This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 318 Homework 1 Solutions February 5, 2008 Problem 1 We choose the z axis to be in the vertical direction. The force on the particle is given by F = sign ( z ) z 2 mg, (1) where > 0 is the constant of proportionality. The factor of sign( z ) in the first term makes the friction force be in the direction opposite to the motion. For the way up, z > 0 and Newtons second law gives m z = z 2 mg (2) m dv dt = v 2 mg = m dv dz dz dt = m 2 d ( v 2 ) dz (3) dv 2 dz = 2 m v 2 2 g (4) v ( z ) = radicalbigg Ae 2 z/m gm , (5) where A is a constant of integration. We use the initial condition v = v at z = 0 to solve for A , giving v ( z ) = radicalbigg ( v 2 + gm ) e 2 z/m gm . (6) We can find the maximum height reached z max by setting v ( z max ) = 0. This gives e 2 z max /m = v 2 t v 2 + v 2 t , (7) where v t = radicalBig mg/ is the terminal velocity (found by setting F = 0). For the trip down the differential equation is m z = z 2 mg , and the same method gives the solution v ( z ) = radicalbigg Be 2 z/m + gm (8) 1 for some constant B . It must be that B = v 4 t /A so that v ( z max ) = 0. Now we can solve for v ( z = 0): v (0) = radicalBig B + v 2 t (9) = v  v t  radicalBig v 2 + v 2 t . (10) [Note there was a typo in the homework set, the absolute value sign was omitted from the final answer (10). The answer given in the homework set is correct if v t is interpreted to be terminal speed instead of terminal velocity.] Problem 2 Spherical polar coordinates are defined by the equations x = r sin( ) cos( ) , y = r sin( ) sin( ) , z = r cos( ) (11) a.) Using the chain rule, we can write d r = d ( r e r ) = dr e r + r e r r dr + r e r d + r e r d (12) To proceed, we will need to know e r for each { r, , } . This is best done with an illustration z x x e r e e r e r r + r r r + r z y x r r + r sin( ) e e r e r + e r r e r r = 0 e r = e e r = sin( ) e (13) Thus using Eqn. (13) we find d r = dr e r + rd e...
View
Full
Document
This homework help was uploaded on 02/24/2008 for the course PHYS 3318 taught by Professor Flanagan during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 FLANAGAN
 mechanics, Force, Work

Click to edit the document details