Mechanics Qualifying Exam Question

In summary: The solution to this isy = A sin(ωt+φ)where φ is the phase at t=0. The initial condition is that at t=0, the wheel is at the peak of the washboard wave, so the initial phase is φ=π/2. At what time t does the wheel lose contact with the washboard surface? At what horizontal location x does this occur? (It should be easy to show that the wheel leaves the surface at the peak of the SHM motion
  • #1
Jeremymu1195
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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.
 
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  • #2
Jeremymu1195 said:

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?
 
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  • #3
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.
 
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  • #4
Jeremymu1195 said:
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.
 
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  • #5
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).
 
  • #6
Jeremymu1195 said:
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?
 
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  • #7
ehild said:
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
 
  • #8
Jeremymu1195 said:
The Normal force is 0 when the wheel loses contact with the surface

You are correct. Any thoughts on how to proceed?
 
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  • #9
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.
 
  • #10
Jeremymu1195 said:
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.
 
  • #11
Jeremymu1195 said:
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 Ny 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, ay = -ω2y.
 
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Related to Mechanics Qualifying Exam Question

1. What is the purpose of a Mechanics Qualifying Exam?

The purpose of a Mechanics Qualifying Exam is to assess a student's understanding and knowledge of fundamental principles and concepts in mechanics, including kinematics, dynamics, and energy. It is typically taken by graduate students in engineering or physics programs as a requirement for advancing to candidacy for a graduate degree.

2. What topics are typically covered in a Mechanics Qualifying Exam?

The specific topics covered in a Mechanics Qualifying Exam may vary depending on the institution, but common areas include kinematics, Newton's laws of motion, work and energy, rotational motion, oscillations, and fluid mechanics. Some exams may also include questions on more advanced topics such as Lagrangian and Hamiltonian mechanics.

3. How should I prepare for a Mechanics Qualifying Exam?

Preparation for a Mechanics Qualifying Exam requires a strong foundation in basic principles and concepts. It is important to review fundamental equations, practice problem-solving techniques, and familiarize yourself with the format and types of questions typically asked on the exam. Some students find it helpful to form study groups or seek out additional resources such as practice exams or review books.

4. Are calculators allowed on a Mechanics Qualifying Exam?

The use of calculators on a Mechanics Qualifying Exam may vary depending on the institution. Some exams may allow the use of calculators, while others may require students to solve problems by hand. It is important to check with the exam administrator or review the exam guidelines to determine if calculators are allowed.

5. How are Mechanics Qualifying Exams graded?

The grading system for Mechanics Qualifying Exams may also vary depending on the institution. Some exams may be graded on a pass/fail basis, while others may assign numerical scores. In most cases, a passing grade is required to advance to candidacy for a graduate degree. It is important to familiarize yourself with the specific grading system for your exam and the passing requirements set by your institution.

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