Answer:
ΔL = MmRgt / (2m + M)
Explanation:
The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.
ΔL = L − L₀
ΔL = Iω − 0
ΔL = ½ MR²ω
To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.
For the block, two forces act: the weight force mg downward and tension force T upward.
For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.
For the sum of forces in the -y direction on the block:
∑F = ma
mg − T = ma
T = mg − ma
For the sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a/R)
T = ½ Ma
Substituting gives:
mg − ma = ½ Ma
2mg − 2ma = Ma
2mg = (2m + M) a
a = 2mg / (2m + M)
The angular acceleration of the pulley is:
αR = 2mg / (2m + M)
α = 2mg / (R (2m + M))
Finally, the angular velocity after time t is:
ω = αt + ω₀
ω = 2mg / (R (2m + M)) t + 0
ω = 2mgt / (R (2m + M))
Substituting into the previous equations gives:
ΔL = ½ MR² × 2mgt / (R (2m + M))
ΔL = MmRgt / (2m + M)
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Answer:
17.35 × 10^(-6) m
Explanation:
Mass; m = 50 kg
Weight; W = 554 N
From the formula:
W = mg
This simplifies to; 554 = 50g
g = 554/50
g = 11.08 m/s²
Also, using the formula;
mg = GMm/r²
hence; g = GM/r²
Rearranging gives;
r = √(GM/g)
With G as a known constant of 6.67 × 10^(-11) Nm²/kg²
r = √(6.67 × 10^(-11) × 50/11.08)
r = 17.35 × 10^(-6) m
