Car rolling down a hill

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1. Apr 14, 2017

gelfand

[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 realised 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: Apr 14, 2017
2. Apr 14, 2017

Staff: Mentor

Looks good to me.

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

3. Apr 14, 2017

scottdave

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. Apr 14, 2017

gelfand

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. Apr 14, 2017

gelfand

All good, thanks :)

6. Apr 14, 2017

Staff: Mentor

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.

7. Apr 14, 2017

Greg Bernhardt

It has 4 replies. Only unanswered homework threads are in the practice problem forums.

8. Apr 14, 2017

gelfand

Ah no I didn't realise - cheers.

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

no worries.

9. Apr 14, 2017

gelfand

Right fair. I wish I'd never unearthed it now ha .

10. Apr 14, 2017

Staff: Mentor

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!

11. Apr 14, 2017

Greg Bernhardt

Yes I see the confusion, when I say unanswered I mean without a reply.

12. Apr 14, 2017

gelfand

Yeah I wouldn't fancy it :P

I only wrote it because it popped up whilst I was asking a related question. Cheers