AQA PHYA4 ~ 16th January 2013 ~ A2 Physics

Original post by UnvesedSplash

Can anyone help on this question please i have absolutely no idea of how they get b.

Thanks

Jan 2003.JPG

pi/2 x 6 = 3pi

so you know it goes around 3pi... think about it thats 1 and a half turn so it would be opposite to point P & they are asking for DISPLACEMENT

Double the radius :smile:

Reply 382

amish123

Original post by posthumus

Thank you very much ! :biggrin:

& yes that's a much better way of thinking about it... when I think satellites I immediately think of the T^2 equation :tongue:

Do you have any ideas on the 2nd question ? :smile:

its just the Vp & Vq stuff I don't get

Right, so we have relative motion between the rod and the magnetic field which will induce an emf, and therefore a current, in the wire. Thus, we must use the RHR for this. We are talking about conventional current so using the RHR, conventional current is from P to Q. As electrons flow in the opposite direction to this, the force of the electrons is from Q to P. Now because there is more electrons at P than Q, and emf is going to be induced because one end is now positive and the other is negative. If there are more electrons are P, Q is going to have a higher potential. That leaves you with D as your answer :smile:

(edited 11 years ago)

Reply 383

Original post by amish123

Right, so we have relative motion between the rod and the magnetic field which will induce an emf, and therefore a current, in the wire. Thus, we must use the RHR for this. We are talking about conventional current so using the RHR, conventional current is from P to Q. As electrons flow in the opposite direction, the force of the electrons is from Q to P therefore. Now because there is more electrons at P than Q, and emf is going to be induced. If there are more electrons are P, Q is going to have a higher potential. That leaves you with D as your answer :smile:

I'm still confused about the potential bit :frown:

Can't seem to find an equation that works
I mean V is proportional to Q?
So according to that shouldn't P, if it has the most charges have the highest voltage also... this is the bit I'm confused with.

Reply 384

Just wondering if anyone can help me on this question, its from the Nelson Thornes textbook, topic 2.4:

"The wheel of the London Eye has a diameter of 130m and takes 30minutes to complete 1 revolution. Calculate the change due to rotation of the wheel of the support force on a person of weight 500N in a capsule at the top of the wheel"

The answer is supposed to be -0.04N
Thanks :smile:

Reply 385

amish123

Original post by posthumus

I'm still confused about the potential bit :frown:

Can't seem to find an equation that works
I mean V is proportional to Q?
So according to that shouldn't P, if it has the most charges have the highest voltage also... this is the bit I'm confused with.

At the ends of the wire, in order for charges to flow, there must be different potentials at each end. The level of potential at any point depends on the deficiency of electrons at the point. So if Q has a lower amount of electrons, it has a higher potential compared to the other end P.

I will try and find an equation way to show this.

Reply 386

amish123

Original post by adam17

Just wondering if anyone can help me on this question, its from the Nelson Thornes textbook, topic 2.4:

"The wheel of the London Eye has a diameter of 130m and takes 30minutes to complete 1 revolution. Calculate the change due to rotation of the wheel of the support force on a person of weight 500N in a capsule at the top of the wheel"

The answer is supposed to be -0.04N
Thanks :smile:

Okay, the centripetal force on the person inside the capsule is m(w^2)r yeah? At the top, the equation for resolving F=ma would be mg-S=m(w^2)r where S is the support force. We are asked to find the support force so re-arrange this to get S=mg-m(w^2)r.

w can be worked out using w=2pi x f which is the same as w=2pi/T and in this case T= 30 x 60 (remember T is always in seconds).

mg is given in the question as the weight of the person, 500N. Therefore m=500/9.81.

r is half the diameter so 65m.

Plug in those values to find S and you should get 499.959...N

To find the change of the support force, takeaway the weight of the person from the value of S and you'll get -0.04N.

Hope that helps :smile:

From this you can work out m which is 500/9.81.

