A thin rod (length = 1.05 m) is oriented vertically, with its bottom end attache

Question

A thin rod (length = 1.05 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

Explanation / Answer

a) Apply conservation of energy

initial potentail energy = final kinetic energy

m*g*L = 0.5*I*w^2

m*g*L = 0.5*m*L^2*w^2

==> w^2 = 2*g/L

w = sqrt(2*g/L)

= sqrt(2*9.8/1.05)

= 4.32 rad/s

b) Let alfa is the angular acceleration.

Apply Torque = I*alfa

m*g*L = m*L^2*alfa

g = L*alfa

==> alfa = g/L

= 9.8/1.05

= 9.33 rad/s^2

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