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{"id":25322,"date":"2018-04-05T11:27:50","date_gmt":"2018-04-05T04:27:50","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=25322"},"modified":"2018-04-05T11:43:40","modified_gmt":"2018-04-05T04:43:40","slug":"chuong-1-bai-4-dong-nang-cua-vat-ran-quay-quanh-mot-truc-co-dinh","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-1-bai-4-dong-nang-cua-vat-ran-quay-quanh-mot-truc-co-dinh\/","title":{"rendered":"Ch\u01b0\u01a1ng 1 – B\u00e0i 4: \u0110\u1ed9ng n\u0103ng c\u1ee7a v\u1eadt r\u1eafn quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh"},"content":{"rendered":"

C\u00c2U H\u1eceI<\/strong><\/p>\n

C\u00e2u 1 (trang 20 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Vi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9ng n\u0103ng c\u1ee7a m\u1ed9t v\u1eadt r\u1eafn quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Momen qu\u00e1n t\u00ednh v\u1edbi tr\u1ee5c quay \u0111\u00e3 cho:<\/p>\n

\"\"<\/p>\n

C\u00e2u 2 (trang 20 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0N\u00eau nh\u1eadn x\u00e9t v\u1ec1 vai tr\u00f2 c\u1ee7a momen qu\u00e1n t\u00ednh trong c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9ng n\u0103ng quay.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Momen qu\u00e1n t\u00ednh trong c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9ng n\u0103ng quay \u0111\u00f3ng vai tr\u00f2 gi\u1ed1ng nh\u01b0 kh\u1ed1i l\u01b0\u1ee3ng trong c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9ng n\u0103ng t\u1ecbnh ti\u1ebfn.<\/p>\n

B\u00c0I T\u1eacP<\/strong><\/p>\n

B\u00e0i 1 (trang 21 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t b\u00e1nh \u0111\u00e3 c\u00f3 momen qu\u00e1n t\u00ednh 2,5 kg.m2<\/sup>\u00a0quay v\u1edbi t\u1ed1c \u0111\u1ed9 g\u00f3c 8900 rad\/s. \u0110\u1ed9ng n\u0103ng quay c\u1ee7a b\u00e1nh \u0111\u00e0 b\u1eb1ng.<\/span><\/p>\n

A. 9,1.108<\/sup>\u00a0J.\u00a0\u00a0\u00a0 B. 11 1125 J.<\/p>\n

C. 9,9.107<\/sup>\u00a0J.\u00a0\u00a0\u00a0 D. 22 250 J.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn C.<\/p>\n

\u0110\u1ed9ng n\u0103ng c\u1ee7a v\u1eadt:<\/p>\n

W\u0111<\/sub>= (1\/2)l\u03c92<\/sup>\u00a0= (1\/2).2,5.(8900)2<\/sup>\u00a0= 9,9.107<\/sup><\/p>\n

B\u00e0i 2 (trang 71 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t \u0111\u0129a tr\u00f2n c\u00f3 momen qu\u00e1n t\u00ednh l, \u0111ang quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh v\u1edbi t\u1ed1c \u0111\u1ed9 g\u00f3c \u03c9o<\/sub>\u00a0. Ma s\u00e1t \u1edf tr\u1ee5c quanh nh\u1ecf kh\u00f4ng \u0111\u00e1ng k\u1ec3. N\u1ebfu t\u1ed1c \u0111\u1ed9 g\u00f3c c\u1ee7a \u0111\u0129a gi\u1ea3m \u0111i hai l\u1ea7n th\u00ec momem \u0111\u1ed9ng l\u01b0\u1ee3ng v\u00e0 \u0111\u1ed9ng n\u0103ng quanh c\u1ee7a \u0111\u0129a \u0111\u1ed1i v\u1edbi tr\u1ee5c quay thay \u0111\u1ed5i th\u1ebf n\u00e0o?<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn D<\/p>\n

– Khi \u03c9 gi\u1ea3m 2 l\u1ea7n: Momen \u0111\u1ed9ng l\u01b0\u1ee3ng L = l\u03c9 gi\u1ea3m 2 l\u1ea7n<\/p>\n

\u0110\u1ed9ng n\u0103ng W\u0111<\/sub>\u00a0= (1\/2)l\u03c92<\/sup>\u00a0gi\u1ea3m 4 l\u1ea7n<\/p>\n

