What is the Force and Speed at Different Points on a Rollercoaster?

  • Thread starter Thread starter Jacobpm64
  • Start date Start date
  • Tags Tags
    Rollercoaster Type
Jacobpm64
Messages
235
Reaction score
0

Homework Statement


A block of mass [tex]m = 1.62 kg[/tex] slides down a frictionless incline as in the figure(link at the bottom). The block is released at a height of [tex]h = 3.91 m[/tex] above the bottom of the loop.
(a) What is the force of the inclined track on the block at the bottom (point A)?
(b) What is the force of the track on the block at point B?
(c) At what speed does the block leave the track?
(d) How far away from point A does the block land on level ground?
(e) Sketch the potential energy [tex]U(x)[/tex] of the block. Indicate the total energy on the sketch.

Homework Equations


[tex]E = K + U[/tex]
[tex]a = m \frac{v^2}{R}[/tex]

The Attempt at a Solution



So, since the track is frictionless, total energy is constant. Therefore, we can calculate the energy at the top of the track (and this has to be the same at every part throughout the trip).

[tex]E = K_{top} + U_{top}[/tex]
[tex]E = 0 + U_{top}[/tex]
[tex]E = mgh[/tex] Which I can find (62.14 J).

Now, this must be the energy everywhere during the trip. Therefore, at the bottom (point A):
[tex]E = K_{A} + U_{A}[/tex]
[tex]mgh = \frac{1}{2}mv_{A}^2 + mgh_{A}[/tex]
[tex]mgh = \frac{1}{2}mv_{A}^2 + 0[/tex] Since [tex]h_{A} = 0[/tex]
[tex]mgh = \frac{1}{2}mv_{A}^2[/tex]
[tex]v_{A} = \sqrt{2gh}[/tex] Which I find to be ( 8.76 m/s ).

Now, to answer part (a), at point A:
[tex]N - mg = ma = m\frac{v_{A}^2}{R}[/tex]
[tex]N = mg + m \frac{v_{A}^2}{R}[/tex] Which is ([tex]\frac{124.31}{R} + 15.89[/tex] Joules)

Is there any way to find what [tex]R[/tex] is?

Part (b):
When the block is at point B,
[tex]N - mgcos(45^{\circ}) = m \frac{v_{B}^2}{R}[/tex]
[tex]N = mgcos(45^{\circ}) + m \frac{v_{B}^2}{R}[/tex]

The problem here is, I don't know [tex]v_{B}[/tex], and I think I can only figure it out if I know [tex]h_{B}[/tex], which I cannot see how to find.

Any help would be appreciated! (Then I can continue working the other parts)

http://img27.imageshack.us/img27/3923/physicsg.jpg"

Note: In the figure, [tex]\alpha = 45^{\circ}[/tex].
 
Last edited by a moderator:
Since R was not given, your answers will be in terms of R.
 
Okay, I see. All of my answers will be in terms of [tex]R[/tex].

Can anyone give me any hints on how to find [tex]v_B[/tex] and [tex]h_B[/tex]?

Thanks!
 
You know the angle and the radius R. So you can find the tangential velocity and the height using trigonometric relations. Again, these answers will be in terms of R.
 

Similar threads

Replies
14
Views
3K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
46
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
10
Views
4K
  • · Replies 28 ·
Replies
28
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K