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

Answer:

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

Explanation:

Using the work energy system

$T_1 + U_{1-2} + T_2$

The initial kinetic energy $T_1$ is ;

$T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}I_o \omega^2$

$T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_1}{r_i})^2$

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

$T_1 = \dfrac{1}{2}(7.7)(0.49)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{0.49}{0.275})^2$

$T_1 = 2.426 \ J$

The final kinetic energy is also calculated as:

$T_2= \dfrac{1}{2}m_cv_2^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_2}{r_i})^2$

$T_2= \dfrac{1}{2}(7.7)(v_2^2)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{v_2}{0.275})^2$

$T_1 =10.10 v_2^2$

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

$U_{1-2} = m_cg(h) - (\dfrac{M}{r_i})h$

where;

g = acceleration due to gravity

h = drop in height of the cylinder

M = frictional moment at O

$U_{1-2} = 7.7*9.81*1.28 - 18.4(\dfrac{1.28}{0.275})$

$U_{1-2} =11.04 \ J$

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 $v_2^2$

$v_2^2$ = $\dfrac{13.466}{10.10}$

$v_2^2$ = 1.333

$v_2 = \sqrt{1.333}$

$v_2 = 1.15 \ m/s$

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