Solve Normal & Tangentiel Co-ordinate System Finding Component of Acceleration

  • #1

Homework Statement

The box slides down the slope described by the equation y = 0.05x^2 m , where x is in meters. If the box has components of velocity and acceleration of vx = 3 m/s and ax = -1.5 m/s^2 at x = 5 m determine the y component of the velocity and acceleration

Homework Equations

The answers to the question were 3 m/s and 0.15 m/s^2 respectively. Getting the velocity was easy enough but I'm not sure how to get acceleration. I'm hoping some of you may be able to help.

The Attempt at a Solution

Getting velocity was simply done by taking the derivative of the slope (dy/dx = 0.1x). At x = 5 the slope is 0.5 so the y component of velocity is equal to vy = vx * 0.5 = 1.5 m/s

In terms of getting acceleration there are two components one parallel to the motion and one perpendicular. There is no indication of how the velocity is changing so I know of no of getting the tangential component.

My first attempt was find an acceleration vector such that is perpendicular to the motion (that is wrong because the path is a parabola). My second attempt was to use dynamics at that particular point to determine the y component of acceleration (assuming that gravity had something to do with it). Again, my answer was wrong.

I would be grateful for an explanation as to how to get to the answer 0.15 m/s^2 for the y-component of acceleration.

Answers and Replies

  • #2

u need the chain rule for the acceleration.

ay=vy'=d/dt(0.05*2x* x')

= d/dt( 0.1 x * x')
= 0.1( x' x' + x x'')
  • #3

The direction of the acceleration vector is going to follow that of the second derivative of the curve (i.e., the slope of the slope). That this is so should be clear by the fact that as the box moves the velocity vector must parallel the slope of the curve, and the curve's slope changes according to how dy/dx changes. Thus the velocity vector changes direction according to the curve d2y/dx2.
  • #4

Thanks Dragon.ENGin & gneill!

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