B\u00e0i 3 (trang 21 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Hai \u0111\u0129a tr\u00f2n c\u00f3 c\u00f9ng momen qu\u00e1n t\u00ednh \u0111\u1ed1i v\u1edbi c\u00f9ng tr\u1ee5c quay \u0111i qua t\u00e2m c\u1ee7a c\u00e1c \u0111\u0129a. L\u00fac \u0111\u1ea7u, \u0111\u0129a 2 (\u1edf ph\u00eda tr\u00ean) \u0111ang \u0111\u1ee9ng y\u00ean, \u0111\u0129a 1 quay v\u1edbi t\u1ed1c \u0111\u1ed9 g\u00f3c \u03c9o<\/sub>\u00a0. Ma s\u00e1t \u1edf tr\u1ee5c quay nh\u1ecf kh\u00f4ng \u0111\u00e1ng k\u1ec3. Sau \u0111\u00f3, cho hai \u0111\u0129a d\u00ednh v\u00e0o nhau, h\u1ec7 quay v\u1edbi t\u1ed1c \u0111\u1ed9 g\u00f3c \u03c9 . \u0110\u1ed9ng n\u0103ng c\u1ee7a h\u1ec7 hai \u0111\u0129a l\u00fac sau so v\u1edbi l\u00fac \u0111\u1ea7u.<\/span><\/p>\n

A. T\u0103ng ba l\u1ea7n.\u00a0\u00a0\u00a0 B. Gi\u1ea3m 4 l\u1ea7n.<\/p>\n

C. T\u0103ng ch\u00edn l\u1ea7n.\u00a0\u00a0\u00a0 D. Gi\u1ea3m hai l\u1ea7n<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn D.<\/p>\n

\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n Momen \u0111\u1ed9ng l\u01b0\u1ee3ng ta c\u00f3:<\/p>\n

l\u03c9o<\/sub>\u00a0= (l + l) \u03c9 = 2l\u03c9<\/p>\n

suy ra \u03c9 = (1\/2)l\u03c9o<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(1)<\/p>\n

\u0110\u1ed9ng n\u0103ng l\u00fac \u0111\u1ea7u:<\/p>\n

W\u0111<\/sub>\u00a0= (1\/2)l\u03c92<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2)<\/p>\n

\u0110\u1ed9ng n\u0103ng l\u00fac sau:<\/p>\n

W\u0111<\/sub>\u00a0= (1\/2).2.l\u03c92<\/sup>\u00a0= l\u03c92<\/sup>\u00a0= (1\/4).l.\u03c90<\/sub>2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(3)<\/p>\n

V\u1eady \u0111\u1ed9ng n\u0103ng gi\u1ea3m 2 l\u1ea7n.<\/p>\n

\n
\n

B\u00e0i 4 (trang 21 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Hai b\u00e1nh xe A v\u00e0 B c\u00f3 c\u00f9ng \u0111\u1ed9ng n\u0103ng quay, t\u1ed1c \u0111\u1ed9 g\u00f3c \u03c9A<\/sub>\u00a0= 3 \u03c9B<\/sub>\u00a0. T\u1ec9 s\u1ed1 momen qu\u00e1n t\u00ednh lA<\/sub>\/lB<\/sub>\u00a0\u0111\u1ed1i v\u1edbi tr\u1ee5c quay c\u1ee7a A v\u00e0 B c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o sau \u0111\u00e2y?<\/span><\/p>\n

A. 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B. 9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0C. 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0D. 1<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn B.<\/p>\n

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

\"\"<\/p>\n

B\u00e0i 5 (trang 21 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t \u0111\u0129a tr\u00f2n \u0111\u1ed3ng ch\u1ea5t c\u00f3 b\u00e1n k\u00ednh R = 0,5 m, kh\u1ed1i l\u01b0\u1ee3ng m = 1kg quay \u0111\u1ec1u v\u1edbi t\u1ed1c \u0111\u1ed9 g\u00f3c \u03c9 = 6 rad\/s quanh m\u1ed9t tr\u1ee5c vu\u00f4ng v\u1edbi \u0111\u0129a v\u00e0 \u0111i qua t\u00e2m c\u1ee7a \u0111\u0129a. T\u00ednh \u0111\u1ed9ng n\u0103ng c\u1ee7a \u0111\u0129a.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

