Rod's Physics 3/4 Questions Thread

[Rod]

^^


- Another question going all the way back to the start of the year. If I'm on a car and move around a round about I would need friction to provide a centripetal force so I can actually turn instead of moving off the tracks. For banked turns, how come there isn't really a reliance on friction? Apparently it's got something to do with the normal force providing the centrepetal force, I don't fully understand it though, would be nice if someone could please give me some pointers

[lzxnl]

^^


- Another question going all the way back to the start of the year. If I'm on a car and move around a round about I would need friction to provide a centripetal force so I can actually turn instead of moving off the tracks. For banked turns, how come there isn't really a reliance on friction? Apparently it's got something to do with the normal force providing the centrepetal force, I don't fully understand it though, would be nice if someone could please give me some pointers

Look at the direction of the normal force. It's not vertical; it's slanting with some horizontal component. What other forces are there? Well, just gravity really, which acts down. The vertical part of the normal force must balance with gravity and the horizontal part of the normal force must therefore provide the centripetal acceleration.

[Rod]

Look at the direction of the normal force. It's not vertical; it's slanting with some horizontal component. What other forces are there? Well, just gravity really, which acts down. The vertical part of the normal force must balance with gravity and the horizontal part of the normal force must therefore provide the centripetal acceleration.

thanks man

And was just wondering if you had ever used the A+ note study guide for physics. Did you find the exam questions useful? Were the far off from VCAA level?

[Yacoubb]

thanks man

And was just wondering if you had ever used the A+ note study guide for physics. Did you find the exam questions useful? Were the far off from VCAA level?

Hey Rod

I've personally attempted some of the A+ study guide questions for Physics. A lot of them are great to start you off, but you do want to get your hands on some more nitty-gritty questions that like to throw students into pitfalls, just to familiarise yourself with such tricks!

[lzxnl]

thanks man

And was just wondering if you had ever used the A+ note study guide for physics. Did you find the exam questions useful? Were the far off from VCAA level?

I've only used an English Language study guide xP I'm really not the person to ask about these

[Rod]

Hey Rod

I've personally attempted some of the A+ study guide questions for Physics. A lot of them are great to start you off, but you do want to get your hands on some more nitty-gritty questions that like to throw students into pitfalls, just to familiarise yourself with such tricks!

Thanks Yacoubb

*throws away A+ notes, gets out checkpoints*

hehe xD

[Yacoubb]

Thanks Yacoubb

*throws away A+ notes, gets out checkpoints*

hehe xD

Haha, yeah I really like the Physics Checkpoints. Definitely be going through them over the two week break! I think they've got great questions. Just a shame that there aren't 'great' worked solutions to the questions. Also, some of the explain-type questions for VCAA questions aren't the same as those in the VCAA assessment reports,which is kinda annoying!!
 

[Rod]

On the surface of a planet, 20 J of work is done in lifting a mass of 1 kg through a height of 5m. What is the magnitude of gravitational field strength at the surface of the planet?

Since work = gravitational potential energy

Then 20 = 1xgx5

Hence 20/5 g = 4 N Nkg

If the radius of the planet referred in the question before is 1.0 x 10^4, what would be the graviational field strength 1.0x 10^4 km above the surface??

??

g = GM/r2

4= 6.67x10^11 X M/5^2

M = 1.5x10^-11

Using this mass the gravitational field strength turns out to be 9x10^-24 (LOL)

Where am I going wrong for part 2?

[lzxnl]

On the surface of a planet, 20 J of work is done in lifting a mass of 1 kg through a height of 5m. What is the magnitude of gravitational field strength at the surface of the planet?

Since work = gravitational potential energy

Then 20 = 1xgx5

Hence 20/5 g = 4 N Nkg

If the radius of the planet referred in the question before is 1.0 x 10^4, what would be the graviational field strength 1.0x 10^4 km above the surface??

??

g = GM/r2

4= 6.67x10^11 X M/5^2

M = 1.5x10^-11

Using this mass the gravitational field strength turns out to be 9x10^-24 (LOL)

Where am I going wrong for part 2?

Yeah you've messed up here. For starters, you shouldh ave 4 = 6.67x10^-11 M/(1x10^7)^2 (10^7 if your 10^4 is in km)

Then, if you know the gravitational field strength is 4 N/kg at the surface at a distance of 1x10^4 km from the centre, a further 1x10^4 km means you've doubled the distance from the centre. Hence the gravitational field strength should go down by a factor of four. Inverse square relationship?

[Rod]

Yeah you've messed up here. For starters, you shouldh ave 4 = 6.67x10^-11 M/(1x10^7)^2 (10^7 if your 10^4 is in km)

Then, if you know the gravitational field strength is 4 N/kg at the surface at a distance of 1x10^4 km from the centre, a further 1x10^4 km means you've doubled the distance from the centre. Hence the gravitational field strength should go down by a factor of four. Inverse square relationship?

Yeah sorry I did stuff up big time haha

Thanks!

PS - Saw your UNI results congrats man they were epic !

[Rod]

Why is the weight of an object considered to also be the graviational force at surface of the object due to the Earth?

[lzxnl]

Why is the weight of an object considered to also be the graviational force at surface of the object due to the Earth?

Depends on where the object is. If the object is at the surface of the Earth, its weight is the gravitational force on it by the Earth at the surface. However, weight forces can be defined for objects in space as well.

[Rod]

Some of this stuff is really doing my head at the moment

http://www.vcaa.vic.edu.au/Documents/exams/physics/2008physics1-web-copyrgt.pdf

Question 15

I get the first mark, I get how the energy stays the same throughout the orbit

Although, if the meteor is closer to the sun, gravitational potential energy will be higher. If GPE is higher wouldn't the kinetik energy decrease? So less speed at X than y?

thanks

[Rod]

For two marks:
1 mark - The speed of the comet decreases from a maximum value at X to a minimum value at Y.
1 mark - Its total energy remains constant around its orbit.

VCE logic: At X the kinetic energy is maximum and it's gravitational potential energy (GPE) is a minimum. It's closest to the "source" of gravity (the sun) at this point.
Hence at Y since it's the furthest away from the "source" of gravity (the sun) its GPE must be at its maximum and hence its kinetic energy minimum.

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.

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!

[Rod]

A very valid question.

Sure but that gravitational field strength is a tiny amount compared to the distances we're looking at between planets (or a comet and a star in this case).

In VCE you're taught GPE = mgh. Let's say we're looking at the Earth and taking the core of the Earth as our "center" point or "source".

Arguably if we're standing 1m above the Earth's core somehow and your mass is 70 kg. g = 10, m = 70, h = 1 so GPE = 70 * 10 * 1 = 700 J
Let's say we move 6371 km above the core (i.e. on the surface of the earth). Now g will decrease to what we know g to be; 9.8. g = 9.8, m = 70, h = 6371*10^3 so GPE = 70 * 9.7 * 5000*10^3 = 4.3 billion J

4.3 billion >>>>>>>>>>>>> 700

So even though our gravitational field strength is weaker, the huge height we've had to move still makes our GPE increase overall (by a huge amount)

The figures I used above are pretty accurate and non fictional, you can Google all of those yourself if you'd like.

Reiterating:
- Yes, the gravitational field strength does get weaker as you move away from the "source" or "center" but the distances increased are immensely larger than the decrease in gravitational field strength. Which results in an overall huge increase in GPE.

Ahh, awesome explanation Basic Acid. Thank you so much, I fully understand it now !