This question is incomplete, the missing image is uploaded along this answer below.
Answer:
the speed of the 50-kg cylinder after it has descended is 3.67 m/s
Explanation:
Given the data in the question and the image below;
relation between velocity of cylinder and velocity of the drum is;
V = ω
× r
----- let this be equ 1
where V is velocity of cylinder, ω
is the angular velocity of drum C and r
is the radius of drum C
Now, Angular velocity of gear B is;
ω = ω
ω = V
/ r
-------- let this equ 2
so;
V / 0.1 m = 10V
Next, we determine the angular velocity of gear A;
from the diagram;
ω( 0.15 m ) = ω
( 0.2 m )
from equation 2; ω = V
/ r
so
ω( 0.15 m ) = (V
/ r
) 0.2 m
substitutive in value of radius r (0.1 m)
ω( 0.15 m ) = (V
/ 0.1 m ) 0.2 m
ω( 0.15 ) = 0.2V
/ 0.1
ω = 2V
/ 0.15
ω = 13.333V
----- let this be equation 3
To get the speed of the cylinder, we use energy conversation;
assuming that the final position is;
T₁ + ∑ = T₂
0 + mgh =
m
V²
+
ω²
+
ω²
so
mgh =
m
V²
+
(m
k
²)(13.333V
)² +
(m
k
²)(10V
)²
we given that; m = 50 kg, h = 2 m, m
= 10 kg, k
125 mm = 0.125 m, m
= 30 kg, k
= 150 mm = 0.15 m.
we know that; g = 9.81 m/s²
so we substitute
50 × 9.81 × 2 = ( × 50 × V
²) +
( 10 × (0.125)² )(13.333V
)² +
( 30 × (0.15)²)(10V
)²
981 = 25V² + 13.888V
² + 33.75V
²
981 = 72.638V²
V² = 981 / 72.638
V² = 13.5053
V = √13.5053
V = 3.674955 ≈ 3.67 m/s
Therefore, the speed of the 50-kg cylinder after it has descended is 3.67 m/s
