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Mechanics

A bead moves on a smooth wire in the x,y plane (y is vertical). The wire is described by the equation y=f(x). Show that the speed of the bead is

v=dx/dt*sqrt(1+f'(x)^2)

I have so far found the force on the bead in the direct of the line of movement, it was mg/sin(theta) where theta is the angle between the horizontal and the tangent. I then found acceleration=a=g/sin(theta)=g*sqrt(1+f'(x)^2)/f'(x)

What can I do from here (if what I have done is correct)?

Reply 1

ghostwalker

Original post by james22

What can I do from here (if what I have done is correct)?

Forget about F=ma, and g, and all that.

Let s be the displacement along the curve.

So, we're looking for v which is ds/dt

Can you take it from there? I've left a lot out obviously, but it might seem familiar.

Reply 2

atsruser

Original post by james22

A bead moves on a smooth wire in the x,y plane (y is vertical). The wire is described by the equation y=f(x). Show that the speed of the bead is

v=dx/dt*sqrt(1+f'(x)^2)

As ghostwalker has pointed out, you are looking for the rate of change of displacement along the curve, ds/dt. The result is more-or-less trivial if:

1. you can see how a small displacement along the curve is related to small displacements along the x and y axes (these are dx and dy, of course)

2. you know the chain rule.

A further hint would be to consider what

\sqrt{1+(f'(x))^2}

must be if the result is to be true, and to try to figure out how you would show that it has the required value.

Reply 3

james22

Could I say ds/dt=ds/dx*dx/dt

then ds/dx=d/dx(integral of sqrt(1+f'(x)^2))=sqrt(1+f'(x)^2) by the formula for arc length?

Reply 4

atsruser

Original post by james22

Could I say ds/dt=ds/dx*dx/dt

Yes.

then ds/dx=d/dx(integral of sqrt(1+f'(x)^2))=sqrt(1+f'(x)^2) by the formula for arc length?

Not really, though you're thinking along the right lines. Rather, you need to think about how you would derive the formula for arc length from first principles. Draw a diagram that relates ds, dx, and dy and all should be clear.

Reply 5

james22

OK thanks. The next bit of the question is to find d^2x/dt^2

I found that first -mg/sin(theta)=md^s/dt^2 where theta is the angle the tangent makes with the x axis.

I then got tan(theta)=f' so sin(theta)=f'/sqrt(1+(f')^2)

I used the first part to get d^s/dt^2=d/dt(ds/dt)

=d/dt(dx/dt*sqrt(1+(f')^2))=x''(1+(f')^2)+(x')^2(1+(f')^2)f'f''

I combined the 2 to get

-g*sqrt(1+(f')^2)/f'=x''*sqrt(1+(f')^2)+(x')^2*(1+(f')^2)f'f''

but this lead to an incrrect answer