B\u00e0i 8: Th\u1ef1c h\u00e0nh: Kh\u1ea3o s\u00e1t chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do. X\u00e1c \u0111\u1ecbnh gia t\u1ed1c r\u01a1i t\u1ef1 do<\/h2>\n\n\n\n

B\u00e1o c\u00e1o th\u1ef1c h\u00e0nh<\/strong><\/p>\n\n\n\n

1. Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi:<\/strong> S\u1ef1 r\u01a1i t\u1ef1 do l\u00e0 g\u00ec ? N\u1ebfu \u0111\u1eb7c \u0111i\u1ec3m c\u1ee7a chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do v\u00e0 vi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh gia t\u1ed1c r\u01a1i t\u1ef1 do ?<\/ins><\/p>\n\n\n\n

– S\u1ef1 r\u01a1i t\u1ef1 do l\u00e0 s\u1ef1 r\u01a1i ch\u1ec9 d\u01b0\u1edbi t\u00e1c d\u1ee5ng c\u1ee7a tr\u1ecdng l\u1ef1c.<\/p>\n\n\n\n

– \u0110\u1eb7c \u0111i\u1ec3m:<\/p>\n\n\n\n

   + Ph\u01b0\u01a1ng th\u1eb3ng \u0111\u1ee9ng, chi\u1ec1u t\u1eeb tr\u00ean xu\u1ed1ng.<\/p>\n\n\n\n

   + L\u00e0 chuy\u1ec3n \u0111\u1ed9ng nhanh d\u1ea7n \u0111\u1ec1u.<\/p>\n\n\n\n

   + T\u1ea1i m\u1ed9t n\u01a1i nh\u1ea5t \u0111\u1ecbnh tr\u00ean Tr\u00e1i \u0110\u1ea5t v\u00e0 \u1edf g\u1ea7n m\u1eb7t \u0111\u1ea5t, m\u1ecdi v\u1eadt \u0111\u1ec1u r\u01a1i t\u1ef1 do v\u1edbi c\u00f9ng gia t\u1ed1c g.<\/p>\n\n\n\n

– C\u00f4ng th\u1ee9c t\u00ednh gia t\u1ed1c r\u01a1i t\u1ef1 do:<\/p>\n\n\n\n

<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

Trong \u0111\u00f3:    s : qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a v\u1eadt r\u01a1i t\u1ef1 do (m).<\/p>\n\n\n\n

               t : th\u1eddi gian v\u1eadt r\u01a1i t\u1ef1 do (s).<\/p>\n\n\n\n

2. K\u1ebft qu\u1ea3:<\/strong><\/p>\n\n\n\n

B\u1ea3ng 8.1 Kh\u1ea3o s\u00e1t chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do : \u0110o th\u1eddi gian r\u01a1i \u1ee9ng v\u1edbi c\u00e1c kho\u1ea3ng c\u00e1ch s kh\u00e1c nhau.\n<\/p>\n\n\n\n

V\u1ecb tr\u00ed \u0111\u1ea7u c\u1ee7a v\u1eadt r\u01a1i: s_{0<\/sub> = 0 (mm).\n<\/p>\n\n\n\n}

V\u1ebd \u0111\u1ed3 th\u1ecb: D\u1ef1a v\u00e0o k\u1ebft qu\u1ea3 trong B\u1ea3ng 8.1, ch\u1ecdn t\u1ec9 l\u1ec7 th\u00edch h\u1ee3p tr\u00ean c\u00e1c tr\u1ee5c tung v\u00e0 tr\u1ee5c ho\u00e0nh \u0111\u1ec3 v\u1ebd \u0111\u1ed3 th\u1ecb s = s(t2).\n<\/p>\n\n\n\n

a)\tTa c\u00f3: s = (gt^{2<\/sup>)\/2 = s(t). Nh\u01b0 v\u1eady s ph\u1ee5 thu\u1ed9c v\u00e0o th\u1eddi\n gian l\u00e0 h\u00e0m b\u1eadc 2 \u1ea9n t, do v\u1eady n\u1ebfu v\u1ebd \u0111\u1ed3 th\u1ecb bi\u1ec3u di\u1ec5n s qua t th\u00ec n\u00f3 \nc\u00f3 d\u1ea1ng m\u1ed9t \u0111\u01b0\u1eddng cong Parabol.\n<\/p>\n\n\n\n}

