Uniform Circular Motion Hills and Cars

  • #1
1a. Homework Statement
You are riding in a car over a hill that has a radius of curvature R. What speed should you be going if you want to feel one-third of your normal weight. Express your answer in terms of the radius R and the acceleration due to gravity g.
1b. You continue your journey at the same speed down the other side of the hill until you reach a valley that has a radius of curvature of two-thirds R. In terms of your true weight (mg) what is your apparent weight at the lowest point of the valley?


Homework Equations


F=(GMm)/(r^2)
[tex]\sum[/tex]F=ma
a=(v^2)/(r)
v=(2[tex]\pi[/tex]r)/(T)
v=(2pi r)/(T(period))

The Attempt at a Solution


1a.) I know that the sum of the forces in the y direction are F=(mg)(1/3)-N=mv^2/r
so v=[tex]\sqrt{rg/3-rN/m}[/tex]

1b.) Your weight should be more than your actual weight because mg-N=(mv^2)/r
I just don't know if i should put 2/3r in the equation. Then it would be mg=N-(mv^2)/(2/3r)

Sorry if i used the symbols incorrectly.
Thanks!

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
for the first question, the weight you feel as you go over the curve is the normal reaction N.

So the centripetal force is given by mv2/R=mg-N. You want N to be (1/3)mg, so what is v in terms of R?

For the second part now: If at the lowest point, the weight points downwards and the normal reaction points upwards (your 'apparent' weight). In what direction should the centripetal force point and what is it equal to in terms of your weight and normal reaction?
 
  • #3
So v2=[tex]\sqrt{r(g-1/3g}[/tex] which means v=[tex]\sqrt{r2/3g}[/tex] for 1a.

for 1b.) the centripetal force points in the direction the normal force does, upwards. the equation would be N-mg=mv[tex]^{2}[/tex]/(2r/3)
if i replace v with the equation from 1a.) then the equation would be:

mg=N-(m(2/3)gr)/(2/3 r) which simplifies to

mg=N-m
Is this correct?
Thank you
 
  • #4
rock.freak667
Homework Helper
6,223
31
So v2=[tex]\sqrt{r(g-1/3g}[/tex] which means v=[tex]\sqrt{r2/3g}[/tex] for 1a.

for 1b.) the centripetal force points in the direction the normal force does, upwards. the equation would be N-mg=mv[tex]^{2}[/tex]/(2r/3)
if i replace v with the equation from 1a.) then the equation would be:

mg=N-(m(2/3)gr)/(2/3 r) which simplifies to

mg=N-mg
Is this correct?
Thank you

1a) the 'g' should be in the numerator

b) Make N the subject of the formula and you'll get your answer. (see the part in red where you left out the 'g', I'll assume that is a typo)
 
  • #5
Thanks!
for b i ended up getting mg=N/2
 
  • #6
179
0
Doesn't centripetal acceleration/force always point to the center of the circle? In 1a) I know your apparent weight would be less but wouldn't you have N upwards, mg downwards, and mv2/r downwards also?
 
  • #7
rock.freak667
Homework Helper
6,223
31
Thanks!
for b i ended up getting mg=N/2

well right N=2mg. Which means that you 'feel' as if you weight twice as much.
 
  • #8
To Chris
You are right, centripetal force always point to the center of the circle.
I just chose a coordinate system where the centripetal acceleration is on the positive y-axis.
 

Related Threads on Uniform Circular Motion Hills and Cars

  • Last Post
Replies
7
Views
8K
  • Last Post
Replies
14
Views
2K
Replies
2
Views
4K
  • Last Post
Replies
3
Views
10K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
23
Views
5K
Replies
14
Views
510
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
660
  • Last Post
Replies
3
Views
977
Top