Recent content by series111

Yes my calculations were wrong however I did calculate them again and got the write answers through a little research : K^2 = d^2/8 ( This is for a soild Disk i.e Steel Disk) Where I=mk^2 Where Iw2 - Iw1 ( change in angular momentum) As for change in angular kinetic energy you are...

Thanks for your help i will just battle on and won't bother you again cheers...

So torque =(M) mass x (K^2) radius of gyration x angular acceleration :angular acceleration= (M) mass x (K^2) radius of gyration / torque : angular acceleration= (M) 48.98 kg x (K^2)20x10-3m^2 / (t) 1.5 Nm= 653.06 x 10 -3 rad/s^s final velocity = initial velocity + angular...

k ^2= radius of gyration hence k^2 = d^2/8 The equation = (I) moment of inertia = mass x radius of gyration I calculated the mass by : volume = pie dia^2/4 x length = pie 400 x 10 -3m ^2/ 4 x 50x 10-3m = 6.28 x 10-3m density= mass / volume : mass= density x volume = 7800...

I=MK^2 I = 48.98 x 20 x 10^-3 = 979.6 x 10 ^-3 kgm^2/s Torque = moment of inertia x acceleration : T = 979.6 x 10 ^-3 kgm^2/s x 10.99 = 10.76 Nm Applied torque + frictional torque = 10.76 Nm + 1.5 Nm = 12.26 Nm Bracking Torque... Hope this is correct...

thanks for your reply i will let you no how i get on :redface:

Homework Statement a light shaft carries a steel disk 400m dia, 50mm thick,density 7800kg/m_3 calculate : (a) if the frictional resistance is equivalent to a torque of 1.5Nm, determine the braking torque to bring the disk to rest from 43.98 rad/s in 4 secs. (b) what would be the time...

so I = 3.918kgm_2/s and Δω = 37.7 rad/s and the change in angular momentum is = I * Δω = 3.918 x 37.7 = 147.70 kgm_2/s thanx for putting me right...

Homework Statement a steel disk 400mm in diameter and a mass of 48.98 kg accelerates from 6.28rad/s to 43.98rad/s calculate the change in angular momentum. Homework Equations final angular momentum = intial momentum (I) final x (w) final = (i) initial x (w) intial I= mr _2...

thanks for checking :smile:

Had another look today and relised where i have went wrong here is my new answers : torsion / polar second moment of area = tau/ radius : (t) 122.78 x 10^3/(j) 0.08545 = (tau) 79.95 x 10 ^6/d/2 : (t) 122.78 x 10 ^3/(tau) 79.95 x 10 ^6 = (j) 0.08545 / d/2 D= 3 square root with...

Homework Statement assuming that the maximum shear stress and torsion are the same in both shafts design a hollow shaft to replace the solid one . the ratio diameters is to be 0.6 for the maximum shear stress i calculated 79.95 x 10 ^6 mn/m2 and for the torsion i calculated 122.78x10^3 in the...

when i calculated these i used the diameters and the answers were correct compared to the answers i had.

d is the diameter of each pulley

sorry to reply again as i have misread part (e) in the question it says if the effort needed to raise a load of 400kg is 17N determine : the law of the machine. also should the law of the machine when calculated total 17N thanks again mark