app maths help

Jrembin

excersise 13a

question 8

could someone tell what deltaM is and why its that?

thanks.

Nehpets

Haven't looked at the exercise or anything but is it change in mass?

I could be 100% wrong on that though.

cocoa

delta M is infinitesimally small parts of M, with the summation it just turns into integration, so you needn't worry about its meaning that much.

I am however assuming that your problem was more specific to the question... I'm not counting rigid body rotation so I haven't practised too much but working off the theorem 13.4 with the disk I get this:

M = rho*area = rho * pi (R^2 - r^2)

divide the hoop into more infinitesimally small hoops with radius x and width deltax

if any hoop were cut out and the area measured, it would 2*pi*x*deltax

therefore, deltaM = rho*area = 2*pi*rho*x*deltax

now, I=EdeltaMr^2 = E2*pi*rho*x^3deltax
the summation becomes integration, and as you might have guessed, our limits are R to r

S (R to r) 2*pi*rho*x^3 dx = [2*pi*rho*x^4/4](R to r)

= pi*rho/2 (R^4 - r^4) = pi*rho/2 (R^2 - r^2)(R^2 + r^2)
= pi*rho(R^2 - r^2)*(R^2 + r^2)
= M/2(R^2 + r^2)

i hope that makes it clear and fixes your problem

Jrembin

cocoa wrote:

delta M is infinitesimally small parts of M, with the summation it just turns into integration, so you needn't worry about its meaning that much.

I am however assuming that your problem was more specific to the question... I'm not counting rigid body rotation so I haven't practised too much but working off the theorem 13.4 with the disk I get this:

M = rho*area = rho * pi (R^2 - r^2)

divide the hoop into more infinitesimally small hoops with radius x and width deltax

if any hoop were cut out and the area measured, it would 2*pi*x*deltax

therefore, deltaM = rho*area = 2*pi*rho*x*deltax

thanks a bunch. this is just what i needed. seems so obvious now.