The figure shows a rigid assembly of a thin hoop (of mass m = 0.29 kg and radius

Question

The figure shows a rigid assembly of a thin hoop (of mass m = 0.29 kg and radius R = 0.15 m) and a thin radial rod (of length L = 2R and also of mass m = 0.29 kg). The assembly is upright, but we nudge it so that it rotates around a horizontal axis in the plane of the rod and hoop, through the lower end of the rod. Assuming that the energy given to the assembly in the nudge is negligible, what is the assembly's angular speed about the rotation axis when it passes through the upside-down (inverted) orientation? Number Units

Explanation / Answer

moment of Inertia, I_rod=1/3*m*L^2

I_rod=1/3*m*(2R)^2

I_rod=1/3*0.29*(2*0.15)^2

I_rod=8.7*10^-3 kg.m^2

and

moment of Inertia, I_hoop=Icm+m*x^2

=1/2*m*R^2+m*(L+R)^2

=1/2*m*R^2+m*(2R+R)^2

=1/2*m*R^2+(m*3R)^2

=(1/2+9)*m*R^2

=(1/2+9)*0.29*0.15^2

=61.98*10^-3 kg.m^2

total moment of Inertia, I=Irod+Ihoop

I=8.7*10^-3+61.98*10^-3

I=70.68*10^-3 kg.m^2

by using law of conservation of energy,

1/2*I*w^2=m*g*L+2*m*g*(L+R)

1/2*70.68*10^-3*w^2=m*g*(3L+2R)

1/2*70.68*10^-3*w^2=m*g*(3*2R+2R)

1/2*70.68*10^-3*w^2=0.29*9.8*(8*R)

1/2*70.68*10^-3*w^2=0.29*9.8*(8*0.15)

====> w=9.82 rad/sec

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