TUGAS 3 FISDAS 1 KAMPUS PALEMBANG

NAMA : DADI ISWANTO (03101404065)IIS SETYONINGSIH (03101404032)

NOPRAN KURNIAWAN (03101404007)

10.82 Following Thanksgiving dinner your uncle falls into a deep sleep, sitting straight up facing the television set. A naughty grandchild balances a small spherical grape at thetop of his bald head, which itself has the shape of a sphere. After all the children have had time to giggle, the grape starts from rest and rolls down without slipping. It will ,leave contact with your uncle’s scalp when the radial line joining it to the center of curvature makes what angle with the vertical?Jawab :

anggur

kepala cucu

kepala paman

Penyimpanan dari energi antara puncak dan titik dimana anggur bergerak sebesar :

mg ∆ y=12

mvf2+1

2Iωf

2

mgR ¿

g¿

mg cosθ−n=mv f

2

R

Pada titik dimana anggur bergerak , n=0

mg cosθ=mvf

2

R

g−g cos θ= 710

g cosθ

θ=54,0

R

10.83 (a) A thin rod of length h and mass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. Determine the speed of its center of mass just before it hits the horizontal surface. (b) What If? Now suppose the rod has a fixed pivot at its lower end. Determine the speed of the rod’s center of mass just before it hits the surface.Jawab :

(a) Tidak ada aksi gaya horizontal pada balok, menyebabkan poros dari massa tidak akan bergerak horizontal, poros dari massa menurun dengan balok berotasi kira-kira poros dari massa terjatuh, dari perkiraan energi :

K f +U gf =K i+U gi

12

MvCM2 + 1

2Iω2+0=0+ Mg( h

2 )12

MvCM2 + 1

2 ( 112

Mh2)( vCM

h2 )

2

=Mg ( h2 )

vCM=√ 3 gh4

(b) Di dalam kotak, gerakan rotasi murni mengenai titik poros tetap dengan poros sebuah massa bergerak di bagian sirkuler dari radius h/2. Dari perkiraan energi :

K f +U gf =K i+U gi

12

Iω2+0=0+Mg ( h2 )

12 ( 1

3Mh2)( vCM

h2 )

2

=Mg ( h2 )

vCM=√ 3 gh4

10.84 A large, cylindrical roll of tissue paper of initial radius R lies on a long, horizontal surface with the outside end of the paper nailed to the surface. The roll is given a slight shove (vi ! 0) and commences to unroll. Assume the roll has a uniform density and that mechanical energy is conserved in the process. (a) Determine the speed of the center of mass of the roll when its radius has diminished to r. (b) Calculate a numerical value for this speed at r = 1.00 mm, assuming R = 6.00 m. (c) What If? What happens to the energy of the system when the paper is completely unrolled?Jawab :

(a) Massa dari rol berkurang seperti tidak ada rol, m= Mr2

R2 dimana M adalah massa inisial

dari rol, sejak ∆ E=0, ∆ U g+∆ K trans+∆ K rot=0

ketika I=mr2

2

(mgr−MgR )+ mv2

2+[ mr2

2ω2

2 ]=0

ketika ωr=v ,maka v=√ 4 g ( R3−r3 )3 r2

(b) nilai numerical dari kecepata pada r=1,00mm,asumsikan bahwa R=6,00m, Jadi

v=√ 4 g ( R3−r3 )3 r2

v=√ 4 g (6,0 mm3−r3 )3 r2

v=5,31 x 104 m / s(c) kita asumsikan bahwa ∆ E=0, ketika rol berhenti , maka aka mendapatkan benturan

kaku dengan permukaan. Energinya menjadi energi dalam.

11.53. Global warming is a cause for concern because even small changes in the Earth’s temperature can have significant consequences. For example, if the Earth’s polar ice caps were to melt entirely, the resulting additional water in the oceans would flood many coastal cities. Would it appreciably change the length of a day? Calculate the resulting change in the duration of one day. Model the polar ice as having mass 2.30 x1019 kg and forming two flat disks of radius 6.00 x105m. Assume the water spreads into an unbroken thin spherical shell after it melts.

Jawab :

Kutub es utara

Kutub es selatan

Bumi mempunyai kutub es utara dan selatan, jika terjadi pemanasan global maka es di kutub itu akan mencair yang menyebabkan air di kota-kota pesisir akan menerima banyak tambahan air, dan lama kelamaan akan menyelubungi bumi

The moment of inertia of the rest of the earth is

I=25

M R2=25

5.98 X 1024 kg (6,37 x106 )2=9.71 x1037kg . m2

For the original ice disks

I=12

M r 2=12

2.30 x 1019 kg (6 x105m )2=4.14 x1030kg .m2

For the final thin shell of water

I=23

M r2=23

2.30 x1019 kg (6.37 x 106 m )2=6.22 x1032 kg . m2

Conservation of angular momentum for the spinning planet is expressed by I iωi=I f ωf

( 4.14 x1030+9.71 x1037) 2 π86400 s

=(6.22 x1032+9.71 x1037 ) 2π(86400 s+δ )

(1+ δ86400 s )(1+ 414 x 1030

9.71 x1037 )=(1+ 6.22 x1032

9,71 x1037 )δ

86400 s=6.22 x 1032

9.71 x 1037−4.14 x1030

9.71 x 1037

δ=0,550 s

11.54. A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis AB (Fig. P11.54). A bullet of mass m and speed v is shot at the face opposite ABCD at a height of 4a/3. The bullet becomes embedded in the cube. Find the minimum value of v required to tip the cube so that it falls on face ABCD. Assume m ≪M.

