A 4-kg block lies on a horizontal table whose coefficient of friction is 0.3. A

Question

A 4-kg block lies on a horizontal table whose coefficient of friction is 0.3. A horizontal rope is redirected by a uniform solid disk (pulley) of mass 2 kg to a mass of 12-kg hanging off the table. The radius of the solid disk is 10 cm and the rope doesn A 4-kg block lies on a horizontal table whose coefficient of friction is 0.3. A horizontal rope is redirected by a uniform solid disk (pulley) of mass 2 kg to a mass of 12-kg hanging off the table. The radius of the solid disk is 10 cm and the rope doesn

Explanation / Answer

inertia of pulley = (0.5)*m*r^2 = 0.5*2*0.1^2 = 0.01 kg m^2

let the linear accelearion be a ..

so... angular acceleration = a / r = a / 0.1 = 10 a

let the tension in rope connecting 4 kg block be T1 and that in 12 kg block be T2

so... (T2 - T1)*radius = inertia * angular acc

so.. ( T2 - T1 )*0.1 = 0.01 * 10 a

so... T2 - T1 = a

for 4 kg block..

T1 - (0.3*4*9.8) = 4*a

so.. T1 = 4a + 11.76

so... T2 = T1 + a = 5a + 11.76

for 12 kg block..

12*9.8 - T2 = 12*a

12*9.8 - 5a - 11.76= 12a

so.. a = ((12*9.8)-11.76)/17 = 6.22588 m/sec2

b) angular accelelraion of disk = 10*a = 62.2588 rad/sec2

c) Tension T1 = 4a = 4*6.22588 = 24.90353 N

Tension T2 = 5a = 5*6.22588 = 31.1294 N

d) for top block

a = 6.22588

u = initial vel = 0

t = 2 sec

so distance s = ut + 0.5*a*t^2 = 0 + (0.5*6.22588*2^2) =12.45176 m

e) revolution = s / (2*pi*rad) = 12.45176 / (2*pi*0.1) = 19.81759 revs

f) distance travelle by a point on the edge of the disc = s = 12.45176 m   

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