Nh\u01b0ng b\u00e0i to\u00e1n h\u1ecfi d\u1ea1ng \u0111\u1ed3 th\u1ecb c\u1ee7a s theo \u1ea9n (t^{2<\/sup>), do v\u1eady ch\u00fang ta pahir c\u1ea9n th\u1eadn.\n<\/ins><\/p>\n\n\n\n}

T\u1eeb s = (gt^{2<\/sup>)\/2 \u2192 s = (g.X)\/2 v\u1edbi X = t2<\/sup>, \u1edf \u0111\u00e2y t l\u00e0 bi\u1ebfn n\u00ean X c\u0169ng l\u00e0 bi\u1ebfn.\n<\/p>\n\n\n\n}

Ta nh\u1eadn th\u1ea5y s\u1ef1 ph\u1ee5 thu\u1ed9c c\u1ee7a s theo \u1ea9n X l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t:\n<\/p>\n\n\n\n

Y = A.X + B (v\u1edbi A = g\/2, B = 0) n\u00ean \u0111\u1ed3 th\u1ecb s = s(t^{2<\/sup>) = s(X) c\u00f3 d\u1ea1ng l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng. Nh\u01b0 v\u1eady chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt r\u01a1i t\u1ef1 do l\u00e0 chuy\u1ec3n \u0111\u1ed9ng th\u1eb3ng nhanh d\u1ea7n \u0111\u1ec1u.\n<\/p>\n\n\n\n}

<\/figure><\/div>\n\n\n\n

b) Khi \u0111\u00e3 x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do l\u00e0 m\u1ed9t chuy\u1ec3n \u0111\u1ed9ng \nnhanh d\u1ea7n \u0111\u1ec1u, ta c\u00f3 th\u1ec3 x\u00e1c \u0111\u1ecbnh c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a g theo c\u00f4ng th\u1ee9c g = \n2S\/t^{2<\/sup> v\u00e0 v\u1eadn t\u1ed1c r\u01a1i t\u1ea1i c\u1ed5ng E theo c\u00f4ng th\u1ee9c: v = 2S\/t \u1ee9ng v\u1edbi m\u1ed7i l\u1ea7n \u0111o. H\u00e3y t\u00ednh c\u00e1c gi\u00e1 tr\u1ecb tr\u00ean v\u00e0 ghi v\u00e0o b\u1ea3ng 8.1.\n<\/p>\n\n\n\n}

c) V\u1ebd \u0111\u1ed3 th\u1ecb v = v(t) d\u1ef1a tr\u00ean s\u1ed1 li\u1ec7u c\u1ee7a b\u1ea3ng 8.1, \u0111\u1ec3 m\u1ed9t l\u1ea7n n\u1eefa nghi\u1ec7m l\u1ea1i t\u00ednh ch\u1ea5t c\u1ee7a chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do.\n<\/p>\n\n\n\n

<\/figure><\/div>\n\n\n\n

\u0110\u1ed3 th\u1ecb v = v(t) c\u00f3 d\u1ea1ng m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng, t\u1ee9c l\u00e0 v\u1eadn t\u1ed1c r\u01a1i t\u1ef1 do \nt\u0103ng d\u1ea7n theo th\u1eddi gian. V\u1eady chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt r\u01a1i t\u1ef1 do l\u00e0 chuy\u1ec3n \n\u0111\u1ed9ng th\u1eb3ng nhanh d\u1ea7n \u0111\u1ec1u.\n<\/p>\n\n\n\n

d) Ta c\u00f3: \n<\/p>\n\n\n\n

<\/figure>\n\n\n\n

<\/figure>\n\n\n\n

e, K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p \u0111o gia t\u1ed1c r\u01a1i t\u1ef1 do l\u00e0: \n<\/p>\n\n\n\n

<\/figure>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 8: Th\u1ef1c h\u00e0nh: Kh\u1ea3o s\u00e1t chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do. X\u00e1c \u0111\u1ecbnh gia t\u1ed1c r\u01a1i t\u1ef1 do B\u00e1o c\u00e1o th\u1ef1c h\u00e0nh 1. Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi: S\u1ef1 r\u01a1i t\u1ef1 do l\u00e0 g\u00ec ? N\u1ebfu \u0111\u1eb7c \u0111i\u1ec3m c\u1ee7a chuy\u1ec3n \u0111\u1ed9ng r\u01a1i t\u1ef1 do v\u00e0 vi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh gia t\u1ed1c r\u01a1i t\u1ef1 do ? – […]<\/p>\n","protected":false},"author":12,"featured_media":44112,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1640],"tags":[],"yoast_head":"\n