Chapter 10 Problem 54

Problem

A thin rod of length stands vertically on a table. The rod begins to fall, but its lower end does not slide.

(a) Determine the angular velocity of the rod as a function of the angle $\phi$ it makes with the tabletop.

(b) What is the speed of the tip of the rod just before it strikes the table?

Solution

(a)

The axis of rotation would be coming out of the paper at the point of contact with the floor. There are contact forces between the rod and the table (the friction force and the normal force), but they act through the axis of rotation and so cause no torque. Thus only gravity causes torque. Write Newton’s second law for the rotation of the rod. Take counterclockwise to be the positive direction for rotational quantities. Thus in the diagram, the angle is positive, but the torque is negative.

$\sum \tau =I\alpha =-mg\left({\frac {1}{2}}l\cos \phi \right)={\frac {1}{3}}ml^{2}{\frac {d\omega }{dt}}$

$\rightarrow -{\frac {3g}{2l}}\cos \phi ={\frac {d\omega }{dt}}={\frac {d\omega }{d\phi }}{\frac {d\phi }{dt}}$ $\rightarrow {\frac {3g}{2l}}\cos \phi d\phi =-\omega d\omega$ $\rightarrow {\frac {3g}{2l}}\int _{\pi /2}^{\phi }\cos \phi d\phi =-\int _{0}^{\omega }\omega d\omega$ $\rightarrow {\frac {3g}{2l}}(\sin \phi -1)={\frac {1}{2}}\omega ^{2}$

$\rightarrow \omega ={\sqrt {{\frac {3g}{l}}(1-\sin \phi )}}$

(b)

The speed of the tip is the tangential speed of the tip, since the rod is rotating. At the tabletop, $\phi =0$

$v=\omega l={\sqrt {3gl(1-\sin \phi )}}$ $\rightarrow v(0)={\sqrt {3gl}}$