a)
let d = 0.4 m
xi = 30 cm = 0.3 m
xf = 0.4 - 0.3 = 0.1 m
let v is the speed of the blocks and w is the angular speed of the
disk after m2 falls 0.4 m.
we know,
workdone by friction = change in mechanical energy
fk*d*cos(180) = (1/2)*(m1 + m2)*v^2 + (1/2)*I*w^2 + (1/2)*k*xf^2 - (m2*g*d + (1/2)*k*xi^2 )
-fk*d = (1/2)*(m1 + m2)*v^2 + (1/2)*(1/2)*M*R^2*w^2 + (1/2)*k*xf^2 - (m2*g*d + (1/2)*k*xi^2 )
-mue_k*m1*g*d = (1/2)*(m1 + m2)*v^2 + (1/2)*(1/2)*M*(R*w)^2 + (1/2)*k*xf^2 - (m2*g*d + (1/2)*k*xi^2 )
-mue_k*m1*g*d = (1/2)*(m1 + m2)*v^2 + (1/2)*(1/2)*M*v^2 + (1/2)*k*xf^2 - (m2*g*d + (1/2)*k*xi^2 )
-mue_k*m1*g*d = (1/2)*(m1 + m2 + M/2)*v^2 + (1/2)*k*xf^2 - (m2*g*d + (1/2)*k*xi^2 )
-0.2*6*9.8*0.4 = (1/2)*(6 + 4 + 3/2)*v^2 + (1/2)*500*0.1^2 - (4*9.8*0.4 + (1/2)*500*0.3^2 )
==> v = 2.32 m/s <<<<<<<<<<<------------------------Answer
b) angular momentum = (m1 + m2)*v*R + I*w
= (m1 + m2)*v*R + (1/2)*m*R^2*w
= (m1 + m2)*v*R + (1/2)*m*R*v (since R*w = v)
= (6 + 4)*2.32*0.1 + (1/2)*3*0.1*2.32
= 2.67 Kg.m^2/s <<<<<<<<<<<------------------------Answer
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