acceleration on a curve

[dibilo]

a smooth wire is shaped in the form of a parabola y=ax^2 on a horizontal plane. a bead slides along the wire. as it passes the origin, velocity is Vo. what is the acceleration.

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i see the parabola as part of a big circle and decided to use a=v^2/r and i know s=r(theta)... then i am lost... help!!! :cry:

[Galileo]

Sounds like you need to know r, which is the radius of curvature of the parabola at the origin.

(...looks up radius of curvature...)

Here it is:

R=\frac{[1+\left(\frac{dy}{dx}\right)^2]^{3/2}}{\frac{d^2y}{dx^2}}

[e(ho0n3]

This is interesting. Would you show us how to derive this radius of curvature?

[faust9]

Here ya go. Wolfram already did.

Radius of curvature
http://mathworld.wolfram.com/RadiusofCurvature.html

curvature
http://mathworld.wolfram.com/Curvature.html

The MathWorks web site is fountain of math information. If you have a math question it has probably been answered there.

Enjoy, good luck and have fun.

[HallsofIvy]

But just knowing the velocity at one point tells you nothing about the acceleration, on a curve or straight line. Is there more information?

[ehild]

But just knowing the velocity at one point tells you nothing about the acceleration, on a curve or straight line. Is there more information?

We can assume that the bead slides on the horizontal wire with constant speed. Otherwise the problem would be undetermined. The acceleration can be calculated without knowing about curvature, just from its definition: that it is the second derivative of the displacement, that is the componts of the acceleration are the second derivatives of the coordinates with respect to time. Choosing the plane of motion as the (x,y) plane, the components of the velocity are
v_x=\dot x,
v_y= \dot y = 2ax \dot x =2axv_x.

The components of the acceleration are the derivatives of those of the velocity,

a_x= \dot v_x ,
a_y=\dot v_y=2a\dot xv_x+2ax\dot v_y = 2a\left({v_x}^2+x\dot v_x\right)

Moreover,

v_x^2+v_y^2=const \rightarrow v_x \dot v_x + v_y\dot v_y = 0

The acceleration is asked at the origin, x=0, y=0. Here

v_y=0\mbox{, so } v_x^2=v_0^2.

Because of v_y=0 at x=0, the acceleration in x direction is also 0: \dot v_x=0.

(The tangent of the curve at x=0 is the x axis, and the tangential acceleration is zero when the speed is constant)
The result is :

a_x=0.
a_y=2av_0^2, the nagnitude of the acceleration is

2av_0^2.

ehild

[dibilo]

thx for all the replies ... today my teacher gave me another way of solving this prob.

s=r\theta=2x (whereby x is the 2 points \theta cuts at the x-axis)
a=v^2/r
y=ax^2
\theta/2= tan\theta/2= dy/dx= 2ax
therefore \theta=4ax
sub \theta=4ax into s=r\theta
r=1/(2a)
a=2av^2