- #1
dustybray
- 9
- 0
Hi,
I’m having a really difficult time in physics, and I need some help on this homework assignment.
Here are the questions and the work I’ve done. I really don’t know how to approach them, and at this point, I’m just kinda throwing equations at them haphazardly…
5. A 175 kg weather surveillance satellite of is launched into orbit with an altitude of
250 km. The mass of the Earth is 5.97 x 1024 kg. The radius of the Earth is
6.38 x 103 km.
a. Calculate the tangential velocity required for this orbit.
mg = mv^2 / r
g = v^2 / r
v = sqrt(gr) = sqrt( 9.8m/s^2 * (6.38*10^6 m + 250000m) )
v = 8060.6m/s
b. Calculate the period of this orbit in minutes.
T^2 = (4PI^2 / GM[E]) r^3
T^2 = (4PI^2 / (6.67 * 10^-11) * 5.97 x 1024 kg) * (6.38 x 106 m + 250000m)^3
T = 3.2*10^28s/rev
3.2*10^28s/rev * (1min / 60s) = 5.33*10^26 min/rev ??
c. What is the weight of the satellite?
mg = G( Mm / r^2 ) I can’t use this formula, there are two m’s…
v^2 = G( m/r )
m = rv^2 / G
m = ( (6.38*10^6 m + 250000m) * (8060.6m/s)^2 ) / (6.67*10^-11)
This isn’t right…
9. In an Atwood machine, one block has a mass of 500g and the other a mass of 460g.
The pulley, which is mounted in horizontal frictionless bearings, has a radius of
5.0cm. When released from rest, the heavier block is observed to fall 75cm in 5.0s
with no slippage of the cord on the pulley.
d = v*t + (1/2)*a*t^2
d-v*t = (1/2)*a*t^2
a = 2*( d -v * t)/ t^2 )
a = 2*( ( .75m -0 * 5s) / (5s)^2 ) = .06m/s^2
a. Calculate the acceleration of the two blocks.
a[M] = -.06m/s^2
a[m] = .06m/s^2
b. Calculate the tension in the cord supporting the heavier block.
T[1] = mg+ma = (.46kg) (9.8m/s^2) + (.46kg) (.06m/s^2) = 4.54N
c. Calculate the tension in the cord supporting the lighter block. Note the answer to b and c is not the same.
T[2] – T[1] = Mg – mg – Ma – ma
T[2] – 4.54N = (.5kg) (9.8m/s^2) – (.46kg) * (9.8m/s^2) – (.5kg) * (.06m/s^2) – (.46kg) * (.06m/s^2)
T[2] – 4.54N = .33N
T[2] = 4.87N
d. Calculate the angular acceleration of the pulley.
α = a / r = .06m/s^2 / .05m = 1.2rad/s^2
e. Calculate the moment of inertia of the pulley.
T[2] – T[1] = I (a / r^2)
I = (T[2] – T[1]) / (a / r^2)
I = (.33N) / (.06m/s^2 / (.05m)^2) = .014 kg*m^2
Any help would be greatly appreciated.
Thanks,
dusty…….
I’m having a really difficult time in physics, and I need some help on this homework assignment.
Here are the questions and the work I’ve done. I really don’t know how to approach them, and at this point, I’m just kinda throwing equations at them haphazardly…
5. A 175 kg weather surveillance satellite of is launched into orbit with an altitude of
250 km. The mass of the Earth is 5.97 x 1024 kg. The radius of the Earth is
6.38 x 103 km.
a. Calculate the tangential velocity required for this orbit.
mg = mv^2 / r
g = v^2 / r
v = sqrt(gr) = sqrt( 9.8m/s^2 * (6.38*10^6 m + 250000m) )
v = 8060.6m/s
b. Calculate the period of this orbit in minutes.
T^2 = (4PI^2 / GM[E]) r^3
T^2 = (4PI^2 / (6.67 * 10^-11) * 5.97 x 1024 kg) * (6.38 x 106 m + 250000m)^3
T = 3.2*10^28s/rev
3.2*10^28s/rev * (1min / 60s) = 5.33*10^26 min/rev ??
c. What is the weight of the satellite?
mg = G( Mm / r^2 ) I can’t use this formula, there are two m’s…
v^2 = G( m/r )
m = rv^2 / G
m = ( (6.38*10^6 m + 250000m) * (8060.6m/s)^2 ) / (6.67*10^-11)
This isn’t right…
9. In an Atwood machine, one block has a mass of 500g and the other a mass of 460g.
The pulley, which is mounted in horizontal frictionless bearings, has a radius of
5.0cm. When released from rest, the heavier block is observed to fall 75cm in 5.0s
with no slippage of the cord on the pulley.
d = v*t + (1/2)*a*t^2
d-v*t = (1/2)*a*t^2
a = 2*( d -v * t)/ t^2 )
a = 2*( ( .75m -0 * 5s) / (5s)^2 ) = .06m/s^2
a. Calculate the acceleration of the two blocks.
a[M] = -.06m/s^2
a[m] = .06m/s^2
b. Calculate the tension in the cord supporting the heavier block.
T[1] = mg+ma = (.46kg) (9.8m/s^2) + (.46kg) (.06m/s^2) = 4.54N
c. Calculate the tension in the cord supporting the lighter block. Note the answer to b and c is not the same.
T[2] – T[1] = Mg – mg – Ma – ma
T[2] – 4.54N = (.5kg) (9.8m/s^2) – (.46kg) * (9.8m/s^2) – (.5kg) * (.06m/s^2) – (.46kg) * (.06m/s^2)
T[2] – 4.54N = .33N
T[2] = 4.87N
d. Calculate the angular acceleration of the pulley.
α = a / r = .06m/s^2 / .05m = 1.2rad/s^2
e. Calculate the moment of inertia of the pulley.
T[2] – T[1] = I (a / r^2)
I = (T[2] – T[1]) / (a / r^2)
I = (.33N) / (.06m/s^2 / (.05m)^2) = .014 kg*m^2
Any help would be greatly appreciated.
Thanks,
dusty…….