复旦大学：《基础物理实验》课程教学资源_刚体转动惯量的不确定度计算

转动惯量不确定度计算公式 弹性系数K [1 Determination of K 2 We get T=4 K K Thus K=4 Uncertainty of k u(K) u(lpo) T-t K T2 1(x2-)-0)+7)&(r2)=27×u(T) Since we measure the time of 10 periods, so for each period, we have 1=.4 and ()=√ln2(1)+n2(t1) (T)=D4().Whe()=,(-4)2 n(n-1) 0.01 lB2(1 √33 2. Uncertainty of different objects l1 Plastic cylinder: u(D) 2× u( A(D)+B2(D) u(m)=√lx2(m)+an2(m) Whe n(m)=d=0.1g u(D) n(n-1) a0.2 g a0.002 √33 12 Metal b &l(l) u(K)「u(x2-n2 K

II-1 转动惯量不确定度计算公式: 1. 弹性系数 K [1] Determination of K. From 0 0 2 I T K   & 0 1 2 PC I I T K    We get 2 2 0 0 4 I T K   & 2 2 0 1 4 PC I I T K    Thus 2 2 2 1 0 4 PC I K T T    [2] Uncertainty of K. 2 2 2 2 1 0 2 2 1 0 ( ) ( ) ( ) PC PC u I u T T u K K I T T                  . Please note: 2 2 2 2 2 2 1 0 1 0 u T T u T u T ( ) [ ( )] [ ( )]    & 2 1 1 1 u T T u T ( ) 2 ( )   . Since we measure the time of 10 periods, so for each period, we have 1 1 1 10 T t  and 1 1 1 ( ) ( ) 10 u T u t  . Where, 2 2 1 1 2 1 2 1 1 1 2 1 ( ) ( ) ( ) ( ) ( ) ( 1) 0.01 ( ) 3 3 A B i A B u t u t u t t t u t n n a u t s         . 2. Uncertainty of different objects: [1] Plastic cylinder: 1 1 2 2 2 8 PC I mr mD   & 2 2 ( ) (D) ( ) 2 PC PC u m u u I I m D                 Where, 1 2 2 2 1 2 ( ) ( ) ( ) ( ) 0.1 0.2 ( ) 3 3 B B B B u m u m u m u m d g a u m g       & 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( 1) 0.002 ( ) 3 3 A B i A B u D u D u D D D u D n n a u D cm         . [2] Metal barrel: 2 2 1 0 2 ( ) 4 MB K T T I    & 2 2 2 2 1 0 2 2 1 0 ( ) ( ) ( ) MB MB u K u T T u I I K T T                

n()=,2m24mD2+D2 ,=m(D2+D2)& n(m)7(2D(D)+(2DD) 3 Iron Ball m=x(2-2 &a(m)=,/4(k) (72-72 K u(m) u(D) 14] Thin Bar K(72-7) l(7B) k) u(t K 1=mL&u(l1) (m) 2 u(L)I L -2

II-2 2 2 1 2 1 ( ) 8 t I m D D   &       2 2 2 2 1 2 2 2 1 2 2 2 2 1 1 1 1 2 2 2 1 2 ( ) ( ) ( ) ( ) 2 ( ) 2 ( ) t t t u m u D D u I I m D D u m D u D D u D I m D D                           [3] Iron Ball 2 2 1 0 2 ( ) 4 IB K T T I    & 2 2 2 2 1 0 2 2 1 0 ( ) ( ) ( ) IB IB u K u T T u I I K T T                 . 2 1 2 2 5 10 t I mr mD   & 2 2 ( ) (D) ( ) 2 t t u m u u I I m D                 . [4] Thin Bar 2 2 1 0 2 ( ) 4 TB K T T I    & 2 2 2 2 1 0 2 2 1 0 ( ) ( ) ( ) TB TB u K u T T u I I K T T                 . 1 2 12 t I mL  & 2 2 ( ) ( ) ( ) 2 t t u m u L u I I m L                