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Unformatted text preview: 90. The speed of the center of mass of the car is v = (40)(1000/3600) = 11 m/s. The angular speed of the wheels is given by Eq. 112: ω = v/R where the wheel radius R is not given (but will be seen to cancel in these calculations). (a) For one wheel of mass M = 32 kg, Eq. 1034 gives (using Table 102(c)) 1 11 = Iω 2 = MR 2 2 22 Krot FG H IJ FG v IJ K H RK 2 = 1 Mv 2 4 which yields Krot = 9.9 × 102 J. The time given in the problem (10 s) is not used in the solution. (b) Adding the above to the wheel’s translational kinetic energy, K wheel =
1 2 Mv 2 , leads to 1 1 3 2 Mv 2 + Mv 2 = ( 32 kg )(11 m/s ) = 3.0 × 103 J. 2 4 4 (c) With Mcar = 1700 kg and the fact that there are four wheels, we have 1 3 M car v 2 + 4 Mv 2 = 12 × 105 J. . 2 4 FG H IJ K ...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Center Of Mass, Mass

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