Solutions to some of Oliver Murphy's exercises

setanta222

Looking for some help.
Does anyone have the method of solving (i) Q 13 in Exercise 7A and (ii) Q 5(b) in Exercise 8F of Oliver Murphy's LC text (Fundamentals of Applied Maths)

breadmonkey

You'd be better off posting the actual questions.

setanta222

setanta222 wrote: »

Looking for some help.
Does anyone have the method of solving (i) Q 13 in Exercise 7A and (ii) Q 5(b) in Exercise 8F of Oliver Murphy's LC text (Fundamentals of Applied Maths)

Q13 is in the area of Work/Energy/Momentum:

Two particles, of masses m and 3m, are connected by a light rigid rod. The system rests on a smooth horizontal table, the heavier mass due East of the lighter mass. The heavier mass is projected with initial velocity uj (vector j). Find the velocities of the particles when the rod runs North-South. (Note: When the rod runs North-South, the j-component of both velocities must be equal. Why?)

Q 5(b) is in the area of statics and moments:

A heavy uniform rod of mass m and length 2l is suspended from a point o by two equal inelastic strings. Each string is fixed to o and to the end points of the rod so that the rod hangs horizontally. If, then, a mass of m/2 is suspended half-way between the centre and one end of the rod, so rod is no longer horizontal, find T1 and T2, the tension in each string?

japester

Hi Setanta,
I've had a look at both questions. The first one I'm not 100% sure on. I do believe though that the principal of conservation of momentum comes into play here. I am thinking that this situation is a bit like a collision where applying a force to the larger mass and setting it off with an initial velocity of u in the j-direction (north) will automatically cause a force to be applied to the mass m, which is attached to it by the rigid rod. I am inclined to think (possibly wrong on this!!) that in the j-direction, the momentum before "impact" is equal to that after "impact" to give m(0) + 3m(u) = -mVSin(theta) + 3mVSin(theta). Just thinking about it intuitively, I would imagine that when the larger mass moves with an initial velocity of u in the j-direction, it will be inclined to rotate on the table in the negative i-direction as it travels further in the j-direction. Likewise, the smaller mass would travel in the positive i-direction and in the negative j-direction, with the angle of motion to the horizontal being theta in both cases, due to the fact that they ar tied by the rod. So, when the rod is in the north-south alignment, I would see the larger mass as being due north of the smaller mass and in this case theta would be 90 degrees. In this situation, 3mu = 2mV and I would say that the speed of the masses at this point is V = 3u/2. I would imagine that the speed of the masses should be the same here after the "impact", as the masses are tied to one another through the rod, and should be moving with the same speed. But perhaps its not as easy as my analysis implies here

In the second question, I consider that, once the m/2 mass is added, the rod is off the horizontal by an angle theta. The angle between the leftmost string and the horizontal is alpha, so the total angle between the string and the rod is (theta + alpha) - this is the same angle for the other string. Now, taking moments about the original centre of gravity (before the extra mass was added) gives:

T1*L*Sin(theta + alpha) + mg/2*L/2*Cos(theta) = T2*L*Sin(theta + alpha) which eventually leaves us with

Sin(theta+alpha) = 7mg/8*Cos(theta)


Now taking moments about the new centre of gravity (having added the new mass) gives:

mg*L/6*Cos(theta) - mg/2*2L/6*Cos(theta) + T1*7L/6*Sin(theta + alpha) - T2*5L/6*Sin(theta + alpha) = 0

The terms involving mg and Sin(theta + alpha) dissapear here and we are left with T1 = 5/7*T2

Using the formula forces up = forces down and taking the "up" direction to be the perpendicular to the rod (rather than the usual vertical direction) we get

T1*Sin(theta + alpha) + T2*Sin(theta + alpha) = mg*Cos(theta) + mg/2*Cos(theta). Since T1 = 5/7*T2, this leaves us with:

12/7*T2*Sin(theta+alpha) = 3mg/2*Cos(theta).

Knowing that Sin(theta+alpha) = 7mg/8*Cos(theta) gives us

12/7*T2*(7mg/8*Cos(theta)) = 3mg/2*Cos(theta)

This leaves us with T2 = 1N
and so T1 = 5/7N

Intuitively, by putting the additional mass to the right of the centre of the rod, we would expect the tension in the right string to be greater than that in the left, so the answer is consistent on that front.

