Recent content by tomcell

Also is it not y = -(50x10^3m - 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa)) instead of y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa)) Im thinking the pressure at the surface is 1.95*10^7 but this is obtained from other sources

SO what did y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa)) change to etc Thanks. and do you just mean ln instead of log

I thought I would have a go at this but am stuck on how you got (y) = 25x10^3 m from y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa)) either way using log of ln If anyone one could solve this cheers Tom

all done to C) feeling confident but don't have a clue about c)

Ok thanks so is use (alpha)A=a(normal a)T/r or alpha = circular acceleration (linear) times Torque divided by Radius I take it i need to work out Linear acceleration which is a=v^2 / r I don't know how to get velocity, is it just 94N times 5 secs to be 470 Rad/sAlso I have found an equations...

OK firstly this B) iv rotational energy is 1/2 I W(omega) = 0.5 x 89.76 x 0.1671 = 1.2533 J Is meant to be B) iv rotational energy is 1/2 I W(omega) = 0.5 x 537.18 x 0.1671 = 7.500128 J C)ii is used L= mvr instead and got L = 184.4784 using L=276(m) x 0.33(v) x 2(r) C)iii 1/2 IW^2 i got 0.5 x...

at this point I don't really now what to do with the 5 secs the force of 94N was applied so i times 0.033 rad/s by 5 to get an angular speed of 0.033 x 5 = 0.1671 then using this to find angular momentum i get L=IW(omega) so L=537.18 x 0.1671 = 89.76 B) ii) force is in the anti clockwise...

finding angular frequency or Omega using Omega(w)=2pi/T so w=2pi/188 i get 0.033 rads/s Am I going along the right lines? Thanks

I seem to have found an equation for Torque which is T=Fr sin (feta) as the force of 94N is applied tangentially to the rim of the disc it must be at 90 degrees to the centre of rotation So T=94N x 2 sin (90) = 188 Rad/s as the force is applies for 5 seconds can i use this to work out the...

Yeah I understand was just looking for some guidance was not at home earlier so can only start working on it again now, was just after pointers, will see how I get on working the above out. The one thing that is holding me back though it think is how do I relate the 95N of force applied for 5...

Here you go Diagram

Yeah i see the logic now i used 2 instead of 1.9 so i did wrong. i can't post the diagram at present but its basically a solid disc with radius of 2m laying flat with an axis of rotation perpendicular to the centre, with I of 400kgm, the disc is stationary at start but a force 95N pushing...

Hi this is some coursework i have for an assignment is there anyway you could shed some light on what steps to take? So far i have worked out the inertia with the children to be 552 by working out the mass of the roundabout to be 200KGS plus the weight of the children and using 0.5mv^2, not...