Three small spherical masses are located in a plane at the positions shown below

Question

Three small spherical masses are located in a plane at the positions shown below. The masses are Q = 0.200 kg, R = 0.400 kg, and S = 0.300 kg. Calculate the moment of inertia (of the 3 masses) with respect to an axis perpendicular to the xy plane and passing through x = 0 and y = -2. [Since the masses are of small size, you can neglect the contribution due to moments of inertia about their centers of mass.] A uniform rod of length 1.45 m is attached to a frictionless pivot at one end. It is released from rest from an anqle theta = 19.0 degree above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass.

Explanation / Answer

m1 (Q) = 0.2 kg        r1 = sqrt(3^2 + 4^2) = 5 m

m2 (R) = 0.4 kg        r2 = sqrt(4^2+1^2) = 4.12 m

m3 (S) = 0.3 kg        r3 = 3 m

moment of inertia = I = m1*r1^2 + m2*r2^2 + m3*r3^2

I = (0.2*5^2) + (0.4*4.12^2) + (0.3*3^2) = 14.5 kg m^2 <<-----answer

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moment of inertia = I = (1/3)*M*L^2

angular acceleration = alfa

linear accleration = a = L*alfa

torque exerted by gravity = T = M*g**L/2*cos19

net torque = I*alfa

M*g*L/2*cos19 = (1/3)*M*L^2*a/L

g*(1/2)*cosalfa = (1/3)a

9.8*(1/2)*cos19 = (1/3)*a

a = 14 m/s^2 <<--------answer

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