1 BD 12.03.2013r.

Laboratorium z fizyki

Ćw. nr: 2

Wyznaczanie prędkości lotu pocisku na podstawie badania ruch wahadła balistycznego.

Korga Przemysław

L - 07

3. Obliczenia

1. Obliczamy t_1sr i t_2sr, oraz T_1sr i T_2sr.

$t_{1sr} = \frac{11,341s + 11,253s + 11,258s + 11,253s + 11,287s + 11,276s + 11,257s + 11,262s + 11,261s + 11,282s + 11,257s + 11,268s}{12} = 11,27125\ \ s$

$t_{2sr} = \frac{17,377s + 18,118s + 17,218s + 17,430s + 18,166s + 18,945s + 17,326s + 18,765s + 19,022s + 17,352s + 19,024s + 17,359s}{12} = 18,0085\ s$

$T_{1sr} = \frac{11,27125\ s}{10} = 1,127\ s$

$T_{2sr} = \frac{18,0085s}{10} = 1,8009\ s$

1. Obliczamy moment kierujący D sprężyny oraz moment bezwładności I₀ wahadła.

$D = \frac{8 \times {(3,14)}^{2} \times 0,194\ kg\lbrack\left( 0,02\ m \right)^{2}{- \left( 0,08\ m \right)}^{2}\rbrack}{\lbrack\left( 1,127s \right)^{2}{- \left( 1,801s \right)}^{2}\rbrack} = \frac{8 \times 9,86 \times 0,194\ kg\ \lbrack 0,0004\ m^{2} - 0,0064\ m^{2}\rbrack}{1,270\ s^{2} - 3,244\ s^{2}} = \frac{- 0,092\ kg\ \ m^{2}}{- 1,974\ s^{2}} = 0,0465\ \frac{\text{kg\ }m^{2}}{s^{2}}$

$I_{0} = \frac{2 \times 0,194\ kg\ \lbrack\left( 0,02\ m \right)^{2} \times \left( 1,801\ s \right)^{2} - \left( 0,008\ m \right)^{2} \times \left( 1,127\ s \right)^{2}\rbrack}{\lbrack\left( 1,127s \right)^{2}{- \left( 1,801s \right)}^{2}\rbrack} = \frac{1,88\lbrack(0,0004\ m^{2} \times \ 3,244\ s^{2}) - (0,0064\ m^{2} \times 1,270\ s^{2})\rbrack}{1,270\ s^{2} - 3,244\ s^{2}} = \frac{1,88\ kg\ \lbrack 0,0013\ m^{2}\text{\ s}^{2} - 0,0081\ m^{2}\text{\ s}^{2}\rbrack\ }{- 1,974\ \text{\ s}^{2}} = \ \frac{- 0,0128\ \text{kg\ m}^{2}\text{\ s}^{2}}{- 1,974\ \text{\ s}^{2}} = 0,0065\ \ kgm^{2}$

1. Obliczamy prędkość pocisku v .

$$v = \frac{0,0465\ \frac{\text{kg\ }m^{2}}{s^{2}} \times 0,12\pi\ \times 1,127\ s\ }{2\pi \times 0,006\ kg \times 0,118\ m} = 4,442\ \frac{m}{s}$$

1. Obliczamy niepewności standardowe u(t_1sr) i u(t_2sr) metodą typu A.

Do olbliczenia niepewności standardowej typu A wykorzystam poniższy wzór:

$u(t_{1sr}) = \sqrt{\frac{\left( 11,341 - 11,271 \right)^{2} + ({11,253 - 11,271)}^{2} + \left( 11,258 - 11,271 \right)^{2} + \left( 11,253 - 11,271 \right)^{2} + \left( 11,287 - 11,271 \right)^{2} + \left( 11,276 - 11,271 \right)^{2} +}{12\left( 12 - 1 \right)}}$

$\sqrt{\frac{\left( 11,257 - 11,271 \right)^{2} + \left( 11,262 - 11,271 \right)^{2} + \left( 11,261 - 11,271 \right)^{2} + \left( 11,282 - 11,271 \right)^{2} + \left( 11,268 - 11,271 \right)^{2} + {(11,257 - 11,271)}^{2}}{}}$ =

$\sqrt{\frac{0,004865 + 0,000333 + 0,000176 + 0,000333 + 0,000248 + 0,000023 + 0,000203 + 0,000086 + 0,000105 + 0,000116 + 0,000203 + 0,000011}{132}} =$

