解:取坐标如图 v1=0×r1=Ok×(xi+y;/j) (-yi+x方j) La=∑r1×mvn =2(xi+yij)xm; o(-Vi'i+x;'j) =ok 2m;(xi'+yi2)=Jcok=Lc Jco
11 解:取坐标如图 C x y mi r i vri = r i =k(xi i +yi j) = ( -yi i +xi j ) Lcr = r i mivri = (xi i +yi j)mi( -yi i +xi j ) = k mi (xi 2+yi 2 ) = JCk = Lc Lcz = JC
例题12-4.如图所示,双单摆在 Oxy平面内振动在图示瞬时角 速度O1=5rad/s,o2=10rads 如质点A,B的质量m1=m2=5kg 2 OA=AB=1m求在该瞬时质点系 02 B 对O点的动量矩. (Sin1=0.6;Sin62=0.8) 12
12 例题12-4.如图所示,一双单摆在 Oxy 平面内振动.在图示瞬时角 速度1 = 5rad/s , 2 = 10rad/s. 如质点A,B的质量m1 = m2 = 5kg OA=AB =1m.求在该瞬时质点系 对O点的动量矩. (sin1= 0.6 ; sin2= 0.8) x y O A B 1 2 1 2
解(1)应用LO=∑L0=∑r1×mv 0N0 J r1=0.8i+06j r2=1.4i+14j 02 =01Xn=5k×(0.8i+0.6j) B -3i+4 v=v+v21=01×n1+02×(AB) =-3i+4j+10k×(06i+0.8j)=-11i+10j Lo=r1×m11+P2×m2v2 =(0.8+0.6j×5(-3计+4j(1.4i+1.4j×5(-11+10j =172k
13 x y O A B 1 2 1 2 解:(1)应用LO = Loi= ri mivi r1 = 0.8i + 0.6 j r2 = 1.4i + 1.4 j v1 = 1 r1 v2 = v1 + v21 = 5k (0.8i + 0.6 j) = -3i + 4 j r1 r2 = 1 r1 + 2 (AB) = -3i + 4 j +10k (0.6i + 0.8 j ) = -11i + 10 j LO = r1 m1v1 + r2 m2v2 =(0.8i+0.6 j)5(-3i+4 j)+(1.4i +1.4 j)5(-11i +10 j) = 172 k
(2)应用Lo=r×P+L 0N、O1 r=1.1i+ P=m1v1+m2v2=-70i+70j 2 r1=-0.3i-0.4j 62 B 2=0.3i+0.4 Lc=r1×m1v1+r2×m2v2=25k rcXP=(1.1计j×(70i+70j=147k Lo=25k+147k=172k 14
14 (2)应用LO = rc P + Lc rc = 1.1 i + j P = m1v1 + m2v2 = -70i + 70 j x y O A B 1 2 1 2 C rc r 2 = 0.3i + 0.4 j r 1 = - 0.3i - 0.4 j r1 ´ r2 ´ Lc = r 1 m1v1 + r 2 m2v2 = 25k rcP = (1.1i+ j) (-70i + 70 j) = 147 k LO = 25k + 147 k = 172k
例题12-5.如图所示,双复摆 在Oxy平面内振动在图示瞬时, 01 角速度o1=5rad/s,o2=10rad/s 如杆OAAB的质量m=m2=6kg 2 OA=AB=1m求在该瞬时质点 系对O点的动量矩. B (Sin1=0.6;Sin2=0.8) 15
15 例题12-5. 如图所示,一双复摆 在Oxy平面内振动.在图示瞬时, 角速度1= 5rad/s , 2=10rad/s. 如杆OA,AB的质量m1=m2= 6kg OA = AB =1m.求在该瞬时质点 系对O点的动量矩. (sin1= 0.6 ; sin2= 0.8) x y O A B 1 2 1 2