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)
Density is defined as the mass divided by the volume.
You can alter the density of a substance by adjusting either its mass or volume.
Increasing the volume while maintaining a constant mass will result in a decrease in density (as the denominator of the fraction increases).
Furthermore, reducing the mass while keeping the volume the same will also lower the density (because the numerator is reduced).
Therefore, to achieve a lower density, you should either reduce the mass or increase the volume, keeping the other constant.
I hope this is helpful.
Answer:
x₂=2×1
Explanation:
According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;
mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.
mgx=(kx)²/2
x=2mg/k----------------compression when the object is at rest
However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy
Thus, if kx²=mv² then
v=x *√(k/m) ----------------where v=0
<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁
