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TUGAS PERTEMUAN VII MEKANIKA KLASIK

TUGAS MATAKULIAH

Mekanika Klasikyang dibina oleh Bapak Markus Diantoro, Drs., M.Si., Dr.

Oleh:

Rindu Rahmatiah140321807462

UNIVERSITAS NEGERI MALANG

PASCASARJANA

PROGRAM STUDI PENDIDIKAN FISIKAOktober 2014

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1. Show that the parallel axis theorem works:

a. Calculate inertia for a thin rod about perpendicular to its length tn the center of CM as

well as in its edge. Why it has only one inertia?

b. What is the meaning of m and I?

c. Sebuah bet pingpong memiliki tebal a dengan jari-jari r dilengkapi dengan pegangan

sepanjang L dengan diameter d. Digunakan untuk smash dengan kecepatan translasi

V0dan rotasi . Panjang tangan pemain pingpong h. anggap massa jenis seragam.

Berapa L dan T?

d. Give analyses of the units of m, I, T, L, of translational and rotational system. Give

your opinion.

Answer :

(A)Axis Center Of Rod

Sehingga untik mencari nilai dari center of rod

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(B)

Momen inersia (I) (SatuanSI : kg m2) adalah ukuran kelembaman suatu benda

untukberotasi terhadap porosnya. Besaran ini adalah analog rotasi daripadamassa. Momen

inersia berperan dalam dinamika rotasi seperti massa dalam dinamika dasar, dan menentukan

hubungan antaramomentum sudut dankecepatan sudut,momen gaya danpercepatan sudut,dan beberapa besaran lain,

Sedangkan m adalah massa partikel, karena Untuk menentukan momen inersia suatu

benda tegar, benda ditinjau ketika sedang berotasi karena letak sumbu rotasi

mempengaruhi nilai momen inersia. Selain bergantung pada letak sumbu rotasi, momen

inersia (I) partikel bergantung juga pada massa partikel (m)dan kuadrat jarak partikel

dari sumbu rotasi (r2). Massa semua partikel penyusun benda sama dengan massa

benda tersebut

(C)

Menggunakan sumbu teorema paralel, Karena kita akan menghitung pada posisi inersia

moment di salah satu tepi cylinder. anggap h=0 karena kita akan menghitung nilai L bukan

kecepatan pukulan smash

kita dapat menghitung saat batang ini inersia tentang hal apapun selama kita tahu jarak posisiini (h)dari pusat objek massa. Mari kita berlatih dengan menghitung momen inersia batang

tipis tentang ujungnya kiri.

http://id.wikipedia.org/wiki/SIhttp://id.wikipedia.org/wiki/Rotasihttp://id.wikipedia.org/wiki/Massahttp://id.wikipedia.org/wiki/Momentum_suduthttp://id.wikipedia.org/wiki/Kecepatan_suduthttp://id.wikipedia.org/w/index.php?title=Momen_gaya&action=edit&redlink=1http://id.wikipedia.org/wiki/Percepatan_suduthttp://id.wikipedia.org/wiki/Percepatan_suduthttp://id.wikipedia.org/w/index.php?title=Momen_gaya&action=edit&redlink=1http://id.wikipedia.org/wiki/Kecepatan_suduthttp://id.wikipedia.org/wiki/Momentum_suduthttp://id.wikipedia.org/wiki/Massahttp://id.wikipedia.org/wiki/Rotasihttp://id.wikipedia.org/wiki/SI

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Perhatikan bahwa ini adalah hasil yang sama bahwa kita akan memperoleh telah kami

terintegrasi definisi dasar kita untuk momen inersia.

Jadi L adalah balik tuh rumusnya L2= ???

Untuk menghitung T, disini baru digunakan nilai kecepatan pukulan smash

T = Ma, Thus we stick T = Ma into mg

T = ma, and solve for a, and then T. We geta constant

We get the value

NB : lupa ini saya nmr C hihihihhi

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2. Calculate the moment and product of inertia for rotation about z axis of the following rigid

bodies:

a. Single mass m located at the position (0, y0, z0) as shown in

figure for the single mass of the figure.

Solution:

The sum and each reduce to a single term and we find: and

Thus, the inertia tensor is:

b. The same as in part (a) but with a second equals mass

place cimetrically below the xy plane as in figure:

Solution:

For the two masses of figure each of the sum in the

contains two term and we find:

, because both masses have

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= = = =

= = =

= = =

= = = 0

= =

= 0

= = =

Thus, the inertia tensor is:

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3. A uniform rectangular plate has mass M and sides 2a and 2b shown in figure. Find the

inertia tensor at the point G and C in the coordinate system shown. Use integrals as

shown as Steiners theorem.

Solution:

a. Consider, first the center of mass point G. Then I11is the moment of inertia of the

plate about the axis Gx, which is given in the Appendix to be 1/3 M b2. Similarly, I22

= 1/3 M a2, and by the perpendicular axes theorem I33= I11+ I22= 1/3 M (a2+ b2).

The product of inertia are all zero. The products m x2x3and m x1x2because all

the mass lies in the plane x3= 0. The product m x1x2is zero because of symmetry.

The plate is symmetrical about the axis x1= 0, but the term of the sum are odd

function of x1. The contributions to the sum from the right and the left halves of the

plate therefore cancel. Hence I23= I31= I12= 0

dA = d x1. d x2 x3= 0 = = = = * +=

* +=

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dA = d x1. d x2 x3= 0 = =

= = * +=

* +=

dA = d x1. d x2 x3= 0 = = = * += * +=

*

+

= dA = d x1. d x2 x3= 0 =

= = * += * ( ) ( )+= =

=

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dA = d x1. d x2 x3= 0 = =

= 0

dA = d x1. d x2 x3= 0 = = = 0

Thus, the inertia tensor is:

[

]

[

]

b. Now, consider the corner point C. Then I11is the moment inertia of the plate about

the axis Cx1which is given by the parallel axes theorem to be 1/3 mb2+ mb2= 4/3

mb2. Similarly, I22= 4/3 ma2and by the perpendicular axes theorem I33= I11+ I22=

4/3 mb2 + 4/3 ma2= 4/3 m (a2 + b2). I23= I31= 0

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dA = d x1. d x2 x3= 0 = =

= * += * += * +=

* +=

dA = d x1. d x2 x3= 0 = = = 0

dA = d x1. d x2 x3= 0 =

= = 0

Thus, the inertia tensor is:

[ ]

[

]

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