Answer:
[tex]v=4.07\frac{m}{s^2}[/tex]
Explanation:
According to the law of conservation of energy, when an object falls, its gravitational potential energy is changed to kinetic energy. In this case we have linear as well as rotational motion:
[tex]mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}[/tex]
A solid disk has a moment of inertia equal to [tex]\frac{mR^2}{2}[/tex]. The height of the disk is given by [tex]h=dsin\theta[/tex]. Recall that [tex]\omega=\frac{v}{R}[/tex]. Replacing and solving for v:
[tex]mgdsin(25^\circ)=\frac{mv^2}{2}+\frac{1}{2}\frac{mR^2}{2}\frac{v^2}{R^2}\\mgdsin(25^\circ)=\frac{3mv^2}{4}\\v^2=\frac{4gdsin(25^\circ)}{3}\\v=\sqrt{\frac{4(9.8\frac{m}{s^2})(3m)sin(25^\circ)}{3}}\\v=4.07\frac{m}{s^2}[/tex]
