四、计算(32小题,共395.0分) 1．(5分)[答案]
v (1) tg?????e22OA?e?1
(2)??45
2．(10分)[答案] (1) s=K? 当 s=10,??(2) 即s??30时，K= 3?30? ??当??45?时，s=7.5 mm
4R=r0+s=32.5 mm L=v/??30 ?3．(10分)[答案] (1)?C?0
O1O)?26.57 OA(2) h=60 mm
?D?arctg(4．(10分)[答案]
(1) 从动件推程运动规律为等速运动规律，当???时，s=15 mm 2ds9.55d??(2) tg?? r0?s30?15
??11.98?
5．(10分)[答案]
dshdShhds (1)?????v
t0dTt0dtdtt0td(0v)hdShd2sh(2)2??2??2?a 2dtt0dTt0dtt026．(10分)[答案] (1) 位移方程为
??4h?(2) 速度方程为 v?(???)
??24h?2(3) 加速度方程为 a??
??22s?h?2 h(???)2
7．(10分)[答案]
dshdShht0vds (1)? ????v???dT?dt??d?t0tdS)d(0v)2thdShh?2?dT??h?02a (2) 222dt??dT??dT????t02d(d2s ?2? 2?d?a8．(10分)[答案] (1)
s?h??
vh? ??h?(2) tg???0.285 h??()?r0?49? ??15.9．(10分)[答案] (1)
vdsds? , ?? 线图如图示。 ?d?d??(2) tg?(dsd?)maxdsd? ， tg?max? r0r0?2h? r0? tg?max48 ?max?32.
10．(10分)[答案] (1)
s?2h? ?
ds2h? d??(2)
r0min?dsd??22.06mm
tg?max11．(10分)[答案]
dsh??20 mm ，h?20??20?mm d??dsh20?回程时，??????40mm
?d???2 (1)
(2) r0min?(dsd??e?s)2?e2
tg?max20?e?s)2?e2 tg30
?(12．(10分)[答案] (1) 推程AB段
h2?sin(?? )?2??5?2式中??????
663??????
6 sAB?h???
?为从坐标原点量起的凸轮转过的角度，则
sAB?3h?h?(??)?sin[3(??)] 2p62?6h(2) 回程CD段
h2?sin(??)]
??2???11?7?2??? 式中???6637??????
63h7?h7?(??)?sin[3(??)] 则sCD?h?[2?62?6sCD?h?[???13．(10分)[答案] (1) ??xB??cos?????yB???sin?sin???xB0??ssin?? ???? ??cos???yB0??scos??r02?e2
式中 xB0?e,yB0?整理后，xB?(r0?e?s)sin??ecos?
22yB?r02?e2?s)cos??esin?
(2)
??90,s?20mm
xB?54.64mm
yB??20mm
14．(10分)[答案]
(1) 回程运动角????
vmax?2h??0.4 m/s ?2(2) 加速段a??2vmax?/???8 ms?
??(3) 减速段a?2vmax?/???8 ms2???
15．(10分)[答案] (1)s,ds,?如图示。 d?(2) tg??dsd??er?e?s202
代入
dsd??0.00?21?20.m0 24e?0.002?6?0.012 m
s?0.002?10?0.02 m r002?12?0.024 m 0?0.??17?