Respuesta :
Answer:
[tex]W_s = 0.67 rad/s[/tex]
it is lost the 98.82% of the energy.
Explanation:
for answer this, we will use the conservation of the angular momentum L:
[tex]L_f = L_i[/tex]
so:
[tex]I_1W_1 + I_2W_2 = I_sW_s[/tex]
where [tex]I_1[/tex] is the moment of inertia, [tex]W_1[/tex] is the angular velocity of the cylinder 1, [tex]I_2[/tex] is the moment of inertia of the second cylinder, [tex]W_2[/tex] is the angular velocity of the second cylinder, [tex]I_s[/tex] is the moment of inertia of the cylinders couple and [tex]W_s[/tex] is the angular velocity of the cylinders couple
note: we will take the clockwise rotation as positive.
Replacing the values, we get:
[tex](2 kg*m^2)(5 rad/s) + (1 kg*m^2)(-8 rad/s) = (3 kg*m^2)W_s[/tex]
solving for [tex]W_s[/tex]:
[tex]W_s = 0.67 rad/s[/tex]
for find the kinetic energy lost we must calculate the initial Ei and the final energy Ef as:
[tex]E_i = \frac{1}{2}I_1W_1^2+\frac{1}{2}I_2W_2^2[/tex]
[tex]E_i = \frac{1}{2}(2)(5)^2+\frac{1}{2}(1)(8)^2[/tex]
[tex]E_i = 57 J[/tex]
and,
[tex]E_f = \frac{1}{2}I_sW_s^2[/tex]
[tex]E_f = \frac{1}{2}(3)(0.67)^2[/tex]
[tex]E_f = 0.67335 J[/tex]
now, the percentage of the original kinetic energy that is lost to friction is calculated as:
[tex]\frac{E_i-E_f}{E_i}=\frac{57-0.67335}{57}[/tex] = 0.9882 = 98.82%
that means that was lost the 98.82% of the energy.