Jawab :

Untuk kubus bagian ujung atas, pusat massa (CM) harus naik rotasi sumbu AB selesai. Untuk ini, CM harus dinaikkan jarak dari (√ 2-1)

∴Mga (√2−1 )=12

I cube ω2

Dari kekekalan momentum sudut di dapa

4 a3

mv=( 8 M a2

3 )ω

ω= mv2 Ma

12 ( 8 M a

2

3 ) m2 v2

4 M 2 a2 =Mga (√2−1 )

v=Mm

√3 ga (√2−1 )

12.64 A steel cable 3.00 cm2 in cross-sectional area has a mass of 2.40 kg per meter of length. If 500 m of the cable is hung over a vertical cliff, how much does the cable stretch under its own weight? Ysteel = 2.00 x 1011 N/m2.

Peregangan kabel yang menggantung pada tebing sebanding dengan perubahan panjangnya , sehingga dapat kita rumuskan :

dLL

= FYA

Dengan Y = total gaya di setiap penampangnya, maka dapat kita hitung :

∆ y=∫0

Li

( dLL )dy= μg

YA∫0

Li

y dy=12

μg Li2

YA

∆ y=12

(2,40 ) (9,80 )(500)2

(2×1011)(3×10−4)=0,049 m=4,9 cm

12.65 (a) Estimate the force with which a karate master strikes a board if the hand’s speed at time of impact is 10.0 m/s, decreasing to 1.00 m/s during a 0.002 00-s time-of-contact with the board. The mass of his hand and arm is 1.00 kg. (b) Estimate the shear stress if this force is exerted on a 1.00-cm-thick pine board that is 10.0 cm wide. (c) If the maximum shear stress a pine board can support before breaking is 3,60 x106 N /m2, will the board break?Jawab :

a. F = ma

a= ∆ v∆ t

maka gaya yang diberikan oleh karate master sebesar :

F=m(∆ v∆ t )=(1kg )

(10−1 ) ms

0,002 s=4500 N

b. Tekanan yang diberikan karate master ke papan sebesar

strees= FA

= 4500(0.01 )(0.1)

=4.5×106

c. Jika kemampuan tekanan papan sebesar 3,60 x106 N /m2 , maka papan akan hancur oleh pukulan karate master

13.60 Many people assume that air resistance acting on a moving object will always make the object slow down. It can actually be responsible for making the object speed up. Consider

M kabel = 2.40 kgA kabel = 3.00 cm2

L kabel = 500 m

tebing

a 100-kg Earth satellite in a circular orbit at an altitude of 200 km. A small force of air resistance makes the satellite drop into a circular orbit with an altitude of 100 km. (a) Calculate its initial speed. (b) Calculate its final speed in this process. (c) Calculate the initial energy of the satellite–Earth system. (d) Calculate the final energy of the system. (e) Show that the system has lost mechanical energy and find the amount of the loss due to friction. (f) What force makes the satellite’s speed increase? You will find a free-body diagram useful in explaining your answer.

Jawab :

Total gaya yang bekerja pada satelit sebesar Fgravitasi=F sentripetal satelit

G M r m

r 2 =m v2

rMaka didapat kecepatan putar satelit pada jarak tertentu dari bumi sebesar

v=√ G M r

r

a. v i=√ (6,67 × 10−11 N .m2 kg−2 )(5,98 ×1024 kg)(6,37 ×106 m+2 ×105 m)

=7,79 ×103m / s

b. v i=√ (6,67 × 10−11 N .m2 kg−2 ) (5,98× 1024 kg )(6,47 × 106 m)

=7,85× 103 m /s

c. Total energi mekanik satelit adalah :

Em=K+U s=12

m v2−G M E m

r=

12

m(√ G M E

r )2

−12

m(G M E

r )Em=

G M E m

2 rd. Energi awal satelit pada

r200 kmdari jari− jaribumi=6,57 × 106 m

Ei=−( 6,67× 10−11 N . m2 . kg−2 ) ( 5,98× 1024 )(100 kg )

2(6,57 ×106 m)=−3,04 ×109 J

e. r100 kmdari jari− jaribumi=6,47 × 106 m

E f=−(6,67 × 10−11 N .m2 . kg−2) (5,98 ×1024 kg ) (100 kg )

2 (6,47 × 106 m )=−3,08× 109 J

f. Total energi saat terjadi perpindahan dari orbit 200km ke 100km dari altitude sebesarEi−Ef =−3,04×109 J−(−3,08 ×109 J )=4,67 × 107 J

13.61 Two hypothetical planets of masses m1 and m2 and radii r1 and r2 , respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. (a) When their center-to-center separation is d, find expressions for the speed of each planet and for their relative speed. (b) Find the kinetic energy of each planet just before they collide, if m1=2,00 x1024kg , m2=8,00 x1024 kg . r1=3,00 x106 ,∧r2=5,00 x 106 m (Note: Both energy and momentum of the system are conserved.)

