To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.
Let's start considering that the angular velocity is given by,
[tex]\omega = \frac{v}{R}[/tex]
Where,
V = linear speed
R = the radius
In the case of the initial kinetic energy:
[tex]KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2[/tex]
Where I is the moment of inertia previously defined.
[tex]KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2[/tex]
In the case of the final kinetic energy, we have to,
[tex]KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2[/tex]
[tex]KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2[/tex]
For conservation of Energy we have, that
[tex]KE_f = KE_i[/tex], then (canceling the mass and the radius)
[tex]\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2[/tex]
[tex]8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2[/tex]
[tex]1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)[/tex]
[tex]2.2388= (\frac{7}{5}) (v)^2[/tex]
[tex]v=1.26m/s[/tex]
