In the following figure a step cylinder having radius r1 1

In the following figure a step cylinder having radius r_1 = 1 and r_2 =1.5 feet respectively is connected to an object with weight of W= 100 lb. if the coefficient of static friction mu_s = 0.30 and of Dynamic friction mu_d = 0.20 and radius of gyration k=1 feet for the step cylinder, what is the acceleration of the block when released from rest. The rope can unwind.

Solution

as we know the body moves down the inclined plane then work done is

w =Mg (sin0 - u cos 0) * s read o - as theata here 0= 30 ( A)

here Mg= Ma + T T - Tension of string (1)

T=Ib/r₁ b-angular acceleration produced in the cylinder ₍₂₎

here b=a/r₁

now equation (1) beomes

a=Mg/(M+1/r₁²)

now moment of inertia of cylinder about its axis of rotation is

I =1/2 Mr₁²

now from above a= Mg/(M+1/r₁²)

I/m=1/2r₁²=K²=1 =radius of gyration

hence a=Mg/(M+1) as Mg=100lb

a=100/(M+1)=Mg (sin0 - u cos 0) =100 (1/2-u sqrt(3) /2 ) where u =dynamic friction =0.20

=100(1/2-.20 sqrt(3) /2)=32.7