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Two solid bodies rotate about stationary mutually perpendicular intersecting axes with constant angular velocities omega_1=3.0rad//s and omega_2=4.0rad//s. Find the angular velocity and angular acceleration of one body relative to the other. |
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Answer» Solution :The angular velocity is a vector as infinitesimal rotation commute. Then the relative angular velocity of the body 1 with respect to the body 2 is clearly. `vecomega_(12)=vecomega_(1)-vecomega_2` as for relative linear velocity. The relative acceleration of 1 w.r.t. 2 is `((dvecomega_1)/(dt))_(s')` where S' is a frame corotating with the second body and S is a space fixed frame with ORIGIN COINCIDING with the point of INTERSECTION of the two axes, but `((dvecomega_1)/(dt))_s=((dvecomega_1)/(dt))_s+vecomega_2xxvecomega_1` SINCE S' rotates with angular with angular velocity `vecw_2`. However `((dvecomega_1)/(dt))_S=0` as the first body rotates with constant angular velocity in space, thus `vecbeta_(12)=vecomega_1xxvecomega_2`. Note that for any vector `vecb`, the relation in space forced frame `(k)` and a frame `(k^')` rotating with angular velocity `vecomega` is `(dvecb)/(dt):|:_K=(dvecb)/(dt):|_K+vecomegaxxvecb` |
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