Chapter 8, Problem 60P

To determine

The angular frequency of the combination of uniform disk and rod.

Answer to Problem 60P

The angular frequency of the combination of uniform disk and rod is $1.44\text{rev/s}$

Explanation of Solution

Given:

Initial angular velocity of disk, ${\omega }_1=2.4\text{rev/s}$

Initial angular velocity of rod, ${\omega }_2=0\text{rad/s}$

Mass of disk and rod each, $m_1=m_2=m$

Length of rod, $L$

Diameter of disk, $D=L$

Radius of disk, $R=\frac{D}{2}=\frac{L}{2}$

Formula used:

Moment of inertia of rod, $I=\frac{M{L}^2}{12}$

Moment of inertia of disk, $I=\frac{M{R}^2}{2}$

Law of conservation of angular momentum, $I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega \text{'}$

Calculation:

Substituting values for moment of inertia of rod,

  $I_2=\frac{M{L}^2}{12}$

Substituting values for moment of inertia of disk,

  $I_1=\frac{M{\left(\frac{L}{2}\right)}^2}{2}=\frac{M{L}^2}{8}$

Substituting values in law of conservation of angular momentum,

  $\begin{array}{l}\left(\frac{M{L}^2}{8}\right)(2.4)+\left(\frac{M{L}^2}{12}\right)(0)=\left(\frac{M{L}^2}{8}+\frac{M{L}^2}{12}\right)\omega \text{'}\\ \omega \text{'}=\frac{\left(\frac{M{L}^2}{8}\right)(2.4)}{\frac{5M{L}^2}{24}}\\ \omega \text{'}=\frac{(2.4)(3)}{5}\\ \omega \text{'}=1.44\text{rev/s}\end{array}$

Conclusion:

The angular frequency of the combination of uniform disk and rod is $1.44\text{rev/s}$