Problem 10.10 Part A The rotating systems shown in the figure differ only in tha

Question

Problem 10.10 Part A The rotating systems shown in the figure differ only in that the two identical movable masses, each one of mass m, are positioned a distance r from the axis of rotation (left), or a distance r/2 from the axis of rotation (right). In both cases, r is equal to 3 times the radius of the pulley, and m is equal to the mass of the hanging block. You release the hanging blocks simultaneously from rest, and call ti the time taken by the block on the left and tr the time taken by the block on the right to reach the bottom, respectively. The bar, pulley, and rope have negligible mass, the rope does not slip, and there is no friction in the axle of the pulley. Under these conditions L-38 38 R _38 11 tr. Submit My Answers Give Up Continue

Explanation / Answer

Left system = 1 , Right system=2

I1= mR^2+mR^2 = 2mR^2

I1= m(R/2)^2+m(R/2)^2 = mR^2 /4

t net = I*a

For 1,

t net = I*a1

mgR = (2mR^2)*a1

a1=g/R

For 1,

t net = I*a2

mgR = (mR^2/4)*a2

a2=4g/R

a2/a1=(4g/R)/(g/R) = 4

a2= 4a1

a2/a1=4--------------(1)

Now by kinematic equation

q= wi+1/2*a*t^2

wi=0rad/s

After landing both system same q

Hence

wi+1/2*a1*t1^2 = wi+1/2*a2*t2^2

a1*t1^2=a2*t2^2

t1/t2= sqrt(a2/a1) = sqrt(4)= 2

Thus

t1/t2 = 2

tL=2*tR

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