ch10-p075 - 75. The Hint given in the problem would make...

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() ( ) 2 2 22 0 5 110kg 33.5 rad s 7.80m 4.81 10 N. L MM L Fd F r d m r d r L ω ωω == = = = ³³ ³ (b) About its center of mass, the blade has I ML = 2 12 / according to Table 10-2(e), and using the parallel-axis theorem to “move” the axis of rotation to its end-point, we find the rotational inertia becomes I ML = 2 / 3. Using Eq. 10-45, the torque (assumed constant) is ( ) 2 2 4 1 1 33.5rad/s 110kg 7.8 m 1.12 10 N m. 33 6 . 7 s IM L t τα §· Δ § · = = × ¨¸ ¨ ¸ Δ ©¹ © ¹ (c) Using Eq. 10-52, the work done is ( ) 2 6 11 1 1 0 110kg 7.80m 33.5rad/s 1.25 10 J. 3 6 WK I M L =Δ = − = = = × 75. The Hint given in the problem would make the computation in part (a) very straightforward (without doing the integration as we show here), but we present this
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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.

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