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Answer:
Fa = 57.32 N
Explanation:
given data
mass = 5 kg
acceleration = 4 m/s²
angular velocity ω = 2 rad/s
solution
first we take here moment about point A that is
∑Ma = Iα + ∑Mad ...............1
put here value and we get
so here I = ( ) × m × L² ................2
I = ( ) × 5 × 0.8²
I = 0.267 kg-m²
and
a is = r × α
a = 0.4 α
so now put here value in equation is 1
0 = 0.267 α + m r α (0.4) - m A (0.4)
0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4
so angular acceleration α = 7.5 rad/s²
so here force acting on x axis will be
∑ F(x) = m a(x) ..............3
a(x) = m a - m rα
put here value
a(x) = 5 × 4 - 5 × 0.4 × 7.5
a(x) = 5 N
and
force acting on y axis will be
∑ F(y) = m a(y) .............. 4
a(y) - mg = mrω²
a(y) - 5 × 9.81 = 5 × 0.4 × 2²
a(y) = 57.1 N
so
total force at A will be
Fa = ...............5
Fa =
Fa = 57.32 N
Answer:
a) specific impulse = 203 s
(b) mass flow rate = 2511 kg/s
(c) throat diameter = 0.774 m
(d) exit diameter = 1.95 m
(e) thrust = 5.3*10^6 N/m^2
(f) thrust at sea level = 15.6*10^6 N
(g) thrust with hydrogen = 5*10^6 N
(h) thrust with stagnation = 5*10^6 N
Explanation:
See the attached file for explanation.
