The figure shows a rigid structure consisting of a circular

The figure shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 4.2 s. If R = 0.6 m and m = 2.1 kg, calculate the angular momentum about that axis. Rotation axis CS 2R Number Units

Solution

for the square of rods total moment of inertia about rotation axis will be

Total I = I_up + I_down + I_on axis + I_away

Total I = mR^2/3 + m*R^2/3 + 0 + m*R^2

Total I = I_rod = 5*m*R^2/3

Moment of inertia for the hoop will be, using parallel axis theorem

I_hoop = Icm + m*d^2

Icm = moment of inertia about center of mass

I_hoop = m*R^2/2 + m*R^2 = 3*m*R^2/2

I_rod + I_hoop = 5*m*R^2/3 = 3*m*R^2/2

net I = 19*m*R^2/6

Now angular moemntum is given by:

L = net I*w

w = angular velocity = 1 rev/4.2 sec = 1 rev*(2*pi rad/rev)/4.2 sec

w = 2*pi/4.2 = 1.496 rad/sec

m = 2.1 kg

R = 0.6 m

So,

L = 19*2.1*0.6^2*1.496/6

L = 3.58 kg-m^2/sec

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