Question

Time (s) Problem 4 (16 pts) A sailing ship of mass, m, is initially at rest, i.e. V(O) strong wind arises of magnitude and keep push the ship. 0. At time Vo = 40m/s Assume that the force of the wind on the sails in the direction of travel is given by Fw(t) = Bw [V - v(t)] Assume that the viscous drag of the water on the ship is given by a. (4 pts) Formulate a differential equation that describes the ship's velocity, v(t) b. (8 pts) Solve the differential equation from by Laplace tramsform and Inverse Laplace trandform and write an expression for the ship's velocity, v(t)? c. (4 pts) Write an expression for the steady-state velocity, Vs.s , in terms of system parameters

Answer 1

Applying a Force Balance on the sailing ship, we have:

Assuming the direction of the movement as positive

F1 = m* dv/dt

Substituting Fw on this equations, results:

Fw=m* dv/dt

Ordering the equation, we have:

Bw* (.-v(t)) = m * dv/dt

which is an ordinary differential equation of first order

m * dv/dt + Bw* u(t) = Bw* v,

Appling Laplace transform:

m * L (dv/dt) + Bu * L(u(t)) = Bw *V,* L(1)

Knowing that: L{f'(x)} = s*F(s) - f(0); L{f(x)} = F(s) and L{constant} = constant*L{1} = 1/s

Rearranging the equation we have:

m*(3*V(s) – v(0)) + Bw*V(s) = Bw*v./s

, Knowing that v( t = 0 ) = 0, and dividing all the equation by m:

(ms + B) *V(S) – v(0) = Bu *ve/s

solving for V(s)

(s + Bw/m)*V(s) = Bw*v./(m*s)

, We need to expand in partial fractions, because Laplace transform it is a linear operator

V(s) = (Bu*v./m)/(s*(s + Bu/m))

V(s) = (BW*v/m)/(s*(s+Bw/m)) = A1/s + A2/(s + Bw/m)

Applying Inverse Laplace Transform we get:

1- (V(s)) = A1 *[-+(1/8) + A2 * [-(1/(s +Bq/m))

Knowing that L^-1{1/s} = 1; L^-1{1/(s+a)} = exp(-a*t), the above equation results:

v(t) = A1 + A2 *exp(-Bw/m *t)

All we left to calculate is the constants A1 and A2, so we proceed

We use the original form of the equation before expanding in partial fractions, and the form expanded in partial fractions

(BW*v./m)/(s* (s + Bw/m)) = 41/s + A2/(s+Bw/m)

Let us pass the denominator of the left part to the right part (if you can follow me in paper and pen, it would be great)

(Bu*v./m) = (A1/s + 2/(s + Bw/m))*(s*(s+Bw/m))

(BW*vo/m) = (A1*(s + Bw/m) + A2 *S)

(BW *vo/m) = A1 * 5+ A1 * Bw/m + A2 *S

(Bu *v./m) = (41 + A2) * S + A1 * BW/m

Know, we can form two equations, we notice that there are no terms on the left side with an s, so, (A1+A2) would be equal to cero, but, we also notice that there is a constant term on the left side, so its counterpart on the right side would be equal to them. (You can use this method with any differential equation that you solve by Laplace Transform)

(1)

(Bu *v./m) = 41 * Bw/m

(2)

(A1 + A2) = 0

Solving the system, we obtain that:

A1 = v.

A2 = -A1 = -v.

So, the expression of v(t) results in:

u(t) = A1 + A2 *exp(-Bw/m *t) = V. -V, *exp(-Bw/m*t)

u(t) = v.*(1 - exp(-Bw/m *t))

The steady-state velocity can be calculated evaluating v(t) in time t too hight (When the transient state disappears). In this equation, we notice that the exponential term it's negative, so, its effect on v(t) would be decreasing as time t increase and it will disappear. So, vss would be:

uit = 0) = US5 = V.