A thin rod (length = 2.33 m) is oriented vertically, with its bottom end attache

Question

A thin rod (length = 2.33 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

part a )

From the principle of conservation of mechanical energy

Eo = Ef

Initially the system has only gravitational potential energy. If the level of the hinge is chosen as the zero level for measuring heights

Eo = mgh = mgL

when rod hits the floor there is only kinetic energy = mv^2/2

mgL = mv^2/r

v = sqrt(2gL)

we know angular velocity w = v/r

as the object rotates downwards it move in circle of radius of L

w = sqrt(2gL)/L = sqrt(2g/L)

L = 2.33 m

w = 2.9 rad/s

part b )

a = alpha*r

alpha = a/r

r = L

when rod hits the floor there is only gravitational acceleration

alpha = g/L = 4.2 rad/s^2

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