{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam1_sol

# exam1_sol - PHY 3221 Mechanics I Fall Term 2010 Exam 1 This...

This preview shows pages 1–3. Sign up to view the full content.

PHY 3221: Mechanics I Fall Term 2010 Exam 1, September 29 2010 This is a closed book exam lasting 50 minutes. There are three problems worth a total of 20 pts. Begin each problem on a fresh sheet of paper. Use only one side of the paper. Avoid microscopic handwriting. Put your name, the problem number and the page number in the upper right-hand corner of each sheet. To receive partial credit you must explain what you are doing. Carefully labelled figures are important! Randomly scrawled equations are not helpful. Draw a box around important results (or at least results which you think might be im- portant). Good luck! 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 1. Conservative forces. [4 pts] Consider the following force: F x = y , F y = x , F z = z . Is this a conservative force? If so, find the potential energy U ( x, y, z ) associated with it. Solution To check if the force is conservative, calculate vector ∇ × vector F : ( vector ∇ × vector F ) x = ∂F y ∂z ∂F z ∂y = 0 0 = 0 . ( vector ∇ × vector F ) y = ∂F z ∂x ∂F x ∂z = 0 0 = 0 . ( vector ∇ × vector F ) z = ∂F x ∂y ∂F y ∂x = 1 1 = 0 . Therefore vector ∇ × vector F = 0 and the force is conservative. It can be represented as vector F = −∇ U ( x, y, z ) F x = y = ∂U ∂x = U ( x, y, z ) = yx + C 1 ( y, z ) F y = x = ∂U ∂y = x ∂C 1 ∂y = ∂C 1 ∂y = 0 = C 1 ( y, z ) = C 2 ( z ) = U ( x, y, z ) = yx + C 2 ( z ) F z = z = ∂U ∂z = ∂C 2 ∂z = ∂C 2 ∂z = z = C 2 ( z ) = 1 2 z 2 + const The end result is U ( x, y, z ) = xy 1 2 z 2 + const Check: the negative of the gradient of this function should reproduce the given force.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}