$\sqrt{\frac{0,00670}{132}} = \sqrt{0,000051} = 0,00712\ s$

$u\left( t_{2sr} \right) = \sqrt{\frac{\left( 17,377 - 18,0085\ \ \right)^{2} + \left( 18,118 - 18,0085\ \ \right)^{2} + \left( 17,218 - 18,0085\ \ \right)^{2} + \left( 17,430 - 18,0085\ \right)^{2} + \left( 18,166 - 18,0085\ \ \right)^{2} + \left( 18,945 - 18,0085\ \ \right)^{2} +}{12\left( 12 - 1 \right)}}$

$\sqrt{\frac{\left( 17,326 - 18,0085 \right)^{2} + \left( 19,024 - 18,0085\ \right)^{2} + \left( 18,765 - 18,0085\ \right)^{2} + \left( 19,022 - 18,0085\ \right)^{2} + \left( 17,352 - 18,0085\ \right)^{2} + {(17,359 - 18,0085\ )}^{2}}{}} =$

$\sqrt{\frac{0,3988 + 0,0120 + 0,6249 + 0,3347 + 0,0248 + 0,8770 + 0,4658 + 0,5723 + 1,0272 + 0,4310 + 1,0312 + 0,4219}{132} = \ }\sqrt{\frac{6,2215}{132}} = \sqrt{0,0471} = 0,217\ s$

1. Obliczam u(T_1śr) i u(T_2śr).

$u\left( T_{1sr} \right) = \ \sqrt{\frac{{(1,1341 - 1,1271)}^{2} + {(1,1253 - 1,1271)}^{2} + {(1,1258 - 1,1271)}^{2} + {(1,1253 - 1,1271)}^{2} + {(1,1287 - 1,1271)}^{2} + \left( 1,1276 - 1,1271 \right)^{2} +}{12(12 - 1)}}$

$\sqrt{\frac{{(1,1257 - 1,1271)}^{2} + {(1,1262 - 1,1271)}^{2} + {(1,1261 - 1,1271)}^{2} + {(1,1282 - 1,1271)}^{2} + {(1,1257 - 1,1271)}^{2} + {(1,1268 - 1,1271)}^{2}}{}} =$

$\sqrt{\frac{0,000048 + 0,0000033 + 0,00000176 + 0,0000033 + 0,0000025 + 0,00000023 + 0,0000020 + 0,0000009 + 0,0000011 + 0,0000012 + 0,0000020 + 0,0000001}{132}} =$

$\sqrt{\frac{0,000067}{132}} = \sqrt{0,00000051} = 0,000712\ s\ $

$u\left( T_{2sr} \right) = \sqrt{\frac{{(1,7377 - 1,8009)}^{2} + {(1,8118 - 1,8009)}^{2} + {(1,7218 - 1,8009)}^{2} + {(1,7430 - 1,8009)}^{2} + {(1,8166 - 1,8009)}^{2} + {(1,8945 - 1,8009)}^{2} +}{12(12 - 1)}}$

$\sqrt{\frac{{(1,7326 - 1,8009)}^{2} + {(1,8765 - 1,8009)}^{2} + {(1,9022 - 1,8009)}^{2} + {(1,7352 - 1,8009)}^{2} + {(1,9024 - 1,8009)}^{2} + {(1,7359 - 1,8009)}^{2}}{}} =$

$\sqrt{\frac{0,0040 + 0,0001 + 0,0062 + 0,0033 + 0,0002 + 0,0088 + 0,0047 + 0,0057 + 0,0103 + 0,0043 + 0,0103 + 0,0042}{132}} = \sqrt{\frac{0,0622}{132}} =$

$\sqrt{0,0005} = 0,0217\ s$

1. Obliczam niepewności standardowe u(R₁) i u(R₂) metodą typu B, przyjmując ∆R₁ = ∆R₂=1mm.

Korzystam ze wzoru na niepewności standardowe metodą typu B: $u\left( x \right) = \frac{x}{\sqrt{3}}$

$u\left( R_{1} \right) = u\left( R_{2} \right) = \frac{0,0010\ m}{\sqrt{3}} = 0,000577\ m$

1. Obliczam niepewności standardowe u(D) i u(I₀).

$u\left( D \right) = \left| \frac{\partial D}{\partial R_{1}} \right| \times u\left( R_{1} \right) + \left| \frac{\partial D}{\partial R_{2}} \right| \times u(R_{2}) + \left| \frac{\partial D}{\partial T_{1}} \right| \times u(T_{1}) + \left| \frac{\partial D}{\partial T_{2}} \right| \times u(T_{2}) =$

