Answer to Question #288048 in Classical Mechanics for rosa

Question

A thin spherical shell of mass M = 3.00 [kg] and radius R = 1.50 [m] compresses a spring (k = 500. [N/m]) by 0.400 [m]. After releasing the shell from rest, it rolls without slipping along the floor and up a surface inclined at an angle θ = 30.0◦, as shown in the figure. What is the angular speed of the shell when it reaches a distance L = 2.00 [m] from the base of the incline? (I = 2/3mr^2)

Answer 1

$\frac{I\omega^2}2=mgl\sin\theta+\frac{kl^2}2,$

$\omega=\sqrt{\frac{2mgl\sin\theta+kl^2}I},$

$\omega=1.4~\frac{rad}s.$