Four particles, each of mass 0.23 kg, are placed at the vertices of a square wit

Question

Four particles, each of mass 0.23 kg, are placed at the vertices of a square with sides of length 0.47 m. The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal, as shown in the figure. (a) What is the rotational inertia of the body about axis A? (b) What is the angular speed of the body about axis A at the instant rod AB swings through the vertical position?

Explanation / Answer

(a), though two of them are quite similar.

The obvious solution 9and apparently shortest) is I(A)= 0.2kg (0+0.5^2+0.5^2+2*0.5^2)m^2
= 0.2 kg*m^2
As hinted by you.

2) Let d be the distance between two vertices. Since the vertices of the square trace a circle with radius, r= squareroot(2*(d/2)^2), use the formula:
I(A)= I(CoM)+mh^2
=mr^2+ mh^2, h=r
I(A) = 2mr^2=0.8kg*2*2*(0.25^2)m^2
=0.2 kg*m^2

3) Similiar to above, but proves that I(CoM) of a square with point masses at vertices is similiar to a circular band of negligible thickness.

I(A)= I(CoM)+mh^2
= 4m(particle)*(distance of particle from CoM)^2+mh^2
= 4*0.2kg*(2*0.25^2)m^2 + 0.8kg(2*0.25^3)m^2
Note: same as above
= 0.2kg*m^2

Therefore, the rotational inertia of a square with the only point masses at the vertices over an axis passing through one of the axis perpendicular to the plane the square is drawn in is:
I= Total mass*Length of square^2

And for rotation with the axis passing through the CoM:
I=Total mass*Length of square^2*(1/2)

For a circular band rotating with an axis passing through it's fringes (or a particle at distance r from the center):
I=2mr^2

Now for (b)

Apparently I thought the blasted square has to rotate counter-clockwise despite being able to rotate clockwise...

D C

A B

to

A D

B C

change of height, delta h=h(i)-h(f)=0.5m

(Initial gravitational potential energy)+(Initial rotational kinetic energy)= (Final gravitional potential energy)+(Final rotational kinetic energy)

Since initial rotational velocity=0,
mgh(i)=mgh(f)+0.5I(rotational velocity)^2
rotational speed=squareroot[2mg(delta height)/I]
(speed is positive since it is the magnitude of velocity (in this case)), despite velocity being negative due to clockwise rotation.

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