+ Momen qu\u00e1n t\u00ednh c\u1ee7a \u0111\u0129a:<\/p>\n

l = (1\/2)mR2<\/sup>\u00a0= 0,5.1. 0,25 = 0,125 kg.m2<\/sup><\/p>\n

+ \u0110\u1ed9ng n\u0103ng quay c\u1ee7a \u0111\u0129a:<\/p>\n

W\u0111<\/sub>= (1\/2)l\u03c92<\/sup>\u00a0= 0,5.0,125.36 = 2,25 (J)<\/p>\n

B\u00e0i 6 (trang 21 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t r\u00f2ng r\u1ecdc c\u00f3 momen qu\u00e1n t\u00ednh \u0111\u1ed1i v\u1edbi tr\u1ee5c quay c\u1ed1 \u0111\u1ecbnh l\u00e0 10 kg.m2<\/sup>, quay \u0111\u1ec1u v\u1edbi t\u1ed1c \u0111\u1ed9 60 v\u00f2ng\/ ph\u00fat. T\u00ednh \u0111\u1ed9ng n\u0103ng quay c\u1ee7a r\u00f2ng r\u1ecdc.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

+ T\u1ed1c \u0111\u1ed9 g\u00f3c:<\/p>\n

\u03c9 = 2\u03c0n = 2\u03c0.2 = 2\u03c0 rad\/s<\/p>\n

+ \u0110\u1ed9ng n\u0103ng c\u1ee7a r\u00f2ng r\u1ecdc:<\/p>\n

W\u0111<\/sub>\u00a0= (1\/2)l\u03c92<\/sup>\u00a0= 0,5.10.(2\u03c0)22<\/sup>\u00a0= 5.4.10 = 200 (J)<\/p>\n

B\u00e0i 7 (trang 21 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t b\u00e1nh \u0111\u00e0 quay nhanh d\u1ea7n \u0111\u1ec1u t\u1eeb tr\u1ea1ng th\u00e1i ngh\u1ec9 v\u00e0 sau 5s th\u00ec c\u00f3 t\u1ed1c \u0111\u1ed9 g\u00f3c 200 rad\/s v\u00e0 c\u00f3 \u0111\u1ed9ng n\u0103ng quay l\u00e0 60 kJ. T\u00ednh gi\u00e1 t\u1ed1c g\u00f3c v\u00e0 momen qu\u00e1n t\u00ednh c\u1ee7a b\u00e1nh \u0111\u00e0 \u0111\u1ed1i v\u1edbi tr\u1ee5c quay.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– Gia t\u1ed1c g\u00f3c c\u1ee7a b\u00e1nh \u0111\u00e0:<\/p>\n

\"\"<\/p>\n

– Momen qu\u00e1n t\u00ednh:<\/p>\n

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

C\u00c2U H\u1eceI C\u00e2u 1 (trang 20 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0Vi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9ng n\u0103ng c\u1ee7a m\u1ed9t v\u1eadt r\u1eafn quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh. L\u1eddi gi\u1ea3i: Momen qu\u00e1n t\u00ednh v\u1edbi tr\u1ee5c quay \u0111\u00e3 cho: C\u00e2u 2 (trang 20 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0N\u00eau nh\u1eadn x\u00e9t v\u1ec1 vai tr\u00f2 c\u1ee7a momen qu\u00e1n […]<\/p>\n","protected":false},"author":3,"featured_media":25323,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1304],"tags":[1432,1427],"yoast_head":"\nCh\u01b0\u01a1ng 1 - B\u00e0i 4: \u0110\u1ed9ng n\u0103ng c\u1ee7a v\u1eadt r\u1eafn quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/chuong-1-bai-4-dong-nang-cua-vat-ran-quay-quanh-mot-truc-co-dinh\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 1 - B\u00e0i 4: \u0110\u1ed9ng n\u0103ng c\u1ee7a v\u1eadt r\u1eafn quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh\" \/>\n<meta property=\"og:description\" content=\"C\u00c2U H\u1eceI C\u00e2u 1 (trang 20 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0Vi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ed9ng n\u0103ng c\u1ee7a m\u1ed9t v\u1eadt r\u1eafn quay quanh m\u1ed9t tr\u1ee5c c\u1ed1 \u0111\u1ecbnh. 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