Answer:
[tex]I = 1.6MR^2[/tex]
Explanation:
When the pumpkin rolling down an incline of vertical height H, its potential energy is converted to rotational energy and kinetic energy
[tex]MgH = \frac{Mv^2}{2} + \frac{I\omega^2}{2}[/tex]
where[tex]\omega = \frac{v}{R}[/tex] is the angular speed
[tex]2MgH - Mv^2 = \frac{Iv^2}{R^2}[/tex]
[tex]2MgHR^2 - Mv^2R^2 = Iv^2[/tex]
[tex]I = 2MgH\frac{R^2}{v^2} - MR^2[/tex]
we can substitute [tex]v = \sqrt{\frac{5gH}{4}}[/tex]
[tex]v^2 = \frac{5gH}{4}[/tex]
[tex]I = 8MgH\frac{R^2}{5gH} = 1.6MR^2[/tex]
