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[tex]_D[/tex] = ω[tex]_c[/tex] × r[tex]_c[/tex] ----- let this be equ 1
where V[tex]_D[/tex] is velocity of cylinder, ω[tex]_c[/tex] is the angular velocity of drum C and r[tex]_c[/tex] is the radius of drum C
Now, Angular velocity of gear B is;
ω[tex]_B[/tex] = ω[tex]_C[/tex]
ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex] -------- let this equ 2
so;
V[tex]_D[/tex] / 0.1 m = 10V[tex]_D[/tex]
Next, we determine the angular velocity of gear A;
from the diagram;
ω[tex]_A[/tex]( 0.15 m ) = ω[tex]_B[/tex]( 0.2 m )
from equation 2; ω[tex]_B[/tex] = V[tex]_D[/tex] / r[tex]_c[/tex]
so
ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / r[tex]_c[/tex] ) 0.2 m
substitutive in value of radius r[tex]_c[/tex] (0.1 m)
ω[tex]_A[/tex]( 0.15 m ) = (V[tex]_D[/tex] / 0.1 m ) 0.2 m
ω[tex]_A[/tex]( 0.15 ) = 0.2V[tex]_D[/tex] / 0.1
ω[tex]_A[/tex] = 2V[tex]_D[/tex] / 0.15
ω[tex]_A[/tex] = 13.333V[tex]_D[/tex] ----- let this be equation 3
To get the speed of the cylinder, we use energy conversation;
assuming that the final position is;
T₁ + ∑[tex]U_{1-2[/tex] = T₂
0 + m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}I_A[/tex]ω²[tex]_A[/tex] + [tex]\frac{1}{2}I_B[/tex]ω²[tex]_B[/tex]
so
m[tex]_D[/tex]gh = [tex]\frac{1}{2}[/tex]m[tex]_D[/tex]V²[tex]_D[/tex] + [tex]\frac{1}{2}[/tex](m[tex]_A[/tex]k[tex]_A[/tex]²)(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex](m[tex]_B[/tex]k[tex]_B[/tex]²)(10V[tex]_D[/tex])²
we given that; m[tex]_D[/tex] = 50 kg, h = 2 m, m[tex]_A[/tex] = 10 kg, k[tex]_A[/tex] 125 mm = 0.125 m, m[tex]_B[/tex] = 30 kg, k[tex]_B[/tex] = 150 mm = 0.15 m.
we know that; g = 9.81 m/s²
so we substitute
50 × 9.81 × 2 = ( [tex]\frac{1}{2}[/tex] × 50 × V[tex]_D[/tex]²) + [tex]\frac{1}{2}[/tex]( 10 × (0.125)² )(13.333V[tex]_D[/tex])² + [tex]\frac{1}{2}[/tex]( 30 × (0.15)²)(10V[tex]_D[/tex])²
981 = 25V[tex]_D[/tex]² + 13.888V[tex]_D[/tex]² + 33.75V[tex]_D[/tex]²
981 = 72.638V[tex]_D[/tex]²
V[tex]_D[/tex]² = 981 / 72.638
V[tex]_D[/tex]² = 13.5053
V[tex]_D[/tex] = √13.5053
V[tex]_D[/tex] = 3.674955 ≈ 3.67 m/s
Therefore, the speed of the 50-kg cylinder after it has descended is 3.67 m/s
