Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 19 - Planar Kinetics of a Rigid Body: Impulse and Momentum - Section 19.4 - Eccentric Impact - Problems - Page 551: 42


$\omega_2=57 rad/s$ $U_F=366.66J$

Work Step by Step

The required angular velocity and the work done can be determined as follows: According to the principle of work and energy $T_1+\Sigma U_{1\rightarrow 2}=T_2$ [eq(1)] We know that $T_1=\frac{1}{2}I_{z_1}\omega_1^2$ $\implies I_{z_1}=2(\frac{ml^2}{12}+mr^2)$ $\implies I_{z_1}=2(\frac{10(0.6)^2}{12}+10(0.3+0.1)^2)$ $\implies I_{z_1}=3.8Kg/m^2$ Now $T_1=\frac{1}{2}(3.8)(3)^2=17.1J$ Similarly $\Sigma U_{1\rightarrow 2}=U_W+U_F$ $\implies U_W=2(-mgh)=2[-10(9.81)(0.3)]=-58.86J$ $\implies \Sigma U_{1\rightarrow 2}=-58.86+U_F$ We also know that $T_2=\frac{1}{2}I_{z_2}\omega_2^2$ $\implies I_{z_2}=2(\frac{ml^2}{12}+mr^2)$ $\implies I_{z_2}=2[\frac{10(0)^2}{12}+10(0.1)^2]$ $\implies I_{z_2}=0.2Kg/m^2$ and $T_2=\frac{1}{2}(0.2)(\omega_2)^2$ $\implies T_2=0.1\omega_2^2$ [eq(2)] According to the conservation of angular momentum $I_{z_1}\omega_1=I_{z_2}\omega_2$ We plug in the known values to obtain: $3.8(3)=0.2\omega_2$ This simplifies to: $\omega_2=57 rad/s$ We plug in this value in eq(2) to obtian: $T_2=0.1(57)^2=324.9J$ Now from eq(1), we obtain: $17.1-58.86+U_F=324.9$ This simplifies to: $U_F=366.66J$
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