Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of

Question

Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through its lowest position?

Explanation / Answer

(1) initial total energy = final total energy The initial total energy is zero however so (2) 0 = final total energy The final total energy is the potential energy (PE) of each particle plus the rotational kinetic energy (KE) of each particle thus (3) 0 = PE(1) + PE(2) + KE(1) + KE(2) If particle 1 is at 50cm (=.5m) and particle 2 is at 80cm (=.8m) then (4) PE(1) = -Mg.5m PE(2) = -Mg.8m Notice the minus sign. This is because the particles are BELOW the starting point which we have taken to be zero PE ---> 0 height level. The rotational KE of a particle of mass M at radius r is (5) KE = .5Mr^2 w^2 (note: ^2 means "squared") Where w is the angular rotation rate in radians per unit time. Thus (6) KE(1) = .5M(.5m)^2 w^2 KE(2) = .5M(.8m)^2 w^2 M is the same for both expressions and since both are on the same rigid stick then w is the same for both too. Substituting (4) and (6) into (3) gives (7) 0 = -Mg.5m -Mg.8m + .5M(.5m)^2 w^2 + .5M(.8m)^2 w^2 Dividing through by M gives (8) 0 = -g.5m - g.8m + .5(.5m)^2 w^2 + .5(.8m)^2 w^2 or taking -g.5m - g.8m =-(1.3m)g to the other side of the equation and factoring w^2 gives us (9) (1.3m)g = [.5(.5m)^2 + .5(.8m)^2]w^2 (1.3m)g = (.445m^2) w^2 w^2 = (1.3m)g/(.445m^2) =(1.3m)(9.8m/s^2)/(.445m^2) = 28.6 s^-2 w = sqrt(28.6 s^-2) ~ 5.4 rad/s *

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