# Acceleration on a curve

1. Aug 10, 2004

### 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.

...............................................................................
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!!!

Last edited: Aug 10, 2004
2. Aug 10, 2004

### 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:

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

3. Aug 10, 2004

### e(ho0n3

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

4. Aug 10, 2004

### faust9

5. Aug 10, 2004

### HallsofIvy

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

6. Aug 11, 2004

### ehild

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

7. Aug 11, 2004

### 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

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