A body of radius 1.06 m and mass 2.90 kg is rolling smoothly with speed 2.69 m/s on a horizontal surface. It then rolls up a hill to a maximum height 8.48 m. What is the body's rotational inertia about the rotational axis through its center of mass?
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Total energy = m . g .h = 2.9 . 9.8 . 8.48 = 241 J
This comes from the KE (trans ) plus the KE (rot)
KE trans = 1/2 . 2.9 . 2.69^2 = 10.5 J
Ke (rot ) = 241 – 10.5 = 230.5 J
KE rot = 1/2 . I . w^2
w = v/r = 2.54 rad/s
I = 2 . 230.5 / 2.54^2 = 71.6 kgm^2
This comes from the KE (trans ) plus the KE (rot)
KE trans = 1/2 . 2.9 . 2.69^2 = 10.5 J
Ke (rot ) = 241 – 10.5 = 230.5 J
KE rot = 1/2 . I . w^2
w = v/r = 2.54 rad/s
I = 2 . 230.5 / 2.54^2 = 71.6 kgm^2