Question

A wheel with a weight of 393 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 22.3 rad/s . The radius of the wheel is 0.628 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of habove the bottom of the hill; this work has a magnitude of 3522 J . Calculate h.

Answer 1

Mass of wheel = 393/g = 393/9.8 = 40.1 kg. Velocity at bottom = ωr = 22.3*0.628 = 14 m/s

Total KE at the bottom of the hill = translational KE of CM plus rotational KE about the CM.

= (1/2)mv^2 + (1/2) I ω^2 = (1/2)mv^2 + (1/2)(.8)mv^2 = 0.9mv^2 = 7078.09 Joules.

Subtracting the magnitude of the work done by 'friction' you are left with 7078.09 - 3522 =3556.09 joules that will be converted into gravitational potential energy:

3556.09 = mgh = 391 h

=> 3556.09 = (40.1)(9.8)h

=> h = 9.1m