Question

A pencil (mass 12.0 g and length 18 cm) is initially balanced so that it is standing vertically on a flat table. If the pencil then falls, what is its angular velocity just before it strikes the tabletop? Assume the mass of the pencil is distributed uniformly and the end of the pencil in contact with the table does not slip.

Answer 1

we use methods of conservation of energy to solve for this

one the pencil is upright, it has potential energy of mgL where m is the mass of the pencil and L its length

when it falls, just before it hits the table all the PE has converted to KE; the pencil has both translational and rotational KE:

KE=1/2 I w^2 + 1/2 mv^2

I=moment of inertia = 1/3 mL^2 for a rod pivoted at one end
w=angular velocity
v=linear velocity and is related to angular velocity through v= wL

so we have

PE at beginning = KE at end
mgL = 1/2 (1/3 mL^2) w^2 + 1/2 m(w^2 L^2)

cancel through the m's:

gL = 2/3 L^2 w^2

w=sqrt[3g/2L]

for L=0.18m, w=sqrt[3x9.8/0.36]=9.03rad/s