Thursday, April 28, 2016
The figure shows the position of a car (black circles) at one-second intervals. What is the acceleration at the time T = 4 s?
The figure shows the position of a car (black circles) at one-second intervals. What is the acceleration at the time T = 4 s?
Solution:
t=T-1s
x=v₀t+at²/2
40m=v₀∙4s+a∙(4s)²/2
13m=v₀∙1s+a∙(1s)²/2=(v₀+a∙1s/2)∙1s
v₀=13m/s-a∙1s/2
40m=(13m/s-a∙1s/2)∙4s+a∙(4s)²/2 = 52m-2s²∙a+8s²∙a = 52m+6s²∙a
6s²∙a=-12m
a=-2m/s²
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙
Solution:
t=T-1s
x=v₀t+at²/2
40m=v₀∙4s+a∙(4s)²/2
13m=v₀∙1s+a∙(1s)²/2=(v₀+a∙1s/2)∙1s
v₀=13m/s-a∙1s/2
40m=(13m/s-a∙1s/2)∙4s+a∙(4s)²/2 = 52m-2s²∙a+8s²∙a = 52m+6s²∙a
6s²∙a=-12m
a=-2m/s²
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙
The potential energy of a mass m=10kg is given by U=-3.98 × 10¹⁵Jm⋅ r⁻¹
The potential energy of a mass m=10kg is given by
U=-3.98 × 10¹⁵Jm⋅ r⁻¹ (potential energy decreases as r approaches zero). The force on this mass when it is located at r=6.38 × 10⁶ m is _________
Solution:
F = -dU/dr=
= -d( -3.98 × 10¹⁵⋅ r⁻¹)/dr=
= 3.98 × 10⁻¹⁵⋅ d(r⁻¹)/dr =
= -3.98 × 10⁻¹⁵⋅ r⁻² =
= -3.98 × 10¹⁵/ (6.38 × 10⁶)²
Calculation on the Google Calculator: -3.98E15/ (6.38E6)^2=-97.7781271804
|F|=97.8N, the direction is toward the origin (r=0).
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞
Solution:
F = -dU/dr=
= -d( -3.98 × 10¹⁵⋅ r⁻¹)/dr=
= 3.98 × 10⁻¹⁵⋅ d(r⁻¹)/dr =
= -3.98 × 10⁻¹⁵⋅ r⁻² =
= -3.98 × 10¹⁵/ (6.38 × 10⁶)²
Calculation on the Google Calculator: -3.98E15/ (6.38E6)^2=-97.7781271804
|F|=97.8N, the direction is toward the origin (r=0).
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞
A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At².
A rocket, speeding along toward Alpha Centauri, has an acceleration
a(t) = At².
Assume that the rocket began at rest at the Earth (x = 0) at t = 0. Assuming it simply travels in a straight line from Earth to Alpha Centauri (and beyond), what is the ratio of the speed of the rocket when it has covered half the distance to the star to its speed when it has travelled half the time necessary to reach Alpha Centauri?
Solution:
T - total time necessary to reach Alpha Centauri
D - distance to the star
v - speed
v₁=v(D/2)
v₂=v(T/2)
v(t)=∫At²dt=⅓At³
x(t)=∫v(t)dt=⅓At³dt=⅓¼At⁴
D=AT⁴/12
D/2=AT⁴/24
D/2=At⁴/12; At⁴/12=AT⁴/24; t⁴=T⁴/2; t⁴/T⁴=½; t/T=∜½
v₁/v₂ = v(t)/v(T/2) = {⅓At³} / {⅓A(T/2)³}
=t³ / (T/2)³ =2³ (t/T)³ = 2³ ⨯ (∜½)³ = (2∜½)³ = (∜16⨯∜½)³ = (∜8)³=4∜2
This result is not among the proposed answers to choose from. Check again algebraic transformations.
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙
a(t) = At².
Assume that the rocket began at rest at the Earth (x = 0) at t = 0. Assuming it simply travels in a straight line from Earth to Alpha Centauri (and beyond), what is the ratio of the speed of the rocket when it has covered half the distance to the star to its speed when it has travelled half the time necessary to reach Alpha Centauri?
Solution:
T - total time necessary to reach Alpha Centauri
D - distance to the star
v - speed
v₁=v(D/2)
v₂=v(T/2)
v(t)=∫At²dt=⅓At³
x(t)=∫v(t)dt=⅓At³dt=⅓¼At⁴
D=AT⁴/12
D/2=AT⁴/24
D/2=At⁴/12; At⁴/12=AT⁴/24; t⁴=T⁴/2; t⁴/T⁴=½; t/T=∜½
v₁/v₂ = v(t)/v(T/2) = {⅓At³} / {⅓A(T/2)³}
=t³ / (T/2)³ =2³ (t/T)³ = 2³ ⨯ (∜½)³ = (2∜½)³ = (∜16⨯∜½)³ = (∜8)³=4∜2
This result is not among the proposed answers to choose from. Check again algebraic transformations.
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙
Wednesday, April 27, 2016
What is the location of the center of mass of the rod?
A long, thin rod lies along the x-axis.
One end of the rod is located at x = 2.00 m and the other end of the rod is located at x = 6.00 m.
The linear mass density of the rod is given by ax³+b ,
where a = 0.100 kg/m⁴ and b = 0.200 kg/m.
