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soal-soal fisika

 

RIGID BODIES : ROTATION ABOUT AN AXIS

1.      Sebuah batang yang sangat ringan, panjangnya 140 cm, dan bekerja tiga gaya F1 = 20 N, F2 =10 N, F3 =40 N dengan arah dan posisi seperti gambar. Besar momen gaya yang menyebabkan batang berotasi pada pusat massanya adalah…

 

ANSWER:

 

Know, F1 = 20 N, F2 = 10 N, F3 = 40 N

Pusat rotasi massa = proses ditengah-tengah batang

 

∑Ʈ       = Ʈ1- Ʈ2 + Ʈ3

             = F1l1 – F2 l2  + F3 l3

             = (20 N) (0,7) – (10 N) (0,3) + (40 N) (0,7)

             = 14 Nm – 3Nm + 28 Nm

             = 39 Nm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.      Sebuah katrol dari pejal dengan tali yang dililitkan pada sisi luarnya ditampilkan pada gambar. Gesekan katrol diabaikan. Jika momen inersia katrol I =   dan tali ditarik dengan gaya tetap F, maka nilai F setara dengan…

ANSWER :

I =

a
 
∑Ʈ = I . a

F. R = I . a

F =

F
 
F =

F =  a (R)-1

 

 

 

Description: Pembahasan soal dinamika rotasi 43.  Perhatikan gambar dua bola yang dihubungkan dengan seutas kawat. Panjang kawat = 12 m, l1 = 4 m dan massa kawat diabaikan, maka besarnya momen inersia sistem adalah…

 

 

 

 

 

ANSWER :

 

. mA       = 0,2 kg

. mB       = 0,6 kg

.rA        = 4 meter

.rB           = 12 – 4 = 8 meter

Ditanya : Momen inersia (I) sistem

 

.Momen inersia bola A

 

IA             = (mA)(rA2)

 = (0,2)(4)2 

 = (0,2)(16)

 = 3,2 kg m2

 

.Momen inersia bola B

 

IB             = (mB)(rB2)

 = (0,6)(8)2

 = (0,6)(64)

 = 38,4 kg m2

 

.Momen inersia sistem partikel :

 

I           = IA + IB

            = 3,2 + 38,4

 = 41,6 kg m2

 

 

4.    Dua buah bola yang dihubungkan dengan kawat (massa kawat diabaikan) disusun seperti gambar. Besar momen inersianya adalah…

ANSWER:

 

.mA = 200 gr = 0,2 kg

.mB = 400 gr = 0,4 kg
.rA = 0
.rB = 25 cm = 0,25 meter

Ditanya : Momen inersia (I) sistem=?

. Momen inersia bola A


IA             = (mA)(rA2)

 = (0,2)(0)2 = 0

. Momen inersia bola B


IB             = (mB)(rB2)

 = (0,4)(0,25)2

 = (0,4)(0,0625)

 = 0,025 kg m2

 

.Momen inersia sistem partikel 


I           = IA + IB

 = 0 + 0,025

 = 0,025 kg m2

 = 25 x 10-3 kg m2

 

Description: contoh soal dinamika rotasi5.  Gambar dibawah menunjukkan sebuah silider pejal yang menggelinding turun pada sebuah bidang miring. Kecepatan silinder pejal di ujung lintasan adalah ….

 

 

 

 

 

 

ANSWER :

Menggunakan Hukum Kekekalan Energi :


.momen inersia silinder pejal : I = m R2

                                               

                        Ek1 + EP1 = Ek2 + Ep2

 

                        0 + mgh = Ektrans + Ekrot + 0

 

                        mgh = m v2  +  I ω2

 

                                mgh =  m v2  +  ( m R2 ) (  )2

                       

            mgh=  m v2  +  m v2 

 

            mgh =  m v2 

 

                v2  =  gh  

            v =

            v =

            v = 6 m/s

 

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