Question

A wheel with a weight of 396 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 24.2 rad/s . The radius of the wheel is 0.597 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 h above the bottom of the hill; this work has a magnitude of 3470 J .

a) Use 9.81 m/s2 for the acceleration due to gravity.

Answer 1

Mass of wheel = 396/g = 396/9.8 = 40.40 kg.

Velocity at bottom = ωr =24.2*0.597= 14.44 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)(0.8)mv^2

= 0.9mv^2

=0.9*40.40*14.44²

= 7581.55 Joules.

Subtracting the magnitude of the work done by 'friction' you are left with 7581.55 - 3470 =4111.55 joules that will be converted into gravitational potential energy:

4111.55 = mgh = 396 h

or h = 10.38 meters in height.