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# PHY4 Jun 05 Thread :( watch

1. (Original post by kevla)
Doesnt the centripetal force act towards the centre of the Earth? Also the weight acts towards the centre of the Earth? So the force registered on a balance surly would be the weight plus the centripetal force? If I'm wrong please someone explain (nicely!) because it seems to make sense to me!
im not sure myself now... but i think u r right
nooo...lost a couple of marks...so far i know i got 5 marks wrong now!..that is alot in a 40 mark paper!
2. (Original post by kevla)
Im sure its F=mg +Centripetal force ....but i could be wrong!
3. (Original post by kevla)
Doesnt the centripetal force act towards the centre of the Earth? Also the weight acts towards the centre of the Earth? So the force registered on a balance surly would be the weight plus the centripetal force? If I'm wrong please someone explain (nicely!) because it seems to make sense to me!
come to think of it...wasnt the centrepetal force very small like somethin like 10tothe -5....to 3 significant figures the balance would still be mg right?
4. (Original post by DmAsTeR)
come to think of it...wasnt the centrepetal force very small like somethin like 10tothe -5....to 3 significant figures the balance would still be mg right?
I got a value of the centripetal force to be 2.03N

First you had to work out the angular velocity:
T=(24)(60)(60) =86,400s
ω=2π÷T
ω=7.27x10-5

Next use that value to work out the Centripetal force:

F=mrω2
F=(60)(6400x103 )(7.27x10-5 )2
F=2.03N
5. (Original post by kevla)
I got a value of the centripetal force to be 2.03N

First you had to work out the angular velocity:
T=(24)(60)(60) =86,400s
ω=2π÷T
ω=7.27x10-5

Next use that value to work out the Centripetal force:

F=mrω2
F=(60)(6400x103 )(7.27x10-5 )2
F=2.03N
ok i see.. i never added centrepetal force... i think i just did mg...
6. (Original post by DmAsTeR)
ok i see.. i never added centrepetal force... i think i just did mg...
think you were supposed to take them away, because teh total centrapetal force is the sum of teh mg and cirular motion

i.e. F=mg-R,

but i didn't do that, i added them - shuold et 3/4 marks though

phil
7. Yeah, the force keeping the girl in circular motion (the 2.03N bit) is given by mg - F, where F is the force that the scales show.
8. (Original post by Spenceman_)
Yeah, the force keeping the girl in circular motion (the 2.03N bit) is given by mg - F, where F is the force that the scales show.

meaning i actually got that right?

was convinced there with all the confusions that i messed it
woohee
me happy
9. (Original post by Steve8052)
hey ppl was the wave for that 10^-6 IR or microwave cos i looked in another txtbk and 10^-6 happened to be the boundary for IR and microwave. Do u think mrks will be given 4 both? I put dwn microwave personally but i cudnt fink of a source and still cant. I just said microwave emitter LOL
My text book has the wavelength of microwave as 10cm and the wavelength of IR as 1 µm (>700nm). I can't remember the value in the question but if it is 106 it has to be IR, right?

Btw, thanks for explaining the Recission velocity thing.
10. (Original post by sumitk87)
λ was 390nm
deltaλ was somin like 300nm or somin (range of visible spectrum)
λ was at edge of visible (near UV ) and it could increase to about 700nm and still be visible so deltaλi used as 700nm-390nm or somin like that
i got about 2x10^(i think 7 or somin) as my velocity
not sure cos it seems very fast
anyone kno if that is a reasable value or is it toooo high ?
I think I used 700-390/700 !! I'm loosing more marks !!!!!
11. (Original post by hajira)

meaning i actually got that right?

