the car at the top of the hill, which is converted into motion as the car rolls downwards.

5. Can a car roll up a hill without an engine?

Yes, a car can roll up a hill without an engine if it has enough potential energy at the top of the hill. However, the car will eventually slow down and stop due to the force of gravity pulling it back down the hill. This is why cars need an engine to maintain their speed and overcome the force of gravity when driving up a hill.

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