Dynamics_Part67

Dynamics_Part67 - Thus, in standard vector form, we have HG...

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Hence, the angular momentum vector is H G =[ I ] ω = 1 2 mr 2 00 0 1 4 mr 2 0 00 1 4 mr 2 ω cos β ω sin β 0 = 1 2 mr 2 ω cos β 1 4 mr 2 ω sin β 0 Thus, in standard vector form, we have H G = 1 2 mr 2 ω cos β i I 1 4 mr 2 ω sin β j I In terms of the xyz coordinate system, this becomes H G = 1 2 mr 2 ω cos β (cos β i +sin β j ) 1 4 mr 2 ω sin β ( sin β i +cos β j ) = 1 4 mr 2 ω D 2cos 2 β +sin 2 β i i + 1 4 mr 2 ω sin β cos β j We can simplify this expression using standard half-angle trigonometric identities, viz., cos 2 β = 1 2 (1 + cos 2 β ) , sin 2 β = 1 2 (1 cos 2 β ) , sin β cos β = 1 2 sin 2 β Substituting into the equation above and combining like terms yields H G = 1 8 mr 2 ω (3 + cos 2 β ) i + 1 8 mr 2 ω sin 2 β j (b) The angle between H G and the axle is given by tan θ = ( H G ) y ( H G ) x = 1 8 mr 2 ω sin 2 β 1 8 mr 2 ω (3 + cos 2 β ) Therefore, we conclude that θ =tan 1 w sin 2 β 3+cos2 β W
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