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hw1.solns

# hw1.solns - Physics 318 Homework 1 Solutions February 5...

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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 ) α ˙ z 2 mg, (1) where α > 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 Newton’s second law gives m ¨ z = α ˙ z 2 mg (2) m dv dt = αv 2 mg = m dv dz dz dt = m 2 d ( v 2 ) dz (3) dv 2 dz = 2 α m v 2 2 g (4) v ( z ) = radicalbigg Ae 2 zα/m gm α , (5) where A is a constant of integration. We use the initial condition v = v 0 at z = 0 to solve for A , giving v ( z ) = radicalbigg ( v 2 0 + gm α ) e 2 zα/m gm α . (6) We can find the maximum height reached z max by setting v ( z max ) = 0. This gives e 2 z max α/m = v 2 t v 2 0 + v 2 t , (7) where v t = radicalBig mg/α is the terminal velocity (found by setting F = 0). For the trip down the differential equation is m ¨ z = α ˙ z 2 mg , and the same method gives the solution v ( z ) = radicalbigg Be 2 zα/m + gm α (8) 1

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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 0 | v t | radicalBig v 2 0 + 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( θ ) cos( ϕ ) , y = r sin( θ ) sin( ϕ ) , z = r cos( θ ) (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 ∂θ + r e r ∂ϕ (12) To proceed, we will need to know e r ∂μ for each μ ∈ { r, θ, ϕ } . This is best done with an illustration z x x e r θ e θ e r e r r + r r r + r z y x r r + r sin( ) θ e ϕ e r e r + e r r e r ∂r = 0 e r ∂θ = e θ e r ∂ϕ = sin( θ ) e ϕ (13) Thus using Eqn. (13) we find d r = dr e r + rdθ e θ + r sin( θ ) e ϕ (14) 2
However we can also get this same result from direct differentiation: d r = dx e x + dy e y + dz e z = d ( r sin( θ ) cos( ϕ )) e x + d ( r sin( θ ) sin( ϕ ))

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hw1.solns - Physics 318 Homework 1 Solutions February 5...

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