Question

The velocity of the 7.7-kg cylinder is 0.49 m/s at a certain instant. What is its speed v after dropping an additional 1.28 m? The mass of the grooved drum is m = 10.5 kg, its centroidal radius of gyration is = 300 mm, and the radius of its groove is ri = 275 mm. The frictional moment at O is a constant 18.4 N·m.

Answer 1

the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

Step-by-step explanation:

Using the work energy system

`T_1 + U_(1-2) + T_2`

The initial kinetic energy
`T_1` is ;

`T_1 = (1)/(2)m_cv_1^2 + (1)/(2)I_o \omega^2`

`T_1 = (1)/(2)m_cv_1^2 + (1)/(2)(m_d \overline k^2)((v_1)/(r_i))^2`

where;

`m_c` = mass of the cylinder = 7.7 kg

`v_1 =` initial velocity of the cylinder = 0.49 m/s

`I_o`= moment of inertia of the drum about O

`m_d =`mass of the drum = 10.5 kg

`r_i =` radius of gyration = 0.3 m

`\omega` = angular velocity of the drum

`T_1 = (1)/(2)(7.7)(0.49)^2 + (1)/(2)(10.5*(0.3)^2((0.49)/(0.275))^2`

`T_1 = 2.426 \ J\\`

The final kinetic energy is also calculated as:

`T_2= (1)/(2)m_cv_2^2 + (1)/(2)(m_d \overline k^2)((v_2)/(r_i))^2`

`T_2= (1)/(2)(7.7)(v_2^2)^2 + (1)/(2)(10.5*(0.3)^2((v_2)/(0.275))^2`

`T_1 =10.10 v_2^2`

Similarly, The workdone by all the forces on the cylinder can be expressed as:

`U_(1-2) = m_cg(h) - ((M)/(r_i))h`

where;

g = acceleration due to gravity

h = drop in height of the cylinder

M = frictional moment at O

`U_(1-2) = 7.7*9.81*1.28 - 18.4((1.28)/(0.275))`

`U_(1-2) =11.04 \ J`

Finally, using the work energy application;

`T_1 + U_(1-2) + T_2`

2.426 + 11.04 = 10.10
`v_2^2`

13.466 = 10.10
`v_2^2`

`v_2^2` =
`(13.466)/(10.10)`

`v_2^2` = 1.333

`v_2 = √(1.333)`

`v_2 = 1.15 \ m/s`

Thus, the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s