was convinced there with all the confusions that i messed it
woohee
me happy
whats this? is adding them right? me and taylor had a long concersation about this lol and he sort of convinced me taking away was right, but i still cant get my head around how that quetsion aws phrased- "reading on scales" - at the time, i thought adding them was the logical thing to do, but i think its wrong now - damn taylor
12. (Original post by Phil23)
whats this? is adding them right? me and taylor had a long concersation about this lol and he sort of convinced me taking away was right, but i still cant get my head around how that quetsion aws phrased- "reading on scales" - at the time, i thought adding them was the logical thing to do, but i think its wrong now - damn taylor
lol..going back to all the answers on the thread, im totally unsure
what i did though was take away the centripetal force from mg to get the 'reading on the scale' not sure watsoever still
its just 3 marks..out of 60 - thats a meager 5%
i refuse to ponder over it any further..got more exams to worry about!
13. (Original post by Phil23)
whats this? is adding them right? me and taylor had a long concersation about this lol and he sort of convinced me taking away was right, but i still cant get my head around how that quetsion aws phrased- "reading on scales" - at the time, i thought adding them was the logical thing to do, but i think its wrong now - damn taylor
to b honest edexcel hav once again confused intelligent ppl (such as urself as i gather) with ridiculous question phrasing

damn them

i took it away
14. hmm....
so her downward force is her weight only
i just done mg - i thought it was a simple question , basically askin whats her weight

Edit: how can her downward force be greater than her weight - centripetal is a resultant force and all i kno is if you put 1kg on a newton meter it will say 9.8N (mg) -very confused now-

what did u guys get for proving the value of g using the pendulum , the one where it gave you 2 periods one with length L and one with length L+1??
15. (Original post by sumitk87)
hmm....
so her downward force is her weight only
i just done mg - i thought it was a simple question , basically askin whats her weight

what did u guys get for proving the value of g using the pendulum , the one where it gave you 2 periods one with length L and one with length L+1??
i was tempted to do that..but seemed too trivial for 3 marks

as for the 'g' value..i used the eqn T=2pi √(l/g)
for l and (l-1) equating to the given values, n solving simultaneously
don remember the figures

Edit: o was it L+1..then yeah that
16. The question with the weight was 3 marks!!!! so i added centripetal force
17. if someone has a mass of 50kg and you put him on a newton meter what will it read ?

thats how i thought of it

1 mark for units
1 mark for rounding (sensible accuracy)
1 mark for working out and use of equation (w=mg)

seems unlikely but i hope its like that
18. (Original post by sumitk87)
if someone has a mass of 50kg and you put him on a newton meter what will it read ?

thats how i thought of it

1 mark for units
1 mark for rounding (sensible accuracy)
1 mark for working out and use of equation (w=mg)

seems unlikely but i hope its like that
Does any one remember doing any lift questions in class - its the same sort of thing.

Your standing on some scales in a lift. If the lift is going down the reading on the scales will drop. If the lift is going up the reading on the scales will go up.

(sum)F=ma.

Lift going down acceleration is down. mg-N=ma
so N (reading on the scales) = mg-ma

Lift going up acceleration is up. N-mg=ma
So N (reafing on the scales) = ma+mg

When it comes to woman on the earth spinning it like going down in a lift. Acceleration is downwards.(centripital is towards centre of circle) Here weight is downwards her normal reaction is in the opposite direction

mg-N=ma mg-ma=N

mg is 60*9.81
N is the reading on the scales - we want to work it out.
ma=m.w^2.r=2 (worked out in question above.)

If the earth went round faster scales reading would drop.
If the girl was in a lift going down, the fast the lift went down the more the reading on the scales would drop.
19. Yeah. Basically you just have to first realise that the scales can only measure the force which THEY EXERT...they can't magically measure the mass of the person above them and then mutliply it by 9.81...
20. (Original post by RMIM)
Does any one remember doing any lift questions in class - its the same sort of thing.

Your standing on some scales in a lift. If the lift is going down the reading on the scales will drop. If the lift is going up the reading on the scales will go up.

(sum)F=ma.

Lift going down acceleration is down. mg-N=ma
so N (reading on the scales) = mg-ma

Lift going up acceleration is up. N-mg=ma
So N (reafing on the scales) = ma+mg

When it comes to woman on the earth spinning it like going down in a lift. Acceleration is downwards.(centripital is towards centre of circle) Here weight is downwards her normal reaction is in the opposite direction

mg-N=ma mg-ma=N

mg is 60*9.81
N is the reading on the scales - we want to work it out.
ma=m.w^2.r=2 (worked out in question above.)

If the earth went round faster scales reading would drop.
If the girl was in a lift going down, the fast the lift went down the more the reading on the scales would drop.
I completely agree

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