River crossing and spiral motion in cylindrical coordinates

[supersix2]

For some reason I just don't get my dynamics homework. Here are the few problems I am having.

Problem 1:
A river is flowing north at 3 m/s (uniform current). If you want to travel in a straight line from point C to point D in a boat that moves at a constant speed of 10 m/s relative to the water, in what direction should you point the boat? How long does it take to make the crossing?

Point C is on the west bank of the river and the river is 500 m wide. Point D is 400 m north of point C on the east side of the river. Basicaly points C and D and the east side of the river make a right triangle with a base of 500 and height of 400 and the hypontenuse CD.

I found the angle of CD to be 38.66 degrees.

I set my coordinate system so that north is positive y or j direction and that east is positive x or i direction.

My first step was to find the the velocity of the the boat and to do that i used the following equation.

Velocity of the boat relative to river (10 m/s) = Vcos(38.66)i + Vsin(38.66)j - 3j

V = velocity of the boat

I found the magnitude by taking the square root of all of that and solving for V (i had to use the quadratic formula to get the roots for V). Therefore:
V= 11.62 m/s

Now I'm stuck. I have no idea where to go from here to find the direction. I substituted 11.62 back into my original velocity equation to get:

V wrt river = 9.07i + 4.26j

I'm thinking that that is only the componets of the velocity not the actual direction the boat is pointing maybe I'm wrong I don't know. Please help.

**Forgot to add the other problem I was stuck on

Problem 2:

A particle moves in a spiral path described by r = 1m and θ = 2z rad (where z is in meters) and moves along the z axis at a constant speed of |v | = 1000 m/s. What is the velocity of the particle in terms of cylindrical coordinates?

I have an equatoin that shows the velocity in cylindrical coordinates which is:
v = dr/dt êr + ω êθ + dz/dt êz

And there I am stuck. I think I can find ω by multiplying the the velocity by 2z because that would be ds/dt * dθ/ds. But after that how would i go about finding dr/dt and dz/dt? The ê are the unit vecots in the r, θ, and z directions respectively.

 

[HallsofIvy]

In problem 1 you say "Velocity of the boat relative to river (10 m/s) = Vcos(38.66)i + Vsin(38.66)j - 3j" That, since it take into account the motion of the river, is relative to the shore if the boat were aimed directly at point D (and if the river were flowing south!).

Assume you aim your boat at an angle θ to the shore. Then the velocity of your both, relative to the river, is 10cos(θ)i+ 10sin(θ)j. The velocity of the river, relative to the shore is +3j (NOT -3j) so the velocity of the boat relative to the shore is 10cos(θ)i+ (10sin(θ)+ 3)j and you want that to take you directly to D: you want (10sin(θ)+3)/10cos(θ)= 400/500. Solve that for θ.

2. Well, since r= 1, a constant, dr/dt= 0, doesn't it? ω IS dθ/dt so, no, you don't multiply by 2z : if θ= 2z, then dθ/dt= 2 dz/dt.

Your formula becomes 0 e_r+ 2 dz/dt e_θ+ dz/dt e_z. Set the length of that to 1000 and solve for dz/dt. Then plug that value back into your formula

 

[supersix2]

Awesome thanks for the help. Everything worked out nicely.