There could easily be a flaw (or flaws!) with my logic here so is anyone can see a problem, please let me know

hope this helps you out some bit,
Japes

setanta222

Thank you Japester. I'm a 'mature student' doing App Maths (and Maths) in L Cert next June on my own (i.e. just from the text book). For the most pare I'm doing ok but I've just joined boards for problems I get stuck on but I'm afraid I'm pretty useless on a PC and find it very difficult to follow 'threads', 'tags', posts, etc and all that. Can you advise as to where in Boards I should look for others that are having problems (and getting solutions) in L Cert (HL) Applied Maths? Are there particular search words? S

japester

Hi Setanta, you're more than welcome for any help I could give you. I did LC Applied Maths back in 1990 but I must admit I only got a D in it (honours) at the time (and an A in honours maths) ..... not enough work on my part. I came across some old LC Applied Maths stuff lately and I got hooked - I wanted to see why I didn't understand things properly back then ( I had a decent teacher but I didn't pay enough attention!!) I have since gone on and completed electronic engineering at 3rd level and got much better at maths/applied maths as a result. I contributed to the LC applied maths thread a few weeks ago where I attempted all the questions that came up in this years exam and posted up my answers + some parts of my solutions. Applied Maths does require a good bit of work and understanding and most importantly, thinking logically, but, as a mature student, I think you are in a great position on all fronts, because I am convinced that your sense of logic improves with age

Boards does not appear to have a dedicated applied maths forum (I guess as it is such a minority subject) but, as the syllabus is more closely related to the mechanics area in physics, you would probably have a better chance of getting people to reply to your questions if you posted to the "physics & chemistry" forum instead, as some of the contributors there are very knowledgable indeed in all kinds of areas of physics.

In my travels, I have also come across the following forum which you should find very useful indeed. It is related to a physics examination that is based in India and I reckon (based on population alone) you are almost always likely to get a quick reply to your queries on this site:

http://engg.entrancecorner.com/discussion-forum/120-mechanics.html

I believe you can create an account for yourself for free (I haven't done this myself, as I have only used it briefly to look at some written posts) and I think some of the contributors really know their stuff (whereas I, even after 4 years of third level, really only "half-know" my stuff on mechanics, but I am enjoying the challenge of trying to re-educate myself on some of it, while learning new stuff also from the likes of the this Indian forum ). By the way, there is a very good online e-book available (though not freely downloadable) from the same Indian exam company, which I have found very informative, with oodles of questions/answers available (a good share of which would be beyond the level of honours applied maths for LC). The link for it is:

http://www.docstoc.com/docs/32115784/IIT-PHYSICS-1

There really is a wealth of information available in that book and it may help you out too.

I'll have a look to see what else I can come up with for you as I have some other bits linked that I thought were useful.

Oh yes, one more thing! That first question you asked from O Murphy's book, about the rod with the 2 masses at either end. My thinking was fairly good but the solution I posted for it is very off the mark. having seen a very similar type of question elsewhere, the question involves two kinds of simultaneous motion, translational and rotational. When the rod is given a "nudge" at one end to move it along the table in the j-direction, it undergoes a so-called translational motion in this direction. Of course, the mass at the other end and the rod as a whole also undergo this motion, all therefore moving initially at u m/s in the j-direction. However, the system will also undergo a rotational motion, where the masses at either end will have a certain angular velocity and hence, a certain tangential velocity correspoding to this, based on the radius of curvature for the mass in question. The question which I have seen that is similar to this one is only similar up to a point (it was based on a rod with no masses at either end getting a certain impulse imparted to it 15 cm from the centre of gravity of the rod). I haven't actually worked out a solution to your problem based on this but bear in mind that, in order to deal with the rotational motion, you will have to calculate the moment of inertia of the mass(es) about the centre of gravity (and note that in your case the centre of gravity will not be at the geometric centre of the rod as one mass is three times the other). Once you know what the angular velocity is, you know what the tangential velocity of the mass is and then you will get a resultant velocity vector based on the translational velocity and the tangential velocity. Its no joke this question!!! By the way, did the question state what the length of the rod was???

setanta222

Thank you again Japester; that is very useful
Setanta

japester

You're more than welcome Setanta, and best of luck with your studies