What is the location of the center of mass of the rod?
xₘ={∫x(ax³+b)dx}/{∫(ax³+b)dx}={∫(ax⁴+bx)dx}/{∫(ax³+b)dx}=
=(⅕a(x₂⁵-x₁⁵)+½b(x₂²-x₁²)) / (¼a(x₂⁴-x₁⁴)+b(x₂-x₁))=
={(⅕0.1(6⁵-2⁵)+½0.2(6²-2²))}/{(¼0.1(6⁴-2⁴)+0.2(6-2))}
Calculation:
(0.1*(6^5-2^5)/5+0.2*(6^2-2^2)/2)/(0.1*(6^4-2^4)/4+0.2(6-2))
=4.81951219512
xₘ≈4.82
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞
One end of the rod is located at x = 2.00 m and the other end of the rod is located at x = 6.00 m.
The linear mass density of the rod is given by ax³+b ,
where a = 0.100 kg/m⁴ and b = 0.200 kg/m.
What is the location of the center of mass of the rod?
xₘ={∫x(ax³+b)dx}/{∫(ax³+b)dx}={∫(ax⁴+bx)dx}/{∫(ax³+b)dx}=
=(⅕a(x₂⁵-x₁⁵)+½b(x₂²-x₁²)) / (¼a(x₂⁴-x₁⁴)+b(x₂-x₁))=
={(⅕0.1(6⁵-2⁵)+½0.2(6²-2²))}/{(¼0.1(6⁴-2⁴)+0.2(6-2))}
Calculation:
(0.1*(6^5-2^5)/5+0.2*(6^2-2^2)/2)/(0.1*(6^4-2^4)/4+0.2(6-2))
=4.81951219512
xₘ≈4.82
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞
Tuesday, April 26, 2016
What is the distance of the resulting plate's center of mass from the origin?
A thin rectangular plate of uniform density σ₁ = 5.94 kg/m² has a length a = 0.740 m and a width of b = 0.330 m. The lower left hand corner is placed at the origin, (x, y) = (0, 0). A circular hole is cut in the plate of radius r = 0.064 m with center at (x, y) = (0.084 m, 0.084 m) and replaced with a circular disk with the same radius composed of another material of the same thickness with uniform density σ₂ = 1.67 kg/m². What is the distance of the resulting plate's center of mass from the origin? (in m)
xₘ=(σ₁ab a/2+πr²(σ₂-σ₁) x) / (σ₁ab a/2+πr²(σ₂-σ₁) x)
yₘ=(σ₁ab a/2+πr²(σ₂-σ₁) y) / (σ₁ab a/2+πr²(σ₂-σ₁) y)
d²=xₘ²+yₘ²
xₘ=(5.94*0.74*0.33*0.74/2 + (pi)*0.064^2*(1.67-5.94)*0.084)/(5.94*0.74*0.33+ (pi)*0.064^2*(1.67-5.94))=0.381260099
yₘ=(5.94*0.74*0.33*0.33/2 + (pi)*0.064^2*(1.67-5.94)*0.084)/(5.94*0.74*0.33+ (pi)*0.064^2*(1.67-5.94))=0.381260099=0.16818904902
d=sqrt(0.381260099^2+0.16818904902^2)=0.41670951429≈0.417 (m)
yₘ=(σ₁ab a/2+πr²(σ₂-σ₁) y) / (σ₁ab a/2+πr²(σ₂-σ₁) y)
d²=xₘ²+yₘ²
xₘ=(5.94*0.74*0.33*0.74/2 + (pi)*0.064^2*(1.67-5.94)*0.084)/(5.94*0.74*0.33+ (pi)*0.064^2*(1.67-5.94))=0.381260099
yₘ=(5.94*0.74*0.33*0.33/2 + (pi)*0.064^2*(1.67-5.94)*0.084)/(5.94*0.74*0.33+ (pi)*0.064^2*(1.67-5.94))=0.381260099=0.16818904902
d=sqrt(0.381260099^2+0.16818904902^2)=0.41670951429≈0.417 (m)
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ
#Physics #problem #PhysicsProblem #PhysicalProblem #problems#PhysicsProblems #PhysicalProblems
#Physics #problem #PhysicsProblem #PhysicalProblem #problems#PhysicsProblems #PhysicalProblems
Center of Mass
Find the y-coordinate of the center of mass of a 10-cm square plate whose density (in kg/m³) is ρ = 4.22⋅10⁴kg/m⁶⋅xy². The lower left-hand side of the plate is at the origin and the upper right-hand corner is at (10 cm, 10 cm). The plate is 0.30 cm thick.
(∫y⋅y²dy) / (∫y²dy) = ¾y = ¾⋅10cm = 7.5cm
Horizontal Uniform Rod
A horizontal uniform rod has the mass M. The point masses m₁ and m₂ are located on the left end and the right end of the rod. In what distance from the left end has to be located a support to keep the rod's equilibrium?
#rod #mass #pointmass #left #end #right #distance #leftend #support#equilibrium #Physics #problem #PhysicsProblem #PhysicalProblem#problems #PhysicsProblems #PhysicalProblems
#rod #mass #pointmass #left #end #right #distance #leftend #support#equilibrium #Physics #problem #PhysicsProblem #PhysicalProblem#problems #PhysicsProblems #PhysicalProblems
Support
A thin 2-dimensional object with uniform mass density is in the shape of a square and has a mass M. The sides of the square are L. The coordinate system is drawn with the origin at the center of the square. Point masses m₁, m₂, m₃, m₄ are placed at the corners of the square object and the assembly is held in equilibrium by positioning the support S at point (x,y) as shown in the drawing. Find the location of the support.
Pendulum
On the surface of some planet, a pendulum with the length L oscillates with the period T. What is the period of the oscillation of the pendulum with the length L₂ there?
Monday, April 25, 2016
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