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GoldShadow
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Homework Statement
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 0.800MR^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 3500 J. Calculate h.
Mass "M" = w/g = 392/9.8, or 40 kg.
Homework Equations
[tex]I=cMR^{2}[/tex]
[tex]W_{total}=\Delta K - W_{friction}= \frac{1}{2}I \omega^{2}_{final}-\frac{1}{2}I \omega^{2}_{initial}-W_{friction}[/tex]
[tex]W_{total}=-\Delta U=Mgh_{initial}-Mgh_{final}[/tex]
The Attempt at a Solution
I set the two equations for Work equal to get:
[tex]Mgh_{initial}-Mgh_{final}=\frac{1}{2}I \omega^{2}_{final}-\frac{1}{2}I \omega^{2}_{initial}-W_{friction}[/tex]
Since initial height is 0 and final angular velocity is 0, this simplifies to:
[tex]-Mgh_{final}=-\frac{1}{2}I \omega^{2}_{initial}-W_{friction}[/tex]
Multiplying this equation by -1 to make the negatives a little more friendly and writing out "I" (moment of inertia) yields:[tex]Mgh_{final}=\frac{1}{2}cMR^{2} \omega^{2}_{initial}+W_{friction}[/tex]
Isolating to solve for "h":
[tex]\frac{(\frac{1}{2}cMR^{2} \omega^{2}_{initial}+W_{friction})}{Mg}=h_{final}[/tex]
Plugging in numbers:
[tex]\frac{(\frac{1}{2}(0.800)(40)(0.600)^{2} (25.0)^{2}+3500)}{(40)(9.80)}=h_{final}[/tex]
For h I get 18.1 m, but the answer according to the book is 11.7 m.
I tried several approaches and couldn't get the right answer, so I would appreciate help.