Answer :
Answer:
the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s
Explanation:
Using the work energy system
[tex]T_1 + U_{1-2} + T_2[/tex]
The initial kinetic energy [tex]T_1[/tex] is ;
[tex]T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}I_o \omega^2[/tex]
[tex]T_1 = \dfrac{1}{2}m_cv_1^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_1}{r_i})^2[/tex]
where;
[tex]m_c[/tex] = mass of the cylinder = 7.7 kg
[tex]v_1 =[/tex] initial velocity of the cylinder = 0.49 m/s
[tex]I_o[/tex]= moment of inertia of the drum about O
[tex]m_d =[/tex]mass of the drum = 10.5 kg
[tex]r_i =[/tex] radius of gyration = 0.3 m
[tex]\omega[/tex] = angular velocity of the drum
[tex]T_1 = \dfrac{1}{2}(7.7)(0.49)^2 + \dfrac{1}{2}(10.5*(0.3)^2(\dfrac{0.49}{0.275})^2[/tex]
[tex]T_1 = 2.426 \ J\\[/tex]
The final kinetic energy is also calculated as:
[tex]T_2= \dfrac{1}{2}m_cv_2^2 + \dfrac{1}{2}(m_d \overline k^2)(\dfrac{v_2}{r_i})^2[/tex]
[tex]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[/tex]
[tex]T_1 =10.10 v_2^2[/tex]
Similarly, The workdone by all the forces on the cylinder can be expressed as:
[tex]U_{1-2} = m_cg(h) - (\dfrac{M}{r_i})h[/tex]
where;
g = acceleration due to gravity
h = drop in height of the cylinder
M = frictional moment at O
[tex]U_{1-2} = 7.7*9.81*1.28 - 18.4(\dfrac{1.28}{0.275})[/tex]
[tex]U_{1-2} =11.04 \ J[/tex]
Finally, using the work energy application;
[tex]T_1 + U_{1-2} + T_2[/tex]
2.426 + 11.04 = 10.10 [tex]v_2^2[/tex]
13.466 = 10.10 [tex]v_2^2[/tex]
[tex]v_2^2[/tex] = [tex]\dfrac{13.466}{10.10}[/tex]
[tex]v_2^2[/tex] = 1.333
[tex]v_2 = \sqrt{1.333}[/tex]
[tex]v_2 = 1.15 \ m/s[/tex]
Thus, the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s