Recent content by ingram010

Thanks edgepflow much appreciated

Hi all I have a revision question I am having some trouble with A cold room has a wall measuring 5.2m by 2.5m. The wall is constructed of 120mm thick brick on the inside, a cork layer 80mm and a 30mm layer ow wood in the outside. the inside temperature is -4ºC and the outside temperature...

Hi all Hope some one can help with this Calculate the increase in kinetic energy due to acceleration? 0.6m diameter Solid cylinder with a mass of 120kg, initial velocity of 20.94 rad/s and final velocity of 146.61 rad/s. I have found an equation for the kinetic energy lost at...

Hi Tiny-Tim and Quinzio I am sure that the answer to my question is 17.97Nm Thank you both very much for your help Kindest Regards John p.s If I am wrong please tell me : )

Ahh! Just found this, I really should spend more time reading. Moment of inertia for a solid disk = 1/2 mr^2

this is the extract from the book I am using:- Torque, moment if inertia and angular motion Torque From Newton's third law, we know that to accelerate a mass we require a force such that: F= ma Now in dealing with angular motion, we know that this force would be applied at a radius r, from...

According to the book I have, inertia = mass x radius^2 so I = mass x raduis^2 = 10.8kgm^2 T= I x acc = 10.8 x 3.142 = 33.93 and mass x acc x radius^2 = 120kg x 3.142 x .3^2 = 33.93 it gives the same answer, I am not sure what you are telling me John

Oh yes, whoops. I have used the following equations to solve the problem:- 200rpm x 2 pi /60 = 20.94 rads/s = initial velocity 1400rpm x 2 pi /60 = 146.61 rads/s = final velocity then final velocity - initial velocity/40 seconds = 3.142 rads/s^2 = angular acceleration then mass x...

Great so say the applied torque is 33.93Nm I just simply subtract the frictional torque which is 1Nm giving 32.93Nm? It seems too simple to be true, but Ill go with it : ) Thanks for you help

Hi Tiny-Tim I have solved the problem but I can’t work out how to factor in the loss of torque due to friction. Is it just a case of deducting it from the applied torque? Hope you can help Cheers John

Thanks tiny-tim I will take a look

Hi all getting a bit stuck on this problem:- a solid cylinder rotating about its polar axis has a mass of 120kg and a diameter of 0.6,. If the bearings provide a frictional torque of Nm, find the torque applied to accelerate the cylinder from 200 to 1400rpm in 40 seconds? if anyone can...

voltage calculation Hi, on reflection it isn't very clear. Sorry. how about this- 500v---\/\R1\/\----X----\/\R2\/\---0v If R1 = 1Mohm and R2 = 10Kohm what is the voltage at X? Or this one - 500v---\/\R1\/\--X--\/\R2\/\--Y--\/\R3\/\--Z--\/\R4\/\---0v If R1 = 1Mohm, R2 =...

Hi all, I am looking for a formula to work out the following- 500v---/\/\/\/\/---X---/\/\/\/\/---0v 1megohm 10kohm what is X? cheers