$\left| \frac{16R_{1\ }\pi^{2}M}{T_{1}^{2} - T_{2}^{2}} \right| \times u\left( R_{1} \right) + \left| \frac{- 16R_{2}\pi^{2}M}{T_{1}^{2} - T_{2}^{2}} \right| \times u\left( R_{2} \right) + \left| \frac{- 8\pi^{2}M(R_{1}^{2} - R_{2}^{2})2T_{1}}{{(T_{1}^{2} - T_{2}^{2})}^{2}} \right| \times u\left( T_{1} \right) + \left| \frac{- 8\pi^{2}M(R_{1}^{2} - R_{2}^{2})2T_{2}}{{(T_{1}^{2} - T_{2}^{2})}^{2}} \right| \times u(T_{2}) =$

$0,000179 + 0,000716 + 0,000060 + 0,018449 = 0,019405\ \frac{\text{kg\ }m^{2}}{s^{2}}$

$u\left( I_{0} \right) = \left| \frac{\partial I_{0}}{\partial R_{1}} \right| \times u{(R}_{1}) + \left| \frac{\partial I_{0}}{\partial R_{2}} \right| \times u{(R}_{2}) + 2 \times \left| \frac{\partial I_{0}}{\partial T_{1}} \right| \times u{(T}_{1}) + 2 \times \left| \frac{\partial I_{0}}{\partial T_{2}} \right| \times u{(T}_{2}) =$

$\left| \frac{4MR_{1}(T_{1}^{2} - T_{2}^{2})}{T_{1}^{2} - T_{2}^{2}} \right| \times u{(R}_{1}) + \left| \frac{- 4MR_{2}(T_{1}^{2} - T_{2}^{2})}{T_{1}^{2} - T_{2}^{2}} \right| \times u{(R}_{2}) + 2 \times \left| \frac{- 8MT_{1}^{}(R_{1}^{2} - R_{2}^{2})}{{{(T}_{1}^{2} - T_{2}^{2})}^{2}} \right| \times u\left( T_{1} \right) + 2 \times \left| \frac{- 8MT_{2}^{}(R_{1}^{2} - R_{2}^{2})}{{{(T}_{1}^{2} - T_{2}^{2})}^{2}} \right| =$

0, 0000864 + 0, 0000358 + 0, 0053944 + 0, 0019759 = 0, 0074926 kg m²

1. Obliczam niepewności standardowe u(α_MAX), u(r), u(m).

r=0,0025 m u(r)=$\frac{0,0025}{\sqrt{3}} = 0,00144\ \ m$

m= 0,0001 kg $u\left( m \right) = \frac{0,0001}{\sqrt{3}} = 0,0000577\ kg\ $

α_MAX = 0, 0055 π rad  ${u(\alpha}_{\text{MAX}}) = \frac{0,0055}{\sqrt{3}} = 0,00317\ \pi\ rad\ $

1. Obliczam niepewność standardową prędkości pocisku u(v).

$u\left( v \right) = \left| \frac{\partial v}{\partial D} \right| \times u\left( D \right) + \left| \frac{\partial v}{\partial\alpha} \right| \times u\left( \alpha \right) + \left| \frac{\partial v}{\partial T} \right| \times u\left( T \right) + \left| \frac{\partial v}{\partial m} \right| \times u\left( m \right) + \left| \frac{\partial v}{\partial r} \right| \times u\left( r \right) =$

$\left| \frac{- \alpha T}{2\pi mr} \right| \times u\left( D \right) + \left| \frac{\text{DT}}{2\pi mr} \right| \times u\left( \alpha \right) + \left| \frac{\text{Dα}}{2\pi mr} \right| \times u\left( T \right) + \left| \frac{- DT\alpha}{2\pi m^{2}r} \right| \times u\left( m \right) + \left| \frac{- DT\alpha}{2\pi r^{2}m} \right| \times u\left( r \right) =$

= 0,590292 + 0,037366 + 0,002806 + 0,042713 + 0,054202 = 0,727379 $\frac{m}{s}$

$$v \pm u\left( v \right) = (4,442\ \pm \ 0,727379)\ \frac{m}{s}$$

4. Wnioski

Celem ćwiczenia było wyznaczenie prędkości lotu pocisku na podstawie badania ruchu wahadła balistycznego. Średnio wyniosła ona( $4,442\ \pm \ 0,727379)\ \frac{m}{s}$. Na błędy pomiarowe złożyła się niedokładność ustawienia ciężarków, zważenia masy pocisku, a także odczytania kąta wychylenia wahadła i miejsca uderzenia w nie pocisku.

Moment kierujący wahadła wyniósł $0,0465\ \frac{\text{kg\ }m^{2}}{s^{2}}$ , zaś moment bezwładności 0, 0065 kg m²  .