解：假设A点的速度为vA，B点的速度为vB，AB杆的角速度为ωAB。根据基点法，可知B点的速度和A点的速度为：
vB vA ωAB rAB (a)
将式(a)向vA方向投影，并考虑到 90 ，得：
AB 2rsin30 vA
AB
将式(a)向vB方向投影，得：
r
2rsin30
vB AB 2rcos30
r
OB
1
vB
r
2-11. 半径是R的轮子沿平面滚动而不滑动。轮心O作匀速vo运动。长l 3R的杆AB在点A与轮子铰接。杆的另一端B沿平面滑动，求杆AB在图示位置时的角速度大小、角加速度大小，以及点B的速度、加速度大小。
解：建立如图所示的坐标系。
y
以O点为基点，分析轮子上A点的速度和加速度，得：
vA v0i ω rOA v0i
v013k i Rj) vi0 j0 （1） R22
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