刚体转动惯量的不确定度计算.pdf

转动惯量不确定度计算公式: 1. 弹性系数 K [1] Determination of K. From T0  2 I0 Thus K  4 I PC T12  T02 & T1  2 I PC  I 0 K K  I0 2 2 I 2 2 I We get T0  4 0 & T1  4 PC K K 2 [2] Uncertainty of K. 2 2  u ( I PC )   u (T12  T02 )  u( K )   K .   2 2   I PC   T1  T0  Please note: u(T12  T02 )  [u(T12 )]2  [u(T02 )]2 & u(T12 )  2T1  u(T1 ) . Since we measure the time of 10 periods, so for each period, we have T1  u (t1 )  u A2 (t1 )  u B2 2 (t1 ) 1 u (T1 )  u (t1 ) . Where, u A (t1 )  10 uB 2 (t1 )  2.  (t  t ) 1i 2 1 . n(n  1) a 0.01  s 3 3 Uncertainty of different objects: [1] Plastic cylinder: 2 2 1 1  u (m)   u (D)   2  I PC I PC  mr 2  mD 2 & u ( I PC )   D  2 8  m   u ( D)  u A2 ( D)  u B2 2 ( D) u ( m)  u B 1 ( m)  u B 2 ( m) 2 2 Where, u B1 (m)  d  0.1g u B 2 ( m)   ( D  D) u ( D)  & A a 0.2  g 3 3 uB 2 ( D)  n(n  1) a 0.002  cm 3 3 [2] Metal barrel: 2 2 2 K (T12  T02 )  u ( K )   u (T1  T0 )  I MB   2 I MB . & u ( I MB )   2  4 2  K   T1  T0  2 II-1 2 i . 1 t1 and 10 2 2 2  u (m)   u ( D1  D2 )  u( It )    It  2 2   m   D1  D2  2 1 I t  m( D12  D22 ) & 8  u (m)   2 D1u ( D1 )    2 D1u ( D1 )     It 2 2 2  m   D1  D2  2 2 2 [3] Iron Ball 2 2 2 K (T12  T02 )  u ( K )   u (T1  T0 )  & u ( I IB )   I IB   2 I IB . 2  4 2  K   T1  T0  2 2 2 2 1  u (m)   u (D)   2  It . I t  mr 2  mD 2 & u ( I t )   D  5 10  m   [4] Thin Bar 2 2 2 K (T12  T02 )  u ( K )   u (T1  T0 )  & u ( ITB )   ITB   2 I . 2  TB 4 2  K   T1  T0  2 2 2 1  u (m)   u ( L)   2  It . I t  mL2 & u ( I t )   L  12  m   II-2