Question

Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10 kg and a radius of gyration of 125 mm about its center of mass. Gear B and drum C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm

Answer 1

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$_D$ = ω$_c$ × r$_c$  ----- let this be equ 1

where V$_D$ is velocity of cylinder,  ω$_c$ is the angular velocity of drum C and r$_c$ is the radius of drum C

Now, Angular velocity of gear B is;

ω$_B$ = ω$_C$

ω$_B$ = V$_D$ / r$_c$  -------- let this equ 2

so;

V$_D$ / 0.1 m = 10V$_D$

Next, we determine the angular velocity of gear A;

from the diagram;

ω$_A$( 0.15 m ) = ω$_B$( 0.2 m )

from equation 2; ω$_B$ = V$_D$ / r$_c$

so

ω$_A$( 0.15 m ) = (V$_D$ / r$_c$ ) 0.2 m

substitutive in value of radius r$_c$ (0.1 m)

ω$_A$( 0.15 m ) = (V$_D$ / 0.1 m ) 0.2 m

ω$_A$( 0.15 ) = 0.2V$_D$ / 0.1

ω$_A$ =  2V$_D$  / 0.15

ω$_A$ = 13.333V$_D$   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑$U_{1-2$ = T₂

0 + m$_D$gh = $\frac{1}{2}$m$_D$V²$_D$ + $\frac{1}{2}I_A$ω²$_A$ + $\frac{1}{2}I_B$ω²$_B$

so

m$_D$gh = $\frac{1}{2}$m$_D$V²$_D$ + $\frac{1}{2}$(m$_A$k$_A$²)(13.333V$_D$)² + $\frac{1}{2}$(m$_B$k$_B$²)(10V$_D$)²

we given that; m$_D$ = 50 kg, h = 2 m, m$_A$ = 10 kg, k$_A$ 125 mm = 0.125 m, m$_B$ = 30 kg, k$_B$ = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( $\frac{1}{2}$ × 50 × V$_D$²) + $\frac{1}{2}$( 10 × (0.125)² )(13.333V$_D$)² + $\frac{1}{2}$( 30 × (0.15)²)(10V$_D$)²

981 = 25V$_D$² + 13.888V$_D$² + 33.75V$_D$²

981 = 72.638V$_D$²

V$_D$² = 981 / 72.638

V$_D$² = 13.5053

V$_D$ = √13.5053

V$_D$ = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s