soalan fizik..minta tolong sekejap..
salam..aku nak search subtopik under fizik tapi tak jumpa, so momod, mintak lalu nak bukak thread pasal fizik eh..
oleh kerana dah berbelas tahun meninggalkan dunia fizik (walopun bidang keja aku nie patutnya berkait rapat dgn fizik, tapi aku tak penah jumpa soklan yang mcm nie berkaitan dgn keja aku)
so aku kene siapkan assignment kawan aku..tapi aku stuck kat 2 soklan nie...
https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-ash3/599076_10151515302948147_2127497819_n.jpg
https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash3/555022_10151515303308147_1849851929_n.jpg
Last edited by dauswq on 30-3-2013 01:11 AM
mana tau kalo2 @dauswq leh tolong2 tgkkan? huwa.. tak ingat dah...
minat lagik add math berbanding fizik nih.. aku bagi hint jelar...
solusi @leyapple boleh work out sendiri kan?:)
SOALAN 1
memandangkan soalan berkenaan angular motion (gerakan memusat, bukannya gerakan linear)...
jadi kita selesaikan ngan menggunakan input yg kita tahu...
1. kenalpasti "Boundary conditions / Initial Conditions"
apa berlaku pada halaju kipas, pecutan kipas, halaju sudut kipas & pecutan sudut kipas bila t=0 ?
diberi jejari kipas (radius) adalah 0.5cm .
2. kenalpasti apa yang berlaku selepas t= 0.5 s?
gunakan formula gerakan yang anda tahu!
FORMULA: http://en.wikipedia.org/wiki/List_of_equations_in_classical_mechanics#Equations_of_motion_.28constant_acceleration.29
MAKSUD SIMBOL DALAM FORMULA: http://en.wikipedia.org/wiki/List_of_equations_in_classical_mechanics#Derived_kinematic_quantities
angular velocity = halaju sudut
angular acceleration = pecutan malar sudut
centripedal acceleration = pecutan memusat
tangential acceleration = pecutan malar linear
jangan salah tafsir gerakan linear = gerakan memusat...oleh itu, pecutan linear tidak disamakan ngan pecutan memusat..
tgk formula di atas bagaimana mereka tukar dari linear <--> memusat...
centripedal acceleration tidak sama ngan tangential acceleration& angular acceleration...kena gugel camne 3 komponen ni related..
SOALAN 2
i love Cari/math/e/a/f/eafc5c8c3478c72a17c89e88c03bb5c5.png
i love Cari/math/e/a/f/eafc5c8c3478c72a17c89e88c03bb5c5.png
where
E is the Young's modulus (modulus of elasticity)
F is the force exerted on an object under tension;
A0 is the original cross-sectional area through which the force is applied;
ΔL is the amount by which the length of the object changes;L0 is the original length of the object.
http://en.wikipedia.org/wiki/Young%27s_modulus#Calculation
aku sbnrnye tak ambil tau pun psl modulus young dlm physics sbb basic introductry kiteorg lebih kpd kinematic mechanic / relativistic..
1. kena faham "momen" atau dalam dlm englishnya angular momentum & torque..gunakan konsep ini untuk dapatkan titik pusat di mana kedudukan rod tetap statik (1st trick)dan tetap dalam "keadaan mengufuk" (2nd trick) ...
2. fahamkan hukum newton ke -3...gunakan gambarajah daya (force diagram) utk menunjukkan daya2 yang bertindak ke atas objek
3. apply modulus young untuk dapatkan daya dikenakan ke atas keluli & kuprum selepas di"stress"kan oleh beban...
Konsep mudah:
(1)
angular momentum dan torque adalah terabadi (conserved)...
di sini, jumlah torque & jumlah angular momentum adalah sifar kerana rod statik...
oleh itu, torque dihasilkan keluli= torque dihasilkan kuprum ---- (1st equation)
USE THIS FORMULA: (lebih baik memahami bagaimana momen diaplikasikan ke sini)
i love Cari/math/3/4/b/34be615483d01fea0d42822a7010abc7.png
i love Cari/math/b/3/7/b37cbcae0e83b25b1f26bbdab44f2e32.pnghttp://en.wikipedia.org/wiki/Torque
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied),
and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector. (dalam kes ini angle adalah 90 sbb keluli/kuprum berserenjang ngan rod)
take "pusat" tempat beban bergantungan sbg reference point
gunakan (1st equation)
Jadi anda boleh dpt hubungan antara jarak dari pusat (tmpt beban bergantungan) ke keluli denganjarak dari pusat (tmpt beban bergantungan) ke kuprum. -----(2nd equation)
gunakan modulus young tuk dptkan ---(3rd equation) & ---(4th equation)
unknown variable di sini adalah daya dikenakan ke atas keluli, F1 , daya dikenakan ke atas kuprum, F2 dan jarak keluli & besi dari pusat.....dapat diselesai daripada 1st, 2nd, 3rd & 4th equations
Last edited by dauswq on 19-3-2013 04:40 PM
dauswq posted on 19-3-2013 11:42 AM static/image/common/back.gif
aku bagi hint jelar...
solusi @leyapple boleh work out sendiri kan?
dauswq dah bagi hint...
aku pulak saja nak try jawab cepat2.... benda yang lama dah tinggal...
so aku kene siapkan assignment kawan aku..tapi aku stuck kat 2 soklan nie...
