Rod's Physics 3/4 Questions Thread

[lzxnl]

Thanks so much

But I just don't understand your last bit.

'Think about how the closer we are to the earth the less GPE we have, while the further up we move into the sky (the further we move away from the "source" of gravity i.e. the earth) the more GPE we have.'

The closer we get to the sun, the more gravitational field strength? Isn't that correct? If the gravitational field strength is higher, then so would gravitational potential energy, hence kinetik energy will be at  minumum

Please correct my reasoning!

OK, I'm going to address your statement about increasing gravitational field strength.
It actually depends. In this case, the increasing gravitational field strength is associated with getting closer to the sun, and the force is directed towards the sun. Now imagine a spring. If you stretch the spring more, the restoring spring force directed inwards increases. However, obviously the potential energy increases. Hence stronger force isn't always more potential energy.

Now, potential energy is actually the negative integral of the force with respect to distance (again not mentioned in VCE because they can't tolerate having year 11 maths in year 12 physics for some reason). In another words, integrate -GMm/r^2 (negative as it's directed inwards) and put a negative sign in front to get -GmM/r
This decreases in value as r decreases. Insert reasoning here

[Rod]

OK, I'm going to address your statement about increasing gravitational field strength.
It actually depends. In this case, the increasing gravitational field strength is associated with getting closer to the sun, and the force is directed towards the sun. Now imagine a spring. If you stretch the spring more, the restoring spring force directed inwards increases. However, obviously the potential energy increases. Hence stronger force isn't always more potential energy.

Now, potential energy is actually the negative integral of the force with respect to distance (again not mentioned in VCE because they can't tolerate having year 11 maths in year 12 physics for some reason). In another words, integrate -GMm/r^2 (negative as it's directed inwards) and put a negative sign in front to get -GmM/r
This decreases in value as r decreases. Insert reasoning here

Thanks Lxnl . So just for the scope of the physics course, when finding out how much gravitational potential energy an orbiting object has around a mass such as the sun, only look at the distance (gp=mgH) and not the gravitational field strength for all your reasons. Then, by analysing whether gpe is gained or lost, we can find what happens to the kinetik energy because Ek + Egp + Es = conserved/constant

Thanks everyone

[lzxnl]

Nope. mgh only works on the surface of the Earth. Above the surface of the Earth, the gravitational potential energy is given by -GmM/r. You'd have to use some reasoning on the lines of closer to the sun => pulled more strongly towards the sun, object moves faster with less potential energy.

Imagine just letting an object go 300 million km from the sun. It's going to accelerate towards the sun. As it gets closer and is more strong attracted, its kinetic energy increases. The gravitational force is conservative so the potential energy must decrease as the kinetic energy increases.

[Rod]

In VCE physics is it okay to always assume the normal reaction force = apparent weight?

[lzxnl]

Yes. Make sure you know which normal force

[Rod]

Yes. Make sure you know which normal force

One more thing, just been doing some questions and some of them assume when there is tension, there is no normal reaction force. Do you have an explanation for this?

[PB]

One more thing, just been doing some questions and some of them assume when there is tension, there is no normal reaction force. Do you have an explanation for this?

Hey Rod, could you please give an example of the question? I am not quite sure what your question is asking.

[Rod]

Hey Rod, could you please give an example of the question? I am not quite sure what your question is asking.

Sure PB!
http://www.vcaa.vic.edu.au/Documents/exams/physics/2012/2012physics1-w.pdf
Q 7/c

Just stuff I don't understand with the question:

1. WHy there is no normal reaction force acting on the ball
2. Why at the top of the vertical circle, it's tension + weight = Fnet. So why tension goes down at the top of the vertical circle.

For 2, is it just commons sense? The fact that since the ball is going up, tension would try and pull it back down, and hence tension will be going downwards?

thnx

[Rod]

Can someone please take a look at my working out and point out what I've done wrong:

http://www.vcaa.vic.edu.au/Documents/exams/physics/2009physics1-w.pdf
Q 10

Work out:
.Find time take for ball to go 45 metres up the cliff and then down again
x=90
t=?
v=0
u=30

x=(u+v)/2x2
t=6c

Therefor e it takes 9-6 = 3 seconds for the ball to go from base of the cliff to the water

x=vt-0.5at^2
x = 45 m

The actual answer is triple (135m)

thanks

[Conic]

When the ball reaches the edge of the cliff again it has a downward velocity of 30 m/s (same as initial upward speed as it must have the same kinetic energy as it did initially), a downward acceleration of 10 m/s^2, and it takes 3 seconds to reach the bottom of the cliff. Take downwards to be positive. We have

so the cliff is 135m high.

[PB]

Sure PB!
http://www.vcaa.vic.edu.au/Documents/exams/physics/2012/2012physics1-w.pdf
Q 7/c

Just stuff I don't understand with the question:

1. WHy there is no normal reaction force acting on the ball
2. Why at the top of the vertical circle, it's tension + weight = Fnet. So why tension goes down at the top of the vertical circle.

For 2, is it just commons sense? The fact that since the ball is going up, tension would try and pull it back down, and hence tension will be going downwards?

thnx

1. what is a normal force? It is a reaction force that acts on an object perpendicular to the surface of contact. Is the ball in contact with anything? (wall?floor?) Nope. It is on a string doing loopdeloops so no normal force in this case.

2. You kind of answered your own question lol.   Your formula is right. So if you rearrange for T, you get Fnet-W.    At the bottom of the circle, T= Fnet+W.    Fnet alllways stays the same if the ball is moving with constant speed around in a circle, and so does W...obviously. So tension is greater at the bottom.

[Rod]

thank you

[Rod]

Can someone please help me with these two multi choice:

Use the information below to answer questions 6 to 8:
Sam is pouring water into a PVC pipe (shown in the diagram below). At the mouth of the pipe is placed a
machine that produces sound at variable frequencies.

(DIAGRAM WAS JUST A SIMPLE OPEN TO CLOSED TUBE WITH WATER, THAT'S IT)
 

Question 7
Sam now sets the frequency at a constant 800 Hz. How far from the mouth of the pipe would you
expect the water level to be when there is an increase in volume?
 8.00 cm
 800 cm
 10.6 cm
 34 cm

Question 8
Which of the following statements is true?
 At the surface of the water there is a displacement node and a pressure node
 At the surface of the water there is a displacement node and a pressure antinode
 At the mouth of the pipe there is a displacement node and a pressure antinode
 At the mouth of the pipe there is a displacement antinode and a pressure antinode