A 3 kg mass and a 6 kg mass are attached to either end of a 3 m long massless ro

Question

A 3 kg mass and a 6 kg mass are attached to either end of a 3 m long massless rod.

a.) Find the center of mass of the system.
m, from the 3 kg mass.

Find the rotational inertia (I) of the system when rotated about:
b.) the end with the 3 kg mass.
kg m2
c.) the end with the 6 kg mass.
kg m2
d.) the center of the rod.
kg m2
e.) the center of mass of the system.
kg m2
(Compare this to parts b-d. Is this what you expect?)

f.) Suppose a frictionless pivot is attached at the center of mass, such that the system is free to rotate about this point in the horizontal plane. A force of 9 N is exerted perpendicular to the rod, pushing directly on the 6 kg mass. What is the size of the system's angular acceleration that would result?
rad/s2

Explanation / Answer

a) for center of mass m1*x1=m2*x2 and x1+x2=L=3m, where x1 is arm from center of mass to m1=3kg, and x2 is arm from center of mass to m2=6kg, then,
m1*x1=m2*(L-x1), hence x1=L*m2/(m1+m2) = 3*6/(3+6) = 2 m;

b) I=m2*L^2 = 6*3^2 =54 kg·m^2

c) I=m1*L^2 = 3*3^2 =27 kg·m^2;

d) I=m1*(L/2)^2 + m2*(L/2)^2 = (3+6)*(3/3)^2 =9 kg·m^2;
e) I=m1*x1^2 + m2*x2^2 = 3*2^2 + 6*(3-2)^2 =18 kg·m^2;

f) the 2nd newton’s law says: torque T=I*w’, where I=18 is inertia moment of the system, T=F*x2, F=9N, x2=L-x1=1m}, hence angular acceleration
w’=T/I = 9*1/18 = 0,5 rad/s^2;

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