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Tugas 3 fisika teknik
 

Tugas 3 fisika teknik

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    Tugas 3 fisika teknik Tugas 3 fisika teknik Document Transcript

    • TUGAS 3 FISIKA TEKNIK MOMENTUM DAN TUMBUKAN 0LEH NAMA: FAKHRI AMAL NIM: 1129040279 KELAS: 06 PTIK FAKULTAS TEKNIKUNIVERSITAS NEGERI MAKASSAR 2011
    • Soal-Soal Latihan Bab 6: 1. Sistem partikel tersusun oleh 3 buah titik massa. Setiap partikel itu bermassa 1 satuan serta berposisi ( ) dan kecepatan ( ): a. Tentukan posisi dan kecepatan pusat massa sistem 3 partikel itu terhadap titik asal. b. Hitunglah momentum linear pusat massa dari sitem 3 partikel itu. c. Hitung pula tenaga gerak sistem 3 partikel, tenaga gerak pusat massa dan tenega gerak internalnya. 2. Mengacu pada soal nomor 1, hitunglah momentum sudut system 3 partikel itu terhadap titik asal dan momentum sudut internalnya.Jawaban : 1. Diketahui: Ditanyakan: a. pm dan pm =……? b. Ppm = ....? c. Kpm = ....? Penyelesaian: a. pm = = = ( ) = (1( + )+1( + )+1( )) = +++ + = +2 +2 = =
    • = = = = = ( ) = (1( )+1( )+1( + + )) = + + = b. Momentum (P) = =3. ( +4 + ) = + + = = = =3 c. = M pm . = . ( +4 + ).(+4+) = ( Kpm = ( = = = Kpm = - KI = = . (m1v1+m2v2+m3v3)(v1+v2+v3)
    • = . ( 1(2 ))+(1( ))+(1( + + )) (2 + + ) = (2 + + + )( 2 + + ) = (4 +2 +2 + +2 + + + + 2 + +2 + + + + + + + KI = (14 +8 + +2 + KI = (14 +8 + +2 + = = KI =2- = = (m1r1v1 + m2+r2+v2 + m3r3v3) = (1( ) 2 ) + (1( ) ) + (1( ) + + ) = (2 +2 )+( )+( + + ) = (2 +3 + + + ) = (2 +3 + + + ) = =- = = (m1r1v1 + m2+r2+v2 + m3r3v3) = (1( ) 2 ) + (1( ) ) + (1( ) + + ) = (2 +2 )+( )+( + + ) = (2 +3 + + + ) = (2 +3 + + + ) =
    • =