A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800?MR?2. Friction does work on the wheel as it rolls up the hill to a stop, a height ?h? above the bottom of the hill; this work has absolute value 3500 J. Calculate ?h?.
Solution to 25E Step 1 Weight of the wheel =392N Mass of the wheel =40 kg (mg=392, m=392/9.8) Velocity at the bottom of the hill =25 rad/s Radius of the wheel =0.6m 2 2 2 Moment of inertia of the wheel =0.8MR =0.8x40(0.6) =11.52kgm Work done by the friction =3500J Step 2 2 Kinetic energy of rotation =K =(½)Ir 2 Kr(½)x11.52x(25) =3600J Translational kinetic energy =K l Kl½) x 40 x(0.6x25)=4500J Total kinetic energy =K +K=K r l K=4500+3600=8100 J
