## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_2=9.22ft/s$, $\Sigma U_{1\rightarrow}2=3.04ft\cdot lb$
We can determine the required speed and work done as follows: According to the principle of conservation of angular momentum $m_B(v_B)r_1=m_B(v_{\theta})_2r_2$ $\implies 6\times 3=2(v_{\theta})_2$ $\implies (v_{\theta})_2=9ft/s$ Now we can calculate the speed as $v_2=\sqrt{(v_{\theta})_2^2+(v_r)_2^2}$ We plug in the known values to obtain: $v_2=9.22ft/s$ Now according to the principle of work and energy $\frac{1}{2}m_Bv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}m_Bv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{4}{32.2})(6)^2+\Sigma U_{1\rightarrow 2}=(\frac{4}{32.2})(9.22)^2$ This simplifies to: $\Sigma U_{1\rightarrow}2=3.04ft\cdot lb$