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Planar Kinetics Of A Rigid Body (Work And Energy) 25:06
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$$
\begin{array}{l}{\text { 18-25. The } 30 \text { -kg disk is originally at rest, and the spring is }} \\ {\text { unstretched. A couple moment } M=80 \mathrm{N} \cdot \mathrm{m} \text { is then applied }} \\ {\text { to the disk as shown. Determine how far the center of mass }} \\ {\text { of the disk travels along the plane before it momentarily }} \\ {\text { stops. The disk rolls without slipping. }}\end{array}
$$

$$
m=30 \mathrm{kg}
$$

$$
M=80 \mathrm{N} \cdot \mathrm{m}
$$

$$
\text { kinetic energy }
$$

$$
T_{1}=T_{2}=0
$$

$$
\theta=\frac{S_{G}}{r}=\frac{S_{G}}{0.5}=2 S_{G}
$$

$$
U_{M}=M \theta=80(2S_G) \mathrm{s} 160 \mathrm{S}_{G}
$$

$$
U_{F{sp}}=-\frac{1}{2} k x^{2}=-\frac{1}{2} 200\left(S_ G^{2}\right)=-100{S}_ G^{2}
$$

$$
T_{1}+\Sigma U_{1 \rightarrow 2}=T_{2} \Rightarrow 0+\left(160S_G-100 S_G^{2}\right)=0
$$

$$
160S_G-100S_G^{2}=0
$$

$$
S_G(160-100S_G)=0
$$

$$
160-100S_G=0
$$

$$
S_{G}=1.6 \mathrm{m}
$$

$$
\begin{array}{l}{\text { 18-39. The spool has a mass of } 50 \mathrm{kg} \text { and a radius of gyration }} \\ {\text { of } k_{O}=0.280 \mathrm{m} \text { . If the } 20-\mathrm{kg} \text { block } A \text { is released from rest, }} \\ {\text { determine the velocity of the block when it descends } 0.5 \mathrm{m} .}\end{array}
$$

$$
m_ s=50 k g
$$

$$
k_{0}=0.28 \mathrm{m}
$$

$$
m _b=20 k g
$$

$$
s=0.5 \mathrm{m}
$$

$$
v_{1}=\left(v_{g}\right)_1=w_{A} y_{1}=20(9.81) 0=0
$$

$$
v_{2}=\left(v_{g}\right)_{1}=-w_{A}  y_{2}=-20(9.81)(0.5)=-98.1 \mathrm{J}
$$

$$
\omega=\frac{v_{A}}{r_{A}}=\frac{v_{A}}{0.2}=5 v_{A}
$$

$$
I_{0}=m k_0^{2}=5{0}(0.28)^{2}=3.92 \mathrm{kg}.\mathrm{m}^{2}
$$

$$
T=\frac{1}{2} I_{0} \omega^{2}+\frac{1}{2} m_{A} V_{A}^{2}
$$

$$
=\frac{1}{2}(3.92)(5 v_A)^{2}+\frac{1}{2}(20) v_{A}^{2}
$$

$$
=59 V_{A}^{2}
$$

$$
T_{1}=0
$$

$$
T_{1}+v_{1}=T_{2}+v_{2}
$$

$$
0+0=59 v _A^{2}+(-98.1)
$$

$$
V_{A}=1.289 \simeq 1.29 \mathrm{m} / \mathrm{s}
$$

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