Mechanics Qualifying Exam Question

[Jeremymu1195]

Homework Statement

A wheel is pulled over a frictionless washboard surface with a constant horizontal velocity component. The equation of the surface is y=Acos(2πx/λ) where A is the height of the surface and x is the horizontal distance along the surface. At what value v_x does the wheel begin to lose contact with the surface, and at what point on the surface will this occur? (assume the radius of the wheel <<λ)

Homework Equations

not sure

The Attempt at a Solution

I wasn't quite sure where to start on this one. I played around with the chain rule (dy/dx) (dx/dt) but didn't get any promising results. If anybody can provide insight on this one, I'd certainly appreciate it.

 

[ehild]

Homework Statement

A wheel is pulled over a frictionless washboard surface with a constant horizontal velocity component. The equation of the surface is y=Acos(2πx/λ) where A is the height of the surface and x is the horizontal distance along the surface. At what value v_x does the wheel begin to lose contact with the surface, and at what point on the surface will this occur? (assume the radius of the wheel <<λ)

Homework Equations

not sure

The Attempt at a Solution

I wasn't quite sure where to start on this one. I played around with the chain rule (dy/dx) (dx/dt) but didn't get any promising results. If anybody can provide insight on this one, I'd certainly appreciate it.

Welcome to PF!
Why don't you show what you tried? The chain rule is a good start.
What is the direction of the velocity at a point with coordinate x? What are the x, y components?
What are the components of the acceleration? And keep in mind Newton's second equation.
What is he condition that the wheel loses contact with the surface?

 

[Jeremymu1195]

By statement of the problem, the velocity in the x-direction is constant, call it v. Since y=A cos(2πx/λ), The velocity in the y-direction (call it u) is given by u=dy/dt. By the chain rule, v=(dy/dx)(dx/dt)= -A(2π/λ)sin(2πx/λ)(dx/dt). But dx/dt = v. So, u = -A(2π/λ)v sin(2πx/λ).

The sum of the forces in y, where vertically upward is taken to be positive, N-mg=ma_y. As long as the wheel is in contact with the surface, N = du/dt = du/dx dx/dt = -A(2π/λ)^2 v^2 cos(2πx/λ). So that a_y = -mA(2π/λ)^2 v^2 cos(2πx/λ)-mg. That is where I get stuck. The condition for the wheel to leave the surface is N=0, a_y = -g. Obvious N→0 as x→n(λ/4). But how does that depend on v_x? I feel like this is a good place to stop and see what comes back to me.

 

[ehild]

By statement of the problem, the velocity in the x-direction is constant, call it v. Since y=A cos(2πx/λ), The velocity in the y-direction (call it u) is given by u=dy/dt. By the chain rule, v=(dy/dx)(dx/dt)= -A(2π/λ)sin(2πx/λ)(dx/dt). But dx/dt = v. So, u = -A(2π/λ)v sin(2πx/λ).

The sum of the forces in y, where vertically upward is taken to be positive, N-mg=ma_y. As long as the wheel is in contact with the surface, N = du/dt = du/dx dx/dt = -A(2π/λ)^2 v^2 cos(2πx/λ). So that a_y = -mA(2π/λ)^2 v^2 cos(2πx/λ)-mg. That is where I get stuck. The condition for the wheel to leave the surface is N=0, a_y = -g. Obvious N→0 as x→n(λ/4). But how does that depend on v_x? I feel like this is a good place to stop and see what comes back to me.

The equation N=du/dt is wrong. N is force, du/dt is acceleration.
The normal force is perpendicular to the surface, it is not vertical. It has both horizontal and vertical components. And there is also the pulling force which ensures the constant horizontal velocity component. It is not clear if that force is horizontal, but we can assume it, as it is not stated otherwise.
The acceleration multiplied by the mass is equal to the sum of forces. Collect all forces and write Newton's second law for both the horizontal and vertical components.

 

[Jeremymu1195]

In the x-direction:

∑F=Fpull-Nx=0

In the y-direction:

∑F=Ny-mg=ma

where a=(du/dx)(dx/dt)=-A(2π/λ)^2 v^2 cos(2πx/λ).

 

[ehild]

In the x-direction:

∑F=Fpull-Nx=0

In the y-direction:

∑F=Ny-mg=ma

where a=(du/dx)(dx/dt)=-A(2π/λ)^2 v^2 cos(2πx/λ).

Correct. What is the normal force when the wheel looses contact with the surface?

 

[Jeremymu1195]

Correct. What is the normal force when the wheel looses contact with the surface?

The Normal force is 0 when the wheel loses contact with the surface

 

[AlephNumbers]

The Normal force is 0 when the wheel loses contact with the surface

You are correct. Any thoughts on how to proceed?

 

[Jeremymu1195]

My attempt,

N2 for the y-direction reads,

Ny-mg=ma_y or,

Ny-mg=-mA(2π/λ)^2 v^2 cos(2πx/λ).

When the wheel loses contact, Ny=0,

-mg=-mA(2π/λ)^2 v^2 cos(2πx/λ)

so,

v^2=g/(A(2π/λ)^2 cos(2πx/λ)) (*)

but when x=nλ, cos(2πx/λ)=1 and a this corresponds to local maxima on the cosine curve, it is the most likely place for the wheel to lose contact.

So, v^2=g/(A(2π/λ)^2 ).

When the wheel loses contact a_y = -g.

so,

a_y=-A(2π/λ)^2 g/(A(2π/λ)^2 ) cos(2πx/λ) = -gcos(2πx/λ)

and for x=nλ, a_y=-g

So that checks, but the general form of v^2 (*) bothers me. (*) violates the assumption that v is constant.

 

[haruspex]

So that checks, but the general form of v^2 (*) bothers me. (*) violates the assumption that v is constant.

You are told the horizontal component is constant.

 

[TSny]

v^2=g/(A(2π/λ)^2 cos(2πx/λ)) (*)

... the general form of v^2 (*) bothers me. (*) violates the assumption that v is constant.

v is constant. Your starred equation tells you what v must be in order for N_y to equal zero when the wheel is at horizontal location x.

An interesting approach to this problem is to go to the frame of reference moving horizontally with speed v so that the wheel only has vertical SHM motion in this frame. So, a_y = -ω²y.