Aaiittt apasal kena siapkan assignment kawan?
Ni saja-saja try.....
Soalan 7
Given
radius, r = 0.5 m
angular acceleration, α = 6π rad/s²
initial angular velocity, ω₁ = 0 rad/s
t = 5 s ( t₁ = 0, t₂ = 5 )
a) In general α = (ω₂ - ω₁)/(t₂ - t₁)
Given α is a constant, after 5 s
ω₂ = αt₂ = (6π rad/s²)(5 s) =30π rad/s
b)In general: ω = (θ₂ - θ₁)/(t₂ - t₁)
Initiallyθ₁ = 0 rad, after 5 s, ω = ω₂
θ₂ = ω₂t₂ = (30π rad/s)(5 s) = 150π rad
c) linear velocity (after 5 s), v = rω = (0.5 m)(30π rad/s) = 15π m/s
linear acceleration, a = αr = (6π rad/s²)(0.5 m) = 3π m/s²
d) radial acceleration, aᵣ
aᵣ = v²/r = (15π m/s)² /(0.5 m) = 450π² m/s²
or
aᵣ = rω² = (0.5 m)(30π rad/s)² = 450π² m/s²
e) magnitude 450π² m/s² in direction towards centre radially
alamak.... subscripts & superscripts tak menjadi....masa taip jadi, bila post tak jadik!
test.... edit balik subscripts & superscripts...
Last edited by mnm77 on 20-3-2013 10:37 AM
Ehh satu lagi...
Soalan 9
Given:
Young modulus of steel, E₁ = 2x10¹¹ Pa
**NOTE: look like the question has typo error on the given data for Young modulus of steel
Young modulus of copper, E₂ = 1.2x10¹¹ Pa
Diameter of steel wire, d₁ = 0.001 m
Diameter of copper wire, d₂= 0.002 m
Length of rod = 0.2 m
Length of both wires is same, say, L
Note: unit for Young modulus, Pa = N/m²
Rod is negligibly light, a weight is placed such that both wires produce similar strain under stressed condition so that the rod is maintained horizontally.
Where should be the weight placed?
Let x is the distance from steel wire to the point P where the weight is attached,
so the distance from P to copper wire is (0.2 - x)
Steel wire Copper wire
M₁ M₂ (moment of force)
F₁ F₂ (stress force)
d₁ d₂ (diameter)
|________P______|
x | 0.2 - x
weight
We expect stress Force F₁ & F₂ should be DIFFERENT in magnitude so that each stretched wire produces the same extra length ΔL to maintain horizontal, since each wire has different Young modulus.
About Young modulus:
http://en.wikipedia.org/wiki/Young%27s_modulus
i love Cari/math/e/a/f/eafc5c8c3478c72a17c89e88c03bb5c5.pngwhere
E is the Young's modulus (modulus of elasticity)
F is the force exerted on an object under tension;
A0 is the original cross-sectional area through which the force is applied;
ΔL is the amount by which the length of the object changes;
L0 is the original length of the object.
Rearrange the above equation we have stress force for steel and copper wire:
F₁ = E₁A₁ΔL/L
F₂ = E₂A₂ΔL/L
Now, the moment of force (or torque), should be equal in magnitude, to have the rod stay in balance horizontally about point P
So, M₁ = M₂
(x)F₁ = (0.2-x)F₂
(x)E₁A₁ΔL/L = (0.2-x)E₂A₂ΔL/L
(x)E₁A₁ = (0.2-x)E₂A₂
knowing cross sectional area of wire, A = πd²/4, hence
(x)E₁d₁² = (0.2-x)E₂d₂²
substitung values for 'E' and 'd' in the above equation we can solve for x
x = 0.1412 m = 14.12 cm (from the steel wire to the point P)
WAllahu a'lam
Last edited by mnm77 on 20-3-2013 05:49 PM
dauswq posted on 19-3-2013 11:42 AM static/image/common/back.gif
aku bagi hint jelar...
solusi @leyapple boleh work out sendiri kan?
dauwsq..
otak ai dah krem....
mnm77 posted on 20-3-2013 05:21 PM static/image/common/back.gif
Ehh satu lagi...
Soalan 9
wah..kaka mnm77!!!
tengkiu so much!!!!!!!!{:1_149:}
leylapple posted on 21-3-2013 08:58 AM static/image/common/back.gif
wah..kaka mnm77!!!
tengkiu so much!!!!!!!!
jap2..
nak checking...@dauswq blh tolong cekkan juga??
soklan nya mcm nie...
purata masa tindakbalas bagi pemandu kereta ialah 0.7s.9masa tindakbalas ialah sela masa antara pemandu melihat isyarat lampu merah dan kakinya menekan brek).