Reply 387

Original post by adam17

Just wondering if anyone can help me on this question, its from the Nelson Thornes textbook, topic 2.4:

"The wheel of the London Eye has a diameter of 130m and takes 30minutes to complete 1 revolution. Calculate the change due to rotation of the wheel of the support force on a person of weight 500N in a capsule at the top of the wheel"

The answer is supposed to be -0.04N
Thanks :smile:

Use F=ma

m= 500/9.8

a=(2pi x f)^2 x r

f = 1/30 x 60 ( mins x 60 = seconds )
r = 130/2 ( diameter/ 2 = r )

Reply 388

Original post by amish123

At the ends of the wire, in order for charges to flow, there must be different potentials at each end. The level of potential at any point depends on the deficiency of electrons at the point. So if Q has a lower amount of electrons, it has a higher potential compared to the other end P.

I will try and find an equation way to show this.

Oh I get it ! If electrons are on one side there is negative charges there so potential becomes negative ??? and on the other side there will be positive charges so... potential is positive ?

Reply 389

Original post by posthumus

Use F=ma

m= 500/9.8

a=(2pi x f)^2 x r

f = 1/30 x 60 ( mins x 60 = seconds )
r = 130/2 ( diameter/ 2 = r )

That seems so easy now that you've said it, I was trying to do the whole
S + mg = mv^2/r at the top thing etc etc...
Thank you!

Reply 390

Original post by adam17

That seems so easy now that you've said it, I was trying to do the whole
S + mg = mv^2/r at the top thing etc etc...
Thank you!

Its still good to use that :smile:

The magnitude of the centrifugal force would be the same as the centripetal force, so I guess its easier just easier to think of it as F=ma

EDIT: Oh right yh, I thought initially that they wanted the tension too, then I read they said supportive force :tongue:

(edited 11 years ago)

Reply 391

Original post by amish123

Okay, the centripetal force on the person inside the capsule is m(w^2)r yeah? At the top, the equation for resolving F=ma would be mg-S=m(w^2)r where S is the support force. We are asked to find the support force so re-arrange this to get S=mg-m(w^2)r.

w can be worked out using w=2pi x f which is the same as w=2pi/T and in this case T= 30 x 60 (remember T is always in seconds).

mg is given in the question as the weight of the person, 500N. Therefore m=500/9.81.

r is half the diameter so 65m.

Plug in those values to find S and you should get 499.959...N

To find the change of the support force, takeaway the weight of the person from the value of S and you'll get -0.04N.

Hope that helps :smile:

From this you can work out m which is 500/9.81.

With the risk of looking like a complete idiot: why is the equation mg-S= mw^2r at the top?
I thought at the top mg+s=mw^2r because both the weight and the support force are acting in the same direction, which is downwards..?

Reply 392

Original post by adam17

With the risk of looking like a complete idiot: why is the equation mg-S= mw^2r at the top?
I thought at the top mg+s=mw^2r because both the weight and the support force are acting in the same direction, which is downwards..?

Supportive force ! :tongue:

Not tension, I initially thought that too :smile:

Reply 393

Original post by posthumus

Supportive force ! :tongue:

Not tension, I initially thought that too :smile:

haha hmmmmm, with the risk of looking like a complete idiot again, what exactly is the supportive force, if it isn't the same as the tension? I feel stupid now

Reply 394

UnvesedSplash

Original post by posthumus

pi/2 x 6 = 3pi

so you know it goes around 3pi... think about it thats 1 and a half turn so it would be opposite to point P & they are asking for DISPLACEMENT

Double the radius :smile:

legend i keep getting half way on the questions then not know what to do :s-smilie:

Reply 395

Original post by adam17

haha hmmmmm, with the risk of looking like a complete idiot again, what exactly is the supportive force, if it isn't the same as the tension? I feel stupid now

no its okay :tongue:

Supportive force is erm the same as the reaction force :smile:

which I'm sure you've come across...

As you know if I have something on a table, the table has a reaction force which is the same as the force of the object (F=ma), This reaction force is Supporting the object... therefore its also called the supportive force :smile:

hope that helps.

Reply 396

Original post by UnvesedSplash

legend i keep getting half way on the questions then not know what to do :s-smilie:

I didn't have a clue about this question either when I did it, PinkHead & Amish123 helped me with it :P

honestly you have to just read the question carefully in the exam (which I'm working on myself) I would suggest you underline the main points in the question in the exam :smile:

Reply 397