HW8-phy2048-Spring-2014-solution

5 z 20 fx 20 fy 0 and fz 30 then we obtain

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Unformatted text preview: (squared, and using d for the horizontal displacement) gives v2 = gd2/2h. Now, to find the speed vp at point P, we apply energy conservation, that is, mechanical energy on the plateau is equal to the mechanical energy at P. With Eq. 11- 5, we obtain 1 1 1 1 mv2 + Icom ω2 + mgh1 = mvp2 + Icom ωp2 . 2 2 2 2 Using item (f) of Table 10- 2, Eq. 11- 2, and our expression (above) v2 = gd2/2h, we obtain gd2/2h + 10gh1/7 = vp2 which yields (using the values stated in the problem) vp = 1.34 m/s. 11P23. ! We use the notation r ′ to indicate the vector pointing from the axis of rotation ! " directly to the position of the particle. If we write r ′ = x ′ " + y ′ " + z ′ k, then (using Eq. i j ! ! 3- 30) we find r ′ × F is equal to ˆ y...
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