Assuming a successful launch, what is the speed of the small CO2 rocket given that it travels about 10 meters in about 3 seconds
1 answer:
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Answer:
v = 7.67 m/s for L= 1m
Explanation:
Let's use the conservation of mechanical energy, at the highest point and the lowest point
Initial. Vertical ruler
Em₀ = mg h
Final. Just before touching the floor
= K = ½ I w²
Em₀ =
m g h = ½ I w²
The moment of inertia of a ruler that turns on one end is
I = 1/3 m L²
Let's replace
m g h = ½ (1/3 m L²) w²2
g h = 1/6 L² w²
They ask for the speed of the end so the height h is equal to the length of the ruler
g L = 1/6 L² w²
The linear and angular variables are related
v = w r
w = v / r
In this case the point of interest a in strangers r = L
g L = 1/6 L² v² / L²
v = √ 6 g L
Let's calculate
Assume that the length of the meter is L = 1 m
v = √ (6 9.8 1)
v = 7.67 m/s