Question

4. A 6-kg block (my), initially at rest on a rough shelf, is connected to a 4-kg block (m2) that hangs by an inextensible string of negligible mass passing over a pulley. The uniform disk-shaped pulley, having a mass of 3.0-kg and a radius of 10 cm, rotates about the symmetry axis through its center. The 6-kg block, which is attached to the spring, is initially pushed against the spring compressing it a distance of 30 cm from its equilibrium position, and then released. The force constant of the spring is 500 N/m. The coefficient of kinetic friction between the 6-kg block and the shelf surface is 0.20. By applying the conservation of energy or any other suitable methods, (a) Calculate the magnitude of the velocity of the each block at the instant the 4-kg block descends a vertical distance of 0.4 m. (b) Find the total angular momentum of the system at the instant when the vertical block moves through the displacement of 0.4 m.

Answer 1

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