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So, basically the frictional force acts the opposite direction to which the wheel would be accelerating (up the hill, wheel accelerates to the right at the point of contact - if the wheel was rolling to the right), therefore Ff would be to the left at tthe point of contact. in my case, wheel is accelerating down hte hill and at point of contact that point of the wheel would be accelerating to the left, so the Ff is to the right.. My textbook is still wrong in this case... unless im not understanding something.. ______________ Back on topic.. I solved a = (alpha) R for each case, and worked out a = Ff/m and 2Ff/m for the ring and disc respectively. Then integrated with respect to time to get velocity... the nagain to get displacement. Subbed in when s = D.. t = sqrt(2Dm/Ff) and sqrt(Dm/Ff) for the ring and disc. Dividing the two to get the ratio gave me Tring/Tdisc = sqrt(2) : 1 This is the wrong answer , the answer should be Tring/Tdisc = Sqrt(4/3) I'll try again with the frictional force in the opposite direction see if that gives me the right answer.. can anyone notice anything i've done absolutely wrong.. Thank you very much. found it.. went to see tutor today again and this one actually helped me.. Thanks for your help guys also this other textbook i went to get shows the friction in the same way the other textbook had it.. I have no idea what to do wtih friction. The tutor said it doesn't matter what you assume, the answer will tell you if its right or wrong so i suppose it doesn't matter? Cheers all!!

Hi guys Haven't posted in this forum for a long long time.. good to read over some more interesting stuff on here i havent had the time to for a long long time.. anyway here goes Dynamics (mechanical engineering) gave me this assignment question which is more of a study question for our final exam (because it doesn't count towards our marks) Here is the question: A thin ring and a circular disc, each of mass 'm' and radius 'R', are released from rest on an inclined surface and allowed to roll a distance 'D'. Determine the ratio of the times required. I firstly drew my FBD (free body diagram) and this is it: (take a close look at the direction of my frictional force, my tutor said its the other way (to the right) but the textbook says its the way i've drawn it, can someone tell me if i've done it correctly). Now, I know the mass moment of inertia for each object, SOLID DISC: I = 0.5 MR^2 THIN RING: I = MR^2 Now, I know that M = I . alpha that is (sum of moments) = (inertia) x (angular accel) So i know the inertia, I can work out sum of moments by looking at the FBD and the only force that would create a moment about the centre would be the frictional force. So sum of moments = -Ff x R. Then I can find alpha (angular acceleration).. Now I thought I can find the straight line acceleration of the centre of mass by using other equations but I dont know how to? Then i might be able to find the ratio of times required? _________________ I'm stuck on these last 2 lines... what do i do... what do i do once i've found alpha? ALSO, I'm not sure if what i've done is correct... it seemed pretty logical to me but then again i could have easily done it wrong.. Thank you all very much i appreciate it a lot..

Ok thanks for your help athiest.. Heres what i'm trying to do.. I dont HAVE to do this but i thought i may aswell just to give more proof that the clay bricks are good option for a climate control system for this house.. i'm no engineer as of yet, and this enginering subject is just a crap subject for general engineering (no major chosen yet).. I'm trying to find the inside temperature against time because then i can show that it will stay fairly constant and a suitable temperature. I dont know how else to explain this. i guess I could just leave out the section because the tutors at Uni didn't know how to do it and they tried to speak some rubbish to me but its complete crap... he didn't know... Anyway thanks..

I was trying to get a function of the inside temperature against time... say a sin wave of the yearly temperature changes against time for the inside of the house, and then I can compare the inside temp against the outside temp and hopefully say THEREFORE IT KEEPS IT COOL IN THE SUMMER TIME AND WARM IN THE WINTER TIME (thats what the clay bricks are meant to do ... thats what i'm using in this assingment) Hope you all sort of understand what i'm tryin to do, i cant really explain it that well....

