Question

A wheel with a radius of 1.95 m and a mass of 0.215 kg rolls without sliding down a plane that is inclined at an angle φ = 12.062 °. g = 9.806 m/s². Calculate the kinetic energy of the wheel after the time 0.923s if it has a moment of inertia 1.6 kg · m², starts from rest and a force acts on the wheel so that its angle of rotation varies with time according to θ(t) = 1.6 · t³ rad.

Answer 1

$\omega=\theta'=4.8 t^2,$

$E=\frac{I\omega^2}{2}+\frac{mv^2}{2}=\frac{I\omega^2}{2}+\frac{m\omega^2 r^2}{2}=\frac{\omega^2}{2}(I+mr^2)=\frac{4.8^2\cdot t^2}{2}(I+mr^2),$

$E=\frac{4.8^2\cdot 0.923^2}{2}(1.6+0.215\cdot 1.95^2)=23.7~J.$