Question

Two masses M and m are connected by a rigid rod of length L and of negligible mass, as shown in Figure P10.27. For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axis passes through the center of mass. Show that this moment of inertia is I = µL2, where µ = mM/(m + M).
2. The four particles in Figure P10.21 are connected by rigid rods of negligible mass where y = 6.60 m. The origin is at the center of the rectangle.

(a) If the system rotates in the xy plane about the z axis with a speed of 6.10 rad/s, calculate moment of inertia

Answer 1

Moment of inertia about r, where M is at r=0 and m is at r=L.:

I = M r^2 + m (L-r)^2

min I will be at dI/dr = 0,
2 M r - 2 m (L-r) =0
solve this to get r, which will be at the center of mass [c.m.].

Substitute this r into I to get [messy] moment of inertia around c.m.

Answer 2

Thank you! I still need help with the second one, though.

Answer 3

Q2
"...The origin is at the center of the rectangle..."
So you can easily find the co-ordinates of each mass.

Use -
I = Summation mr^2

Answer 4

but r is the distance from where it's rotating, and it's rotating around the z axis. I tried finding the length of the diagonals symbolically using 4 as the width and y as the height, but the answer is supposed to be only numbers...

Answer 5

"...but r is the distance from where it's rotating, and it's rotating around the z axis..."
r is the perpendicular distance of the point with the axis of rotation.

"... using 4 as the width and y as the height, but the answer is supposed to be only numbers..."
"...where y = 6.60 m..."

Answer 6

ohhhh thank you