Hi everyone Firstly, i'm at UNI doing this assignment, and stuck on this one part. Its only first year Engineering and it's a really easy engineering subject, probably don't need to go into this much detail but I suppose if i can then i may aswell do it We're doing a climate control system for a house in india, i've got an equation that somewhat models the outside maximum and outside minimum temperatures for the city the house is being built in... Tmax = 30 + 5sin(tπ/11) Tmin = 20 + 5sin(tπ/11) Where t = time in months (January = 0, December = 11) T = temperature Also, the rate of heat flow equation is ... Q/t = [kA(Thot - Tcold) ] / d Where, Q/t = rate of heat flow ??? k = thermal conductivity of the barrier (0.653 W/m.K in this case) A = Area of the wall , I'll assume its 3m x 5m T = temperature (hot is outside, cold is inside i'd suppose) d = thickness of barrier, and in this case the bricks are lets say 270mm thick My question is... How on earth do i figure out what the inside temperature could be when i've got two unknowns, the rate of heat flow and the inside temperature... If anyone can help me with these simple equations that'd be heaps appreciated... I'm sure there's a way you can manipulate the equation or even there might be a whole different equation that you could use.. I know everything except inside temperature (thats what I want to find out) and the rate of heat flow... THANKS EVERYONE! John.

ok, but even if i did multiply the top one by .5 it would still give me the same answer in the long run.? 2C2H6 + 2O2 ==> 2C2H2 + 4H2O H = -520 KJ The 2 unknowns i have here.. how can i figure out those?

I have been given a question: Determine the enthalpy of reaction in kJ for the hydrogenation of acetylene to form ethane : C2H2 (g) + 2H2(g) --> C2H6(g) From the following data: 2C2H2 + 5O2 --> 4CO2 + 2H2O H = -2600 KJ 2C2H6 + 7O2 --> 4CO2 + 6H2O H = -3120 KJ H2 + 1/2O2 --> H2O H = -286 KJ So far i know H(h2o) = -286 kJ But i've swapped around reaction 1 (the first one where H = -2600) and added together the first and second reactions. Now i have: 2C2H6 + 2O2 ==> 2C2H2 + 4H2O H = -520 KJ But i dont know where to go from here, because i'm left with two unknowns, the 2c2h6 and 2c2h2. Any help appreciated. thanks guys/girls

If we are travelling through space at speed x m/s....... wouldn't the wheel need to spin at 300,000 km/s - x m/s...... ? Also if this wheel is travelling at 250,000 km/s ... wouldn't a tiny amount of force cause it to increase in speed (i am talking about the points on the circumference).... which means that every tiny amount of force we spin the wheel wiht will always increase the speed, no matter how small the force is. there is no force acting against it.

but there is the same AMOUNT of heat... ? am i wrong?

I have awesome dreams after drinking, around half the time. none are bad or nightmares. also hardly ever have a hangover, unless i drink something i dont like the taste of.

yes i understand the universe is wayyy bigger than before, but isnt the amount of matter inside the universe exactly the same as before (before being just after the BB). So i've read somewhere else off my memory that the average temperature of the universe is like 3 degrees kelvin or something. Mustn't that be measuring all the matter in the universe. Well just after the BB, the average temperature of ALL the matter in the universe was billions of billions of degrees... hope you can understand what i've written heh

I dont know this could you please explain more, do you mean the heat can be given off from objects as radiation. If so, then where would that radiation go, wouldnt it eventually hit other matter, heating that matter up, keeping the overall average temperature the same as in the beginning?

ok, PV = nrT. as volume increases, temperature increases. ? as we keep the pressure and amount of stuff the same?.. anyway, isnt the amount of matter now the same as the amount of matter at the start of the universe..? just that now it is more spread out?

So, I read that at time = 10 billionths of a second after the big bang, the temperature was "billions of billions of billions" of degrees celcius. Now, from what I've learnt in chemistry, heat can only be passed on to other objects, so if there are two objects out in space at 100 degrees touching each other, they will remain at 100 until they touch another piece of matter that is higher/lower than 100degrees? Well, why when every single particle at those early stages of the universe are at roughly "billions of billions of billions" of degrees celcius, the temperature now on earth is only 30 degrees. the temperature of the hottest suns are only in the millions of degrees celcius, where did all that massive temperature go? Thanks people.

haha, one dream i had is where i went to the toilet.. and i woke up and i was actually pissing in my bed. hmm..... I want to learn how to lucid dream