Four particles, each of mass 0.18 kg, are placed at the vertices of a square wit

Question

Four particles, each of mass 0.18 kg, are placed at the vertices of a square with sides of length 0.53 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

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Four particles, each of mass 0.12 kg, are placed at the vertices of a square with sides of length 0.41 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?

Answer

a)

The moment of intertia (or the rotational inertia) is just thesum of mr2 for each of the three particles thatare not at A. For this, r is the distance from the point toA.

   So:   I = 0.12 *0.412   + 0.12 *0.412   + 0.12 * 0.582 =   0.081 kg m2

b)

Now, for the angular speed. Consider the difference in wherethe particles are in the beginning and when AB is vertical. In thebeginning, one particle is level with A, two particles are a height0.41 above A. When AB is vertical, one particle is level withA and two particles are 0.41 BELOW A

Draw pictures of the initial and final positions to seethis.

The square loses potential energy as it drops, and gains KE.How much PE? Compare the two situations. The difference is that theequivalent of two particles have dropped a total of 0.82 m

So lost PE is   2mg?h = 2 * 0.12 *9.8 * 0.82 = 1.93 Joules

So this must be the KE when AB is vertical. We know

    KE = (1/2) I ?2     or

   ? 2    =   2 K / I = 2 * 1.93 / 0.081 =47.65

so    ?      =   6.90 rad/sec

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