Answer: v = 1.41739 [m/s]
Explanation:
Let:
M = shell mass = 4.5 [Kg]
R = shell radius = 0.085 [m]
[tex]I_{p}[/tex] = pulley's rotational inertia = [tex]3*10^{-3}[/tex] [Kg*m^2]
r = radius of the pulley = 0.05 [m]
m = object's mass = 0.6 [Kg]
h = fall distance = 0.82 [m]
v = velocity of the block in [m]
as there is no losses due to friction we are able to start, so as soon as we let the system interact the block will fall a height h. such energy must be equal to the total kinetic energy of the system, let's draw our equation's energy:
potential energy of the block : [tex]mgh[/tex]
kinetical energy of the block : [tex]\frac{1}{2}*m*v^2[/tex]
kinetical energy of the pulley (only rotational) : [tex]\frac{1}{2}*I_{p}*\omega^2[/tex]
now who is the angular velocity omega?, well the only thing that is making the pulley rotate is the block, and in the external part of the pulley, each point is moving with an angular velocity produced by v:
we have the relationships between these two, with the following equation:
[tex]v = \omega * r =>\omega = v/r[/tex]
kinetical energy of the shell (again only rotational ) : [tex]\frac{1}{2}*I*\omega^2[/tex]
the moment of inertia of a spherical shell is [tex]I = \frac{2}{3}*MR^2[/tex]
again, omega is obtained the same way, with the exception of the new radius which is R (instead of r):
[tex]v = \omega * R =>\omega = v/R[/tex]
Finally we replace and solve v:
we have so far:
[tex]mgh = \frac{1}{2} mv^2 + \frac{1}{2} I_{p}*\frac{v^2}{r^2} + \frac{1*2}{2*3}*MR^2*\frac{v^2}{R^2}\\mgh = \frac{1}{2} mv^2 +\frac{1}{2} I_{p}*\frac{v^2}{r^2} +\frac{1}{3}Mv^2\\mgh =(\frac{1}{2} m+ \frac{1}{2*r^2} I_{p}+\frac{1}{3} M)*v^2[/tex]
finally we solve v and replace values:
[tex]v^2 =\frac{mgh}{\frac{1}{2} m+ \frac{1}{2*r^2} I_{p} +\frac{1}{3} M}\\ v=\sqrt{\frac{mgh}{\frac{1}{2} m+ \frac{1}{2*r^2} I_{p} +\frac{M}{3}}}[/tex]
[tex]v= \sqrt{\frac{0.6*9.8*0.82}{\frac{1}{2} 0.6+ \frac{1}{2*(0.05)^2}*3*10^{-3} +\frac{4.5}{3}}}[/tex]
finally v = 1.41739 [m/s]
