course Mth 275
Since the RTF format seemed to copy correctly, I'll stick with that for now.
Section 11.3:This is an interesting section. It took a while to work through the concepts. The only problems that gave me difficulty were 39c and 42. 39c says to show that Rsubm = (vsub0/g)*sqrt(vsub0^2+2g*ssub0) and angle alpha subm = tan(-1) (vsub0/ sqrt(vsub0^2+2g*ssub0). I got nowhere with this. I tried substituting v sub0^2/g for Rsubm, but this left me with vsub0 = sqrt(vsub0^2+2g*ssub0), and squaring both sides and dividing by vsub0^2 left 1 = 2g*ssub0. But I couldn? see where to take that.
For the x and y coordinates we get
x(t) = v0 cos(alpha) * t and
y(t) = s0 + v0 sin(alpha) * t - 1/2 g t^2.
The projectile strikes the ground when y(t) = 0. Using the quadratic formula to solve y(t) = 0 we obtain
t = (-v0 sin(alpha) + - sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / (-g). Since the square root is greater than v0 sin(alpha), only the - of the +- yields a positive result for t. Thus
t = (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g.
Substituting this into the expression for x(t) we obtain
x( (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g )
= v0 cos(alpha) (v0 sin(alpha) + sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) / g
= v0^2 cos(alpha)sin(alpha) + v0 / g * cos(alpha) sqrt( v0^2 sin^2(alpha) + 2 s0 g) ) .
To maximize R as a function of alpha we take the derivative with respect to alpha and set it equal to zero:
v0^2 cos(alpha)sin(alpha) + v0 / g * cos(alpha) sqrt( v0^2 sin^2(alpha) + 2 s0 g) )
v0^2 (cos^2(alpha) - sin^2(alpha) ) + v0 / g * ( cos(alpha) * v0^2 * 2 sin(alpha) cos(alpha) / sqrt(v0^2 sin^2(alpha) + 2 s0 g) - sin(alpha) sqrt(v0^2 sin^2(alpha) + 2 s0 g) ) = 0.
Dividing through by v0^2 we get
cos^2(alpha) - sin^2(alpha) + v0 / g sin(alpha) cos^2(alpha) / sqrt( v0^2 sin^2*alpha + 2 s0 g) - sin(alpha) sqrt(sin^2(alpha) + 2 s0 g / v0) = 0.
At this point the algebra is running together on the page and I don't have time to write it down on paper. The identitiy cos^2(alpha) = 1 - sin^2(alpha) can reduce this to an expression in just sin(alpha), and I think it will come out
1 - 2 sin^2(alpha) + v0 / (g sqrt( v0^2 sin^2*alpha + 2 s0 g) )(sin(alpha) - sin^3(alpha)) - sin(alpha) sqrt(sin^2(alpha) + 2 s0 g / v0) = 0, but I'm getting lost in the parentheses etc. and it's late.
So I'll revert to DERIVE to obtain their expression for the derivative, which is
COS(alpha)·(v0·‹(v0^2·SIN(alpha)^2 + 2·g·s0)/g + v0^2·SIN(alpha)/g)COS(alpha)^2·(v0^3·SIN(alpha)/(g·‹(v0^2·SIN(alpha)^2 + 2·g·s0)) + v0^2/g) - v0·SIN(alpha)·‹(v0^2·SIN(alpha)^2 + 2·g·s0)/g - v0^2·SIN(alpha)^2/g
and their solution for alpha, which makes no sense.
In any case you get the idea. The solution should be
alpha = v0 / sqrt( v0^2 + 2 g s0)
and plugging this in to the expression for x(t) gives you the maximum range. You would also need to do a first- or second-derivative test to verify that this is a max, but that's pretty straightforward.
The details of the above aren't quite right--something off in my bookkeeping--but the procedure is as I outline it and if everything is done carefully and correctly the specified results fall out of the equations.
Problem 42 says to use a previous problem to show that r*Fsubtheta(t) = d/dt (mr^2dtheta/dt). I couldn? find any opening to this one.
It appears to me that the equation derived in problem 42 gives a way to calculate the tangential moment acting on the orbiting body.
I'll need my glasses and a piece of paper to keep myself straight on this one. Not as involved algebraically as the preceding, but conceptually more challenging, so I'll be able to give you an an explanation; but I don't want to try to do it in my head right now, and I need to get this file posted. I'll get this for you in the morning, and hopefully also have time to type out my explanation of u x (u x (u x v) ).