Sliding block on a ramp

  • Thread starter Zorodius
  • Start date
  • #1
184
0
Imagine an initially stationary block is placed at some height on a frictionless ramp. The ramp has several hills and valleys in it. Assume the block never flies off the ramp, the hills have identical circular tops, and the hills are of progressively greater heights, all shorter than the height at which the block is first placed.

Why is the normal force on the block different from hilltop to hilltop? I realize that the velocity and kinetic energy of the block varies from hilltop to hilltop as it is transferred to and from gravitational potential energy. I believe this means that the centripetal acceleration is greater on shorter hills, because the kinetic energy and thus the velocity is greater there - right? But how is this related to the normal force acting on the block?
 

Answers and Replies

  • #2
Doc Al
Mentor
44,984
1,248
The centripetal acceleration depends on [itex]v^2[/itex] as well as the radius of curvature: [itex]a = v^2/r[/itex]. The shorter the hill, the greater the kinetic energy (proportional to [itex]v^2[/itex]).

To find the normal force acting on the block at the top of a hill, apply Newton's 2nd Law:
[tex]N - mg = -mv^2/r[/tex], or
[tex]N = mg - mv^2/r[/tex]
thus, the greater the KE (greatest for shorter hills) and the sharper the hill (smaller r), the less normal force between the hill and the block. Make sense?

Of course if the speed is too great, and the curve too sharp, the weight will not be enough to hold the block to the hill--it will go flying off as a projectile. :smile:
 
  • #3
AKG
Science Advisor
Homework Helper
2,565
4
I'm not exactly sure what is meant by "the hills are of progressively greater heights, all shorter than the height at which the block is first placed." I can envision this in a couple of ways, so I diagram might help if my following explanation does not. At a hilltop, the net force is the centripetal force, which is (gravity - normal). The force of gravity is constant, so a larger centripetal force means a smaller normal force. Each hill gets successively larger, but it doesn't say that each hill is at a progressivly shorter height. It could be that the block starts at 100m above ground, all hilltops are 99m above ground, but some hills are made of bigger circles than others if the valley before it was also big (hopefully you can visualize that). The point is that since you have no information about the heights of the hilltops compared to each other, just assume they're constant in this case (although a good question would not have left such an ambiguity). So, all that's changing are the radii. As the radii increase, the centripetal accelerations decrease (because the speed would be the same with constant heights for hilltops), and as they decrease, the normal forces increase. You can even kind of guess this because as we approach infinity, the circles approach flat surfaces, and the normal force approaches its maximum, equal to the force of gravity.

Note, I said to assume that the heights of the hills are equal, because if we didn't, they could change in any which way, and it's even possible that all the heights decrease such that the increase in kinetic energy, and thus speed, are just enough to counterbalance the effects of increasing radii, and so no change in normal force occurs. Or it could be arranged so the normal force goes up and down, or even decreases if the steepness of the overall ramp is great enough, I would think.
 
  • #4
Imagine an initially stationary block is placed at some height on a frictionless ramp. The ramp has several hills and valleys in it. Assume the block never flies off the ramp, the hills have identical circular tops, and the hills are of progressively greater heights, all shorter than the height at which the block is first placed.

Why is the normal force on the block different from hilltop to hilltop?
I wonder if there is something out there you could use to draw an online diagram of this problem or something I could put in a post...
 
  • #5
AKG
Science Advisor
Homework Helper
2,565
4
Tony Zalles said:
I wonder if there is something out there you could use to draw an online diagram of this problem or something I could put in a post...
There is, it's called Paint (or whatever you have). Make a drawing (if you have windows you should have "Paint") and save it as a jpg, then upload it as an attachment, or you can upload it to your own online space if you have some (perhaps a geocities account), then provide a link to there.
 
  • #6
184
0
Thanks for the help, it makes sense now.
 
  • #7
AKG said:
There is, it's called Paint (or whatever you have). Make a drawing (if you have windows you should have "Paint") and save it as a jpg, then upload it as an attachment, or you can upload it to your own online space if you have some (perhaps a geocities account), then provide a link to there.
Oh yea....and to think I have adobe photoshop on this computer...I could have nice little diagram...well at least now I know I can upload diagrams as attatchments...so that's cool...

BTW would you happen to know a good program for typing out mathematical and physics equations and symbols...

Thanks.
 
  • #8
AKG
Science Advisor
Homework Helper
2,565
4
You can use the LaTeX typesetting capabilities right here. For example, the following:

[ tex ] \int _{-a} ^{+a} \pi f(x) \vec{V} (x) dx [ / tex ]

Without the spaces between "[" and "tex" and "/", you get:

[tex] \int _{-a} ^{+a} \pi f(x) \vec{V} (x) dx [/tex]

For more on this, search the site for "LaTeX" and there's a 5-page-or-so thread all about the different things you can do with LaTeX. There's also a little pdf file that has some commonly used stuff, but you often find it's not enough. A lot of the time I go to Mathworld.com and search for something, and a lot of the LaTeX commands show up in the search page.
 

Related Threads on Sliding block on a ramp

  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
15
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
18
Views
874
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
12K
  • Last Post
Replies
7
Views
6K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
14
Views
8K
Top