Assume that the pencil rotates about the point of contact with the floor
without sliding … only two forces act on the pencil … the normal force N
acting vertically upward at the point of contact with the floor and the
weight ( mg ) acting vertically downward at the center of mass of the
pencil … in the absence of friction, no force acts along the horizontal …
according to Newton’s second law for rotational motion …
… torque τ = I α = ( mg ) ( ½ ℓ sin θ ) … where … ℓ = length of the pencil
assumed to be uniform … θ = angle between the vertical line and the pencil
… there is no torque due to the normal force since N passes through the axis
of rotation … with the pencil considered as a thin rod pivoted about one of
its ends, we have … I = ⅓ m ℓ ² … so that … τ = I α = [ ⅓ m ℓ ² ] α … using the
earlier expression for the torque … τ = ( mg ) ( ½ ℓ sin θ ) = [ ⅓ m ℓ ² ] α …
solving for the angular acceleration … α = ( ³ / ₂ ) ( g / ℓ ) sin θ … so that
when θ = 10° … we get … α = ( ³ / ₂ ) ( g / ℓ ) sin 10° …