Injector orifice patterms and size will affect propellant mixing and distribution. a)-True b)-False
1 answer:
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Answer:
a) 2.18 m/s^2
b) 9.83 m/s
Explanation:
The flywheel has a moment of inertia
J = m * k^2
Where
J: moment of inertia
k: radius of gyration
In this case:
J = 144 * 0.45^2 = 29.2 kg*m^2
The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.
T = p * r
r is the radius of the wheel.
T = m1 * g * r
T = 18 * 9.81 * 0.6 = 106 N*m
The torque will cause an acceleration on the flywheel:
T = J * γ
γ = T/J
γ = 106/29.2 = 3.63 rad/s^2
SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:
at = γ * r
at = 3.63 * 0.6 = 2.81 m/s^2
Now that we know the acceleration of the block we can forget about the flywheel.
The equation for uniformly accelerated movement is:
X(t) = X0 + V0*t + 1/2*a*t^2
We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:
X(t) = 1/2*a*t^2
Rearranging
t^2 = 2*X(t)/a
It will reach the 1.8 m in 3.6 s.
Now we use the equation for speed under constant acceleration:
V(t) = V0 + a*t
V(3.6) = 2.81 * 3.6 = 9.83 m/s
