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)
Answer:
19.62 ms
Explanation:
t = Time taken = 2 s
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s² (we take downward direction as positive)
Using the equations of motion
The pebble's speed upon contact with the water is 19.62 ms
Answer:
H = 109.14 cm
Explanation:
Given,
Assume that the total energy equals 1 unit.
Energy remaining after the first collision = 0.78 x 1 unit
Balance after the first impact = 0.78 units
Remaining energy after the second impact = 0.78 ^2 units
Balance after the second impact = 0.6084 units
Remaining energy after the third impact = 0.78 ^3 units
Balance after the third impact = 0.475 units
The height reached after the third collision is equivalent to the remaining energy.
Let H denote the height achieved after three bounces.
0.475 (m g h) = m g H
H = 0.475 x h
H = 0.475 x 2.3 m
H = 1.0914 m
H = 109.14 cm
