Cort's 3/4 Physics Thread

[Cort]

Thanks mate.

I'm now touching on Circular Motion. I'm somewhat confuzzled when it comes to the Loop-de-loop ones; especially when it asks about the 'minimum' speed for car to go around at the top. Is there a reason why you would cancel out or why the reaction force (N) become zero in this instance? It's drilling my head in. Thanks.

I've did some reading and it's something related to being 'minimum' so that it equals the g = 9.81ms^-2 or something. Something like that, right?

Edit 2: Right, it has to do with the speed being equal or greater than 9.81 itself to keep it from falling down. If it's smaller than that, gravity will cause the car to fall down. The problem is..how does THAT relate to the normal force at all?

[PB]

The reason why the normal even exists at the top is because the car is pressing against the rails, and hence, the rails presses back (creating that normal force on the car).

Therefore, the fact that there is zero normal force indicates that the car is not pressing against the rails at all - it is at the verge of falling off. And if it goes any slower, it WILL fall off.

[Cort]

The reason why the normal even exists at the top is because the car is pressing against the rails, and hence, the rails presses back (creating that normal force on the car).

Therefore, the fact that there is zero normal force indicates that the car is not pressing against the rails at all - it is at the verge of falling off. And if it goes any slower, it WILL fall off.

Right, that got some of my concerns off --but WHY on earth would the normal force be zero? I'm quite daft here.

Right, I really am stupid. So it's apparent that when it comes to 'minimum/least speed' it means the verge of it falling off as you said. Bah. That means that Fn=0.

[PB]

precisely.

[Cort]

Gravity confuzzles me.

This is from checkpoints

A Small satellite orbits Mars. It has a kinetic energy of 3.0 * 10^10J, and it is at a constant distsancen of 8.0^10^7from the centre of Mars. What is the weight of the satellite at this height?

The furthest I got to was..
W =m*g = m*v^2/r = ??

[lzxnl]

Gravity confuzzles me.

This is from checkpoints

A Small satellite orbits Mars. It has a kinetic energy of 3.0 * 10^10J, and it is at a constant distsancen of 8.0^10^7from the centre of Mars. What is the weight of the satellite at this height?

The furthest I got to was..
W =m*g = m*v^2/r = ??

mv^2/r = GmM/r^2
1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius

[Cort]

mv^2/r = GmM/r^2
1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius

1/2 mv^2 = GmM/2r  -- how did you go from r^2 to 2r?

[Cort]

Alright..This is really confusing.

Calculate the mass of the Earth from the following data:
radius of moon's orbit= (3.8*10^8); G = 6.67*10^-11, Priod = 28 days
I've done T^2/R^3 = GM/4*pi^2 and it ain't working. Because I assumed that the moon would be a satellite here, hence the m's cancel out.

Similarly,
Calculate the altitude of a satellite in a geo-stationary orbit.
Data: G = 6.67*10^-11, Me = 6.4*10^24, Re = 6.5*10^6m.
Would T^2/R^3 be useful here? Is there any special properties when it comes to satellites in geo-stationary orbit?

[lzxnl]

1/2 mv^2 = GmM/2r  -- how did you go from r^2 to 2r?

Because the kinetic energy term doesn't have a "r" term in the denom but the force term does?

Alright..This is really confusing.

Calculate the mass of the Earth from the following data:
radius of moon's orbit= (3.8*10^8); G = 6.67*10^-11, Priod = 28 days
I've done T^2/R^3 = GM/4*pi^2 and it ain't working. Because I assumed that the moon would be a satellite here, hence the m's cancel out.

Similarly,
Calculate the altitude of a satellite in a geo-stationary orbit.
Data: G = 6.67*10^-11, Me = 6.4*10^24, Re = 6.5*10^6m.
Would T^2/R^3 be useful here? Is there any special properties when it comes to satellites in geo-stationary orbit?

The moon is a satellite, yes. Therefore the M in your first formula is the mass of the Earth.

Geostationary orbit => period = 86400s. That's it. Then, be careful about the wording "altitude", not radius.

[Cort]

Because the kinetic energy term doesn't have a "r" term in the denom but the force term does?

The moon is a satellite, yes. Therefore the M in your first formula is the mass of the Earth.

Geostationary orbit => period = 86400s. That's it. Then, be careful about the wording "altitude", not radius.

Thanks - but I'm afraid I'm still confused. Do you mind if you explained about how Ke is found? 1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius - still don't understand.

For geostationary orbit, do you always assume the period would be 1 day = 86400s?  So every time geostationary is mentioned, just whip down T= 1day? Is there a reason why we might use T =1 day = 86400s?

[lzxnl]

Thanks - but I'm afraid I'm still confused. Do you mind if you explained about how Ke is found? 1/2 mv^2 = GmM/2r = kinetic energy = 1/2 × weight force × radius - still don't understand.

For geostationary orbit, do you always assume the period would be 1 day = 86400s?  So every time geostationary is mentioned, just whip down T= 1day? Is there a reason why we might use T =1 day = 86400s?

I really can't explain it. I've written out all the explaining there is.

mv^2/r = GmM/r^2 => centripetal force requirement
Multiply both sides by r/2 to convert the left hand side to kinetic energy
mv^2/2 = GmM/2r

That's the definition of a geostationary orbit; same period as the Earth's rotation

[rhinwarr]

A Small satellite orbits Mars. It has a kinetic energy of 3.0 * 10^10J, and it is at a constant distsancen of 8.0^10^7from the centre of Mars. What is the weight of the satellite at this height?

w = mg = mv^2 / r      (because the centripetal acceleration is being caused by the weight force)
Ek = 1/2 mv^2
2Ek = mv^2
sub 2Ek into the 1st equation
w = 2Ek / r
w = 750N

For geostationary orbit, do you always assume the period would be 1 day = 86400s?

Geostationary orbit is where the satellite's period of orbit is equal to the period of rotation of the centre of mass. This is so the geostationary satellite stays above the same point on the centre of mass throughout its orbit.
Yes, the period will always be one day IF the centre of mass is the Earth, because the period of rotation of the Earth is one day.
However, if the centre of mass is something else, eg. Mars, then it will be different.

[Cort]

Out of curiousity and to consolidate - can you assume (although you never should!) that everytime it mentions a 'orbit circular path' you always go Fc = Fg? What about Kepler's Third Law? Does it apply in situations where it is an orbital circular path, although it was intended for ellipitical equations?


Thanks,
Cort.

[lzxnl]

Out of curiousity and to consolidate - can you assume (although you never should!) that everytime it mentions a 'orbit circular path' you always go Fc = Fg? What about Kepler's Third Law? Does it apply in situations where it is an orbital circular path, although it was intended for ellipitical equations?


Thanks,
Cort.

A circle IS an ellipse, just that both semi-major and semi-minor axes are equal.

And yes for your question on centripetal forces.

[jesterino]

Hi sorry to derail the thread, but I was hoping someone could help me with this question from the circular motion sections.

"In an adventure park ride, people are strapped in a small cage that is whirled in a vertical circle on cables. The cables make an angle of 30 with the vertical. The radius of the circle is 8.0m. The cage has a total of mass 250kg."
(From the previous question, net force was worked out to be 3125 N.)

Show that the tension in each of the cables at the top of the circle is close to 360 N.

So far I have drawn out a triangle with 30 at the top, T (tension) as the hypotenuse, and mg as the vertical component. I'm quite sure this is incorrect though

This question is killing me, I have no idea on how to reach the end result.