Chapter 10 Problem 68

Problem

A 4.00-kg mass and a 3.00-kg mass are attached to opposite ends of a thin 42.0-cm-long horizontal rod. The system is rotating at angular speed $\omega =5.60$ rad/s about a vertical axle at the center of the rod. Determine

(a) the kinetic energy K of the system, and

(b) the net force on each mass.

(c) Repeat parts (a) and (b) assuming that the axle passes through the CM of the system.

Solution

(a)

Let A represent the heavier mass, and B the lighter mass. The rod has negligible mass thus negligible kinetic energy.

$K={\frac {1}{2}}I_{A}\omega _{A}^{2}+{\frac {1}{2}}I_{B}\omega _{B}^{2}={\frac {1}{2}}m_{A}r_{A}^{2}\omega ^{2}+{\frac {1}{2}}m_{B}r_{B}^{2}\omega ^{2}={\frac {1}{2}}r^{2}\omega ^{2}(m_{A}+m_{B})$

${\frac {1}{2}}(0.210{\text{m}})^{2}(5.60{\text{rad/s}})^{2}(7.0{\text{kg}})=4.84{\text{J}}$

(b)

$F_{A}=m_{A}r_{A}\omega _{A}^{2}=(4.0{\text{kg}})(0.210{\text{m}})(5.60{\text{rad/s}})^{2}=26.3{\text{N}}$

$F_{B}=m_{B}r_{B}\omega _{B}^{2}=(3.0{\text{kg}})(0.210{\text{m}})(5.60{\text{rad/s}})^{2}=19.8{\text{N}}$

(c)

$x_{\text{CM}}={\frac {m_{A}x_{A}+m_{B}x_{B}}{m_{A}+m_{B}}}={\frac {(4.0{\text{kg}})(0)+(3.0{\text{kg}})(0.420{\text{m}})}{7.0{\text{kg}}}}=0.180{\text{m}}$ (from mass A)

$K={\frac {1}{2}}I_{A}\omega _{A}^{2}+{\frac {1}{2}}I_{B}\omega _{B}^{2}={\frac {1}{2}}m_{A}r_{A}^{2}\omega ^{2}+{\frac {1}{2}}m_{B}r_{B}^{2}\omega ^{2}={\frac {1}{2}}\omega ^{2}(m_{A}r_{A}^{2}+m_{B}r_{B}^{2})$

$={\frac {1}{2}}(5.60{\text{rad/s}})^{2}\left[(4.0{\text{kg}})(0.180{\text{m}})^{2}+(3.0{\text{kg}})(0.240{\text{m}})^{2}\right]=4.74{\text{J}}$

$F_{A}=m_{A}r_{A}\omega _{A}^{2}=(4.0{\text{kg}})(0.180{\text{m}})(5.60{\text{rad/s}})^{2}=22.6{\text{N}}$

$F_{B}=m_{B}r_{B}\omega _{B}^{2}=(3.0{\text{kg}})(0.240{\text{m}})(5.60{\text{rad/s}})^{2}=22.6{\text{N}}$

Note that the horizontal forces are now equal, and so there will be no horizontal force on the rod or axle.