# Mechanics! Help!

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#1
Would anyone do the question below?
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15 years ago
#2
(a)
The resisting force is -k(x'^2 + y'^2)w, where k is a positive constant (which we are going to work out) and w is a unit vector in the direction of the particle's movement. Since w = (x', y')/sqrt(x'^2 + y'^2), we can write the resisting force as

-k sqrt(x'^2 + y'^2) * (x', y')

So

m x'' = -k sqrt(x'^2 + y'^2) * x'
m y'' = -mg - k sqrt(x'^2 + y'^2) * y'

When the particle is travelling at terminal velocity, x'' = y'' = 0 and so x' = 0 and y' = -V. Hence -mg + kV^2 = 0 and k = mg/V^2. Putting k = mg/V^2 into the equations of motion,

m x'' = -(mg/V^2)sqrt(x'^2 + y'^2) * x'
m y'' = -mg - (mg/V^2)sqrt(x'^2 + y'^2) * y'

x'' = -(g/V^2)sqrt(x'^2 + y'^2) * x'
y'' = -g - (g/V^2)sqrt(x'^2 + y'^2) * y'

x'' = -g[x' sqrt(x'^2 + y'^2)/V^2]
y'' = -g[1 + y' sqrt(x'^2 + y'^2)/V^2]

(b)
x''
= v (-d(psi)/dt sin(psi)) + v' cos(psi)
= -y' d(psi)/dt + v' cos(psi)

y''
= v (d(psi)/dt cos(psi)) + v' sin(psi)
= x' d(psi)/dt + v' sin(psi)

y'' x' - x'' y'
= (x'^2 + y'^2) d(psi)/dt + v' x' sin(psi) - v' y' cos(psi)
= (x'^2 + y'^2) d(psi)/dt + v' v sin(psi) cos(psi) - v' v sin(psi) cos(psi)
= (x'^2 + y'^2) d(psi)/dt

d(psi)/dt = (y'' x' - x'' y') / (x'^2 + y'^2)

(c)
x'' = -g[x' sqrt(x'^2 + y'^2)/V^2]
y'' = -g[1 + y' sqrt(x'^2 + y'^2)/V^2]

y'' x' - x'' y' = -g x'

d(psi)/dt = -g x'/v^2

(d/dt) (1/x'^2)
= (-2/x'^3)x''
= (-2/x'^3)(-gvx'/V^2)
= 2gv/(x'^2 V^2)

(d/d(psi)) (1/x'^2)
= [(d/dt) (1/x'^2)] dt/d(psi)
= [(d/dt) (1/x'^2)] / [d(psi)/dt]
= [2gv/(x'^2 V^2)] [-v^2/(gx')]
= -2v^3/(x'^3 V^2)
= -2/(cos^3(psi) V^2)
0
#3
Thank you very much
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