The four masses shown in the figure below are connected by massless, rigid rods.

Question

The four masses shown in the figure below are connected by massless, rigid rods. (a) Find the coordinates of the center of gravity of this object if M_A = 80 g and M_B = M_C = M_D = 260 g. x = 0.06050 0.0605 m y = 0.0605 m (b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page. Print out this page, or draw the figure on paper. Put the eraser of a pencil on top of mass A so that the pencil is perpendicular to the piece of paper. Then rotate the paper around the tip of your eraser. The pencil is the axis of rotation. 0.5252 X kg m^2 (c) Find the moment of inertia about a diagonal axis that passes through masses B and D. Lay your pencil along the diagonal that would connect masses B and D. The configuration would spin about this asix if masses B and D remain at their same locations, mass C comes out of the paper, and mass A goes into the paper. How far are masses C and A from this axis of rotation (the pencil)? You'll have to do a bit of geometry to find the distance. 0.5252 kg m^2

Explanation / Answer

a)

Center of mass = (Ma*A + Mb*B + Mc*C + Md*D)/(Ma + Mb + M + Md)

= 260*((0,10) + (10,10) + (10,0))/(3*260 + 80)

= 260*(20,20)/(860)

= 0.302*(20,20)

= (6.05 , 6.05) cm

So, x = 0.0605 m

y = 00605 m

b)

Moment of inertia = (Mb*rB^2 + Mc*rC^2 + Md*rD^2)

= 260*(0.1)^2 + 260*(sqrt(2)*0.1)^2 + 260*(0.1^2)

= 10.4 g/m2

= 0.0104 kg/m2

c)

Moment of inertia = (260+80)*(0.1/sqrt(2))^2

= 1.7 g/m2

= 0.0017 kg/m2

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