Question

A 377-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 0.800MR^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 2700 J .

(a) Calculate h.

Answer 1

Step-by-step explanation:

We shall apply law of conservation of mechanical energy to solve the problem

Initially wheel has rotational kinetic energy + linear kinetic energy

= 1 / 2 I ω² + 1/2 m v² , I is moment of inertia , m is mass , ω is angular velocity and v is linear velocity

= 1 / 2 I ω² + 1/2 x 2 Ix v² / r² ( I = 1/2 m r² )

= 1 / 2 I ω² + Ix ω² ( v / r = ω )

= 3/2 I ω²

Given I = .8MR²

= .8 x 377/9.8 x .6²

= 11.07 ,

ω = 25 rad /s

Total kinetic energy = 3/2 x 11.07 x 25²

= 10378 J

energy lost due to work done by friction

= 10378 - 2700 = 7678 J

This energy will help it to scale height

mgh = 7678 , h is ultimate height achieved

h = 7678 / mg

= 7678 / 377

= 20.36 m .