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

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Homework Help Overview

The discussion revolves around a physics problem involving a block sliding down a frictionless incline and its behavior at various points on a rollercoaster-like track. The subject area includes concepts from mechanics, specifically energy conservation, forces, and motion along a curved path.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to find the speed and forces at different points on the track. They express uncertainty about calculating the radius R and the height at point B.

Discussion Status

Participants are exploring the relationships between the variables involved, with some suggesting that the answers will be expressed in terms of R. Guidance has been offered regarding the use of trigonometric relations to find the necessary values, but no consensus has been reached on specific methods.

Contextual Notes

There is a noted lack of information regarding the radius R, which affects the ability to calculate certain values directly. The problem also includes imposed constraints related to the frictionless nature of the track.

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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:
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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.
 

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