he figure shows a rigid assembly of a thin hoop (of mass m = 0.20 kg and radius

Question

he figure shows a rigid assembly of a thin hoop (of mass m = 0.20 kg and radius R = 0.17 m) and a thin radial rod (of length L = 2R and also of mass m = 0.20 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?

Explanation / Answer

mass of the hoop and rod m=0.2 kg

radius of the hoop R=0.17 m

length of the rod L=2*R =2*0.17 =0.34 m

moment of inertia of the rod Irod=1/3*m*L^2

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

I_rod =4/3*m*R^2

moment of inertia of the hoop is,

I_hoop=Icm+m*h^2

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

=19/2*m*R^2

total moment of inertia I=Irod+Ihoop

I=(4/3+19/2)*m*R^2

I=(65/6)*m*R^2

and

initially,

height of the center of mass h1=m1*y1+m2*y2/(m1+m2)

h1=(m*(L/2)+m(L+R))/(m+m)

h1=(L/2+L+R)/2

h1=(R+2R+R)/2

h1=2R

and

finally,

if the system upside down,

height of the center of mass h2=-2R

by using law of conservation of energy,

1/2*I*w^=2*m*g(h1-h2)

1/2*I*w^=2*m*g(h1-h2)

1/2*(65/6)*m*R^2*w^2=2*m*g(4R)

1/2*(65/6)*R*w^2=2*g(4)

===>

w^2=96*g/(65*R)

w^2=96*9.8/(65*0.17)

=====> w=9.23 rad/sec

angular speed w=9.23 rad/sec

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