Mr M's OCR (not OCR MEI) M2 answers June 2011

You had a completely horizontal reaction force from the cylinder and a reaction force from the cone, which had an upwards component. So the only that upward compinent of that reaction force could have contributed to the the equilibrium in the vertical plane, as the other horizontal reaction force from the cylinder and the horizontal component from the conical shell do not act in the vertical plane, so they must provide the centripetal force required. The reaction force does not have to equal the weight. That is only if you only have a reaction force and a weight on a horizontal surface. If there is circular motion, the centripetal force is provided by the reaction force (or any other force that could provide it e.g. friction, gravity etc) and the reaction force provides what is necessary.

Sorry to keep bothering you chaps, I quite agree with what you say and still can't see the problem, I'll go through my working, since the cylinder provides only a horizontal force all the vertical force is from the reaction which means the normal contact force must be what it was in the previous question (where we got 11.9 rad s^-1). Hence F(c) + Rcos(x) = 6.615 and so F(c) = 6.615 - Rcos(x) which must be smaller than 6.615 since x was between 0 and 90.

(edited 12 years ago)

Reply 21

AmroTT

Original post by Mr M

Well the answer isn't right. That method would work if executed correctly.

Yes, I've just worked it through again and have no idea how I got that. However, in that question there's probably 2 marks available for the correct method, so I'll keep my fingers crossed for those at least :rolleyes:

Other than that, high 50s to low 60s. So maybe high B, low A. Which'd suit me down to the ground :smile:

OP

Original post by wibletg

Can you have a look at 10i, I got 327N but I can see why you got 206N

Also, the reaction force from the cylinder, surely that's smaller than the figure you got?

You mean 7i.

Nothing to check there.

F cos 60 x 1.6 = 550 cos 60 x 0.6

F = 550 x 0.6 / 1.6

Reply 23

wibletg

3

Original post by Mr M

You mean 7i.

Nothing to check there.

F cos 60 x 1.6 = 550 cos 60 x 0.6

F = 550 x 0.6 / 1.6

Ah damn, I did this too then scribbled it out and replaced the weight with sin :frown:

I got 1.52N for the curved cylinder question, by the way.

Reply 24

nikita_atikin

14

Original post by Mr M

You mean 7i.

Nothing to check there.

F cos 60 x 1.6 = 550 cos 60 x 0.6

F = 550 x 0.6 / 1.6

Mr M could you run past 1i please? I calculated the KEs and the PE and the work done against resistance but how did you add/subtract them up to get such a massive answer? I got 344 J but a lot of people got 85J or something.
Also 1ii was right but surely it depended on what you got in part i so is that marks lost too?

Reply 25

pi+e=pie

Hi Mr M, I just have two questions

From 3i it looks like I can't even get the area of an equilateril triangle right. Aside from just getting the triangle's area wrong, how many marks do you think I would have lost? and carrying on from that, I obviously used the wrong length in ii but do you think ecf is likely (for full 2marks?)

and on 5i i left the equation with 'cos30' and 'tan30' in without actually writing rt3 / 2 etc. how many marks dropped there do you think ? it didn't ask for a simplified equation though so i'm not sure myself

Thanks, again another amazing job with your answers well done :smile:

(edited 12 years ago)

Reply 26

nikita_atikin

14

Original post by wibletg

Ah damn, I did this too then scribbled it out and replaced the weight with sin :frown:

I got 1.52N for the curved cylinder question, by the way.

I think I got 1.52N as well.

6. (ii) 1.52 N ?

OP

Original post by Jodin

Sorry to keep bothering you chaps, I quite agree with what you say and still can't see the problem, I'll go through my working, since the cylinder provides only a horizontal force all the vertical force is from the reaction which means the normal contact force must be what it was in the previous question (where we got 11.9 rad s^-1). Hence F(c) + Rcos(x) = 6.615 and so F(c) = 6.615 - Rcos(x) which must be smaller than 6.615 since x was between 0 and 90.

Call the oblique contact force R1 and the horizontal contact force R2.

Vertically:

R_1 \sin 30 = 2.94

R_1 = 5.88

Horizontally:

R_2 + R_1 \cos 30 = \frac{0.3 \times 2.1^2}{0.2}

R_2 = 1.52277 ...

Angle between R1 and R2 is 150 degs so use cosine rule to find the magnitude of the resultant.

Edit: I have worked out the resultant of the two contact forces. The question asks for the CYLINDRICAL contact force only. And I tell students to read the question properly ...!

(edited 12 years ago)

Reply 29

Ranjeet

F=0.3x2.1^2/0.2-0.6gxsin60 = 1.52 N
This is 6 (ii)

(edited 12 years ago)

OP

Original post by Ranjeet

6. (ii) 1.52 N ?

Oh I see I have read the question wrong. We are supposed to be just finding one of the two contact forces!

(edited 12 years ago)

Reply 31

Jodin

2

Oh right, I thought they only wanted the force from the vertical cylinder i.e. the 1.52 N, darn it, ought to have read the question better I guess, a mark or two lost then I suppose, thank you very much Mr M.

(edited 12 years ago)

Reply 32

dknt

12

Original post by Mr M

x

Didn't the question ask for the force which the cylindrical shell exerts on the particle, not the conical shell as well? Making it only 1.52277... N? Or did I misread?

(edited 12 years ago)

OP

Original post by dknt

Does the question not ask for the force which the cylindrical shell exerts on the particle, not the conical shell as well? Making it only 1.52277... N? Or did I misread?

Nope I misread. See above.

OP

Original post by Jodin

Oh right, I thought they only wanted the force from the vertical cylinder i.e. the 1.52 N, darn it, ought to have read the question better I guess, a mark or two lost then I suppose, thank you very much Mr M.

Oh heck, this is confusing for everyone. It is cylindrical contact only. Sorry for the mistake. My imagined version of the question was better in my opinion though!

Reply 35

dknt

12

Original post by Mr M

Nope I misread. See above.

Ah, thanks a bunch! :smile:

OP

Original post by pi+e=pie

Hi Mr M, I just have two questions

From 3i it looks like I can't even get the area of an equilateril triangle right. Aside from just getting the triangle's area wrong, how many marks do you think I would have lost? and carrying on from that, I obviously used the wrong length in ii but do you think ecf is likely (for full 2marks?)

and on 5i i left the equation with 'cos30' and 'tan30' in without actually writing rt3 / 2 etc. how many marks dropped there do you think ? it didn't ask for a simplified equation though so i'm not sure myself

Thanks, again another amazing job with your answers well done :smile:

Leaving the trig in will be Ok.

You will just lose some accuracy marks in 3 but I really don't know how many.

Reply 37

nikita_atikin

14

Original post by Mr M

Nope I misread. See above.

Oh thank you Mr M :smile:

That's reassuring slightly because it would have been way more marks if we had components.
Sorry, also, could run through how you add up/subtract the energies in Q1i please? I'm unsure as to how you got such a high answer.

OP

Original post by nikita_atikin

Oh thank you Mr M :smile:

That's reassuring slightly because it would have been way more marks if we had components.
Sorry, also, could run through how you add up/subtract the energies in Q1i please? I'm unsure as to how you got such a high answer.

yes but I can't hang around too long, have work to do

Diff in K.E. = 0.5 x 70 x (2.1^2 - 1.4^2) = 85.75 J

Diff in P.E. = 3 x 70 x 9.8 = 2058 J

Add together.

Reply 39

pi+e=pie

Original post by Mr M

Leaving the trig in will be Ok.

You will just lose some accuracy marks in 3 but I really don't know how many.