Answer:
ΔL = MmRgt / (2m + M)
Explanation:
The pulley starts at rest, so the change in angular momentum is equal to the final angular momentum.
ΔL = L − L₀
ΔL = Iω − 0
ΔL = ½ MR²ω
To find the angular velocity ω, first draw a free body diagram for each the pulley and the block.
For the block, there are two forces: weight force mg pulling down, and tension force T pulling up.
For the pulley, there three forces: weight force Mg pulling down, reaction force pulling up, and tension force T pulling down.
Sum of forces in the -y direction on the block:
∑F = ma
mg − T = ma
T = mg − ma
Sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a/R)
T = ½ Ma
Substitute:
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))
The angular velocity after time t is:
ω = αt + ω₀
ω = 2mg / (R (2m + M)) t + 0
ω = 2mgt / (R (2m + M))
Substituting:
ΔL = ½ MR² × 2mgt / (R (2m + M))
ΔL = MmRgt / (2m + M)
The change in angular momentum of the pulley-block system from the instant that the block is released from rest is
[tex]\dfrac{M \times m \times R \times gt}{2m + M}[/tex].
When an object tends to rotate around an axis, then the momentum possessed by the object is known as angular momentum.
Given data:
The rotational inertia of the pulley is, [tex]I=\dfrac{1}{2}MR^{2}[/tex].
The time interval for the falling of block is, [tex]t_{0}[/tex].
The given problem is based on the concept and fundamental of change in angular momentum.
The pulley starts at rest, so the change in angular momentum is equal to the final angular momentum.
ΔL = L − L₀
ΔL = Iω - 0
ΔL = ½ MR²ω ......................................................(1)
For the block, there are two forces: weight force mg pulling down, and tension force T pulling up.
And three forces: weight force Mg pulling down, reaction force pulling up, and tension force T pulling down.
Sum of forces in the y-direction on the block:
∑F = ma mg − T = maT = mg − ma
Sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a/R)
T = ½ Ma
Solve by substituting the values as,
mg − ma = ½ Ma
2mg − 2ma = Ma
2mg = (2m + M) a
a = 2mg / (2m + M)
Now, the expression for the angular acceleration of the pulley is:
αR = 2mg / (2m + M)
α = 2mg / (R (2m + M))
So, the angular velocity after time t is:
ω = αt + ω₀
ω = 2mg / (R (2m + M)) t + 0
ω = 2mgt / (R (2m + M))
Substituting the value of ω in equation (1) as,
ΔL = ½ MR² × 2mgt / (R (2m + M))
ΔL = M × m × R × g t / (2m + M)
Thus, we can conclude that the change in angular momentum of the pulley-block system from the instant that the block is released from rest is [tex]\dfrac{M \times m \times R \times gt}{2m + M}[/tex].
Learn more about the angular momentum here:
https://brainly.com/question/15104254
