A wheel with a weight of 396N comes off a moving truck and rolls without slippin

Question

A wheel with a weight of 396N 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 23.9rad/s . The radius of the wheel is 0.572m 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, at a height of h above the bottom of the hill; this work has a magnitude of 3484J . Calculate h. Use 9.81 m/s^2 for the acceleration due to gravity. ___m A wheel with a weight of 396N 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 23.9rad/s . The radius of the wheel is 0.572m 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, at a height of h above the bottom of the hill; this work has a magnitude of 3484J . Calculate h. Use 9.81 m/s^2 for the acceleration due to gravity. ___m A wheel with a weight of 396N 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 23.9rad/s . The radius of the wheel is 0.572m 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, at a height of h above the bottom of the hill; this work has a magnitude of 3484J . Calculate h. Use 9.81 m/s^2 for the acceleration due to gravity. ___m Calculate h. Use 9.81 m/s^2 for the acceleration due to gravity. ___m

Explanation / Answer

using change in eneggy= net work done between bottom and top of the mountain

Initial velocity= omega*r=23.9*.572 m/s=13.6708 m/s

intial kinetic energy= roational + translational= (1+.8)* (1/2)*m*v^2=6789.793208 Joule

thus 6789.793208- 3484 (friction work)= weight*h

thus h=8.3479 meters

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