Question

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical? A typical pencil has an average length of 15.0 cmand an average mass of 10.0 g . Assume the tip of the pencil does not slip as it falls.

What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical? Express your answer numerically in radians per second squared to three significant figures.

Answer 1

The formula for moment of inertia of a thin rod about an end is , where is the mass and is the length. You have the massand length in grams and centimeters, respectively. Be sure to convert these quantities to the appropriate units before entering them intothe formula.

$I= 7.5\times 10^{-5}~ kg.m^2$

Recall that the magnitude of a torqueit s defined by

$\tau=rFsin\phi$,

where r is the distance from the pivot point to the point of application for the force, F is the magnitude of the force that exerts the torque, and $\phi$ is the angle between the force and the lineconnecting the pivot to the point of application

$F_G=9.8\times 10^{-2}~N$

$\tau_{net}= -1.276\times 10^{-3}~N.m$

The mathematical representation is based on Newton’s second law for rotational motion:

$\tau_{net}= I\alpha\\ \\ \Rightarrow \alpha=\frac{\tau_{net}}{I}= -17.0~rad/s^2$