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Curvilinear Motion

by mathmannn
Tags: 2-d motion, curvilinear motion, dynamics
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mathmannn
#1
Jan27-12, 11:07 PM
P: 15
1. The problem statement, all variables and given/known data

A stunt car is driven off a cliff with a speed of 110 ft/s. What is the gravitational acceleration of the car normal to its path after falling for 3 seconds?

2. Relevant equations

The kinematic equations...?

I'm pretty sure that this should be done in Normal and tangential components, so with that said:

[itex] s = \theta r [/itex]
[itex] a_t = \dot{v} = v \frac{dv}{ds} = \alpha r [/itex]
[itex] v = \dot{s} = \omega r [/itex]
[itex] a_n = \frac{v^2}{\rho} = \omega^2 r [/itex] Where [itex] \rho [/itex] is the radius of curvature.

3. The attempt at a solution

For the x-direction:
[itex] (v_0)_x = 110 [/itex]

[itex] t = 3 [/itex]

[itex] \Delta x = (v_0)_x t = (110)(3) = 330 [/itex]

For the y-direction:
[itex] y = y_0 + (v_0)_y t + \frac{1}{2} a t^2 [/itex]

Solving for distance in the y-direction:

[itex] y = \frac{1}{2}(-g)t^2 \quad (t = 3) [/itex]

[itex] y = -144.9 ft [/itex]


But I really have no idea if any of that is necessary, or if it is where do I go from there?
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ehild
#2
Jan27-12, 11:36 PM
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Hi mathmann,

You got the position of the car, and you know that the acceleration is g, downward. You need the normal component of the acceleration. The normal of a curve is perpendicular to its tangent. And you also know that the velocity is tangent to the path at any point.

Find the velocity vector at the point (330, -144.9) first.

ehild
mathmannn
#3
Jan27-12, 11:52 PM
P: 15
Quote Quote by ehild View Post
Hi mathmann,

You got the position of the car, and you know that the acceleration is g, downward. You need the normal component of the acceleration. The normal of a curve is perpendicular to its tangent. And you also know that the velocity is tangent to the path at any point.

Find the velocity vector at the point (330, -144.9) first.

ehild


Ok so for the x-direction since there is no acceleration then [itex] v_x = 110 [/itex] and for y-direction [itex] v^2 = v_0^2 + 2(-g)(-144.9) [/itex] to get [itex] v_y = 96.6 [/itex], this is my guess on what to do next.

[itex] v = \sqrt{v_x^2 + v_y^2} = 146.39 [/itex]

But that is a scalar..? So again I'm stuck. Thank you for your first post though

ehild
#4
Jan27-12, 11:58 PM
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Curvilinear Motion

The velocity is a vector. Write it out with its horizontal and vertical components. I see, you have set the coordinate system that the y axis points upward.

ehild
mathmannn
#5
Jan28-12, 12:38 AM
P: 15
Quote Quote by ehild View Post
The velocity is a vector. Write it out with its horizontal and vertical components. I see, you have set the coordinate system that the y axis points upward.

ehild
So [itex] v = (110, 96.6) [/itex] ft/s. I know this is a very basic question I just am so lost on what to do.
mathmannn
#6
Jan28-12, 12:40 AM
P: 15
I don't know if this will be any use.. But its the picture for the question.
Attached Thumbnails
Screen Shot 2012-01-28 at 12.39.04 AM.png  
ehild
#7
Jan28-12, 12:46 AM
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The car moves downward. What is the sign of the y component of velocity? Could you show the velocity vector in the picture?

You need the direction perpendicular to the velocity. What do you know about the components of the vectors which are perpendicular to each other?


ehild


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