Jika kesan brek memberikan nyahpecutan 5 ms-2,kira jarak brek memberikan nyahpecutan 5 ms-2, kira jarak yang akan ditempuh oleh kereta sebelum ia berenti jika semasa isyarat lampu isyarat lampu merah dilihat kereta sedang bergerak pada kelajuan
a- 5o kmj-1
b-90kmj-1kan ada 4 formula tu kan...
1-> v= u + at
2-> v2=u2+2as
3-> s=ut+1/2 at2
4-> s=(1/2)(u+v)tmula2 ai tukarkan 50kmj-1 tu kepada ms-1, = 50 x 1000m / (60 x 60) = 13.88ms-1
knowing that halaju akhir akan 0 ms-1 -->unmovable state->berhenti sepenuhnya,
and kita tau v, u, a (nyapecutan) dan t..and unknown sekarang adalah s
and t adalah masa tindak balas...
thus, ai kena pakai 2 formula kan, 1 utk tahu kesemua jarak kereta sebelum membrek , 1 utk jarak masa tindakbalas...betul?
leylapple posted on 21-3-2013 05:14 PM static/image/common/back.gif
jap2..
nak checking...@dauswq blh tolong cekkan juga??
mula2 ai tukarkan 50kmj-1 tu kepada ms-1, = 50 x 1000m / (60 x 60) = 13.88ms-1
knowing that halaju akhir akan 0 ms-1 -->unmovable state->berhenti sepenuhnya,
and kita tau v, u, a (nyapecutan) dan t..and unknown sekarang adalah s
and t adalah masa tindak balas...
thus, ai kena pakai 2 formula kan, 1 utk tahu kesemua jarak kereta sebelum membrek , 1 utk jarak masa tindakbalas...betul?
Betul tu... kena guna 2 formula:
1. kira jarak ketika halaju seragam (masa tindakbalas)
2. kira jarak ketika nyahpecutan, a = -5 m/s2
Jika nak guna formula dari 4 yang atas tu:
1) dalam masa 0.7 s tu, berapa jarak kereta tu bergerak?
guna:s= ut +1/2 at2
di mana pecutan 0 pada halaju seragam (50 km/j),
maka s = ut = ( 13.889 m/s) *0.7 s= 9.7 m
2) jarak yang diambil semasa kereta sedang memberhenti
guna: v2= u2+2as
0 = (13.889)^2 + 2(-5)s
s = 19.3 m
Jumlah jarak = 9.7 + 19.3 = 29 m
***********
Tanpa formula yang 4 kat atas tu pun boleh kira juga, dengan syarat kita faham konsep halaju dan pecutan
Halaju, v = kadar perubahan jarak = ds/dt = (s2-s1)/(t2-t1) -------(A)
Pecutan, a = kadar perubahan halaju = dv/dt = (v2-v1)/(t2-t1) --------(B)
Masa tindakbalas, t2 - t1 = 0.7- 0 = 0.7s
v = 13.889 m/s
maka jarak s2-s1 = v(t2-t1) ----------- dari (A)
=( 13.889 m/s)(0.7 s)
= 9.7 m
Nyahpecutan ketika memberhenti
a=-5
-5 = (0-13.889)/Δt----------- dari (B)
Δt=2.78 s
Maka untuk kira jarak dalam masa 2.78s itu.....
Dari definisi pecutan, a= dv/dt = d/dt(ds/dt)
Integrate w.r.tt gives: at = ds/dt (limit from t= 0 to t = t)
Integrate again w.r.t. t gives:0.5 at² = s2-s1 (limit from t= 0 to t = t, s = s1 to s = s2)
kira perubahan jarak: 0.5(-5)(2.78)² = s2-s1
s2-s1 = - 19.3 m
Δs = 19.3 m
Jumlah jarak = 9.7 + 19.3 = 29 m
Last edited by mnm77 on 21-3-2013 06:56 PM
leyapple...sori tak dpt tolong sepenuhnya...
sbb my policy is to provide hint, not to give solutions...
ini saya belajar mase buat degree dulu...:P
abg mnm77 dah bg penerangan & make some advice...i think you get the idea to solve the questions...{:1_152:} dauswq posted on 21-3-2013 11:23 PM static/image/common/back.gif
leyapple...sori tak dpt tolong sepenuhnya...
sbb my policy is to provide hint, not to give solution ...
takper...
otak nie dah berkarat2 jugak sebenarnya...
tapi cukup tabik lah dgn kaka @mnm77 sebab masih ingat....{:1_152:}
leylapple posted on 22-3-2013 09:03 AM static/image/common/back.gif
takper...
otak nie dah berkarat2 jugak sebenarnya...
entah..maybe mcm bro@sis kot?
ai tgh nak tgk2 lagi soklan fizik nie...
tuan punya assignment senang hati kat rumah, soklan dah hampir siap~
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