Jawab: Pada pemisahan yang tak terbatas U=0 dan total energimekanik (K )=0. Karena energi

kedua planet konservatif,maka 0=12

m1v12+ 1

2m2 v2

2−G m1 m2

dMomentum awal sistem adalah 0 dan momentum tersebut kekal, oleh karena itu 0=m1m2−m1 m2

penggabungan persamaan 1 dan 2 v1=m2√ 2Gd (m1+m2)

∧v2=m1 √ 2 Gd (m1+m2)

kecepatan relatif vr=v1−(−v2)=√ 2G(m1+m2)d

b. subtitusi memberikan nilai numerik ke dalam persamaan untuk mencari v1 dan v2 pada bagian (a) untuk menemukan

v1=1.03× 104 m s−1∧v1=2,58×103 m s−1

Oleh karena itu, K1=12

m1 v12=1,07 ×1032J∧K2=

12

m2 v22=2,67 × 1031 J

14.62. In about 1657 Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemispheres. Two teams of eight horses each could pull the hemispheres apart only on some trials, and then “with greatest difficulty,” with the resulting sound likened to a cannon firing (Fig. P14.62). (a) Show that the force F required to pull the evacuated hemispheres apart isπ R2 ( P0−P ), where R is the radius of the hemispheres and P is the pressure inside the

hemispheres, which is much less than P0. (b) Determine the force if P=0.100 P0 and R=0.300 m.

Penyelesaian :(a) Tekanan pada permukaan dua setengah bola sama pada seluruh dan gaya pada setiap

elemen luas permukaan diarahkan sepanjang radius gaya yang diterapkan setengah bola. Sepanjang sumbu harus menyeimbangkan gaya pada "efektif "daerah, yang merupakan proyeksi dari permukaan sebenarnya ke pesawat tegak lurus terhadap sumbu x. A=π R2

Oleh karena itu F=( P0−P ) π R2

(b) Untuk membrikan nilai F=( P0−0.100 P0 ) [ π (0.300 )2 ]=0.254 P0=2.58 x 104 N

15.60. A particle with a mass of 0.500 kg is attached to a spring with a force constant of 50.0 N/m. At time t=0 the particle has its maximum speed of 20.0 m/s and is moving to the left. (a) Determine the particle’s equation of motion, specifying its position as a function of time. (b) Where in the motion is the potential energy three times the kinetic energy? (c) Find the length of a simple pendulum with the same period. (d) Find the minimum time interval required for the particle to move from x=0¿ x=1.00 m.

Penyelesaian :

(a) Pada x=A cosωt+ϕ v=−ωAsin (ωt+ϕ )Kita memiliki t=0 v=−ωAsin ϕ=−vmax

Membutuhkan ϕ=900 , so x=A cos ( ωt+900 )dan persamaan ini ekuivalen x=−A sin ωt

Secaranumerik kita mendapatkan ω=√ km

=√ 50 N /m0,5 kg

=10 s−1

dan vmax=ω Α 20ms=(10 s1 ) Α A=2 m

So x=(−2 m )sin [ (10 s1 ) t ](b)

¿ 12

m v2+ 12

k x2=12

k A2

12

k x2=3( 12

m v2)Sehingga menjadi

13

12

k x2+ 12

k x2=12

k A2

43

x2=A2

x=±√ 34

A=± 0.866 A=±1.73m

15.61 A horizontal plank of mass m and length L is pivoted at one end. The plank’s other end is supported by a spring of force constant k (Fig P15.61). The moment of inertia of the plank about the pivot is mL2. The plank is displaced by a small angle . from its horizontal equilibrium position and released. (a) Show that it moves with simple harmonic motion with an angular frequency . (b) Evaluate the frequency ifthe mass is 5.00 kg and the spring has a force constant of 100 N/m.

a. Pada titik keseimbangan, kita mendapatkan

Στ=0−mg( L2 )+k x0 L

Dimana, x0 komposisi titik keseimbangan

Setelah pergeseran sudut sedikit,

Στ=0−mg( L2 )+k x0 L=−mg( L

2 )+k ( x0−Lθ ) L=−kθ L2

tetapi Στ=Ia=13

m L2 d2θd t 2 jadi

d2θd t 2 =−3 k

mθ

Percepatan anguler berlawanan arah pada pergeserannya, jadi kita mendapatkan gerak

harmonik sederhana ω2=3 k

m

b. Dengan menggunakan rumus diatas kita dapat mencari rumus menghitung frekuensi getar pegas:

f =ω2π

=1

2 π √ 3km

=1

2π √ 3(N m−1)5 kg

=1,23 Hz