(Solved): We can calculate the angular momentum and kinetic energy of this object in two different ways, by ...

We can calculate the angular momentum and kinetic energy of this object in two different ways, by treating the object as two separate balls, or as one barbell. I: Treat the object as two separate balls (a) What is the speed of ball 1 ? $?\vec{v}?=\mathrm{m}/\mathrm{s}$ (b) Calculate the translational angular momentum ${\vec{I}}_{\text{trans, }},A$ of just one of the balls (ball 1 ). $?{\vec{I}}_{\text{trans, }},A?=\mathrm{kg}\hat{A}?{\mathrm{m}}^2/5$ zero magnitude; no direction into page out of page (c) Calculate the translational angular momentum ${\vec{I}}_{\text{trans, }},A$ of the other ball (ball 2). $?{\vec{I}}_{\text{trans, }},A?=\mathrm{kg}\hat{A}?{\mathrm{m}}^2/5$ out of page zero magnitude; no direction into page (d) By adding the translational angular momentum of ball 1 and the translational angular momentum of ball 2 , calculate the total angular momentum of the barbell, ${\vec{L}}_{\text{tot, }}$ A. $?{\vec{L}}_{\text{tot, },A}?=\mathrm{kg}\mathrm{A}?{\mathrm{m}}^2/\mathrm{s}$ O out of page zero magnitude; no direction $?$ into page (e) Calculate the trans/ational kinetic energy of ball 1 . $K_{\text{trans, }1}=\frac{1}{2}m?\vec{v}{?}^2=$ (f) Calculate the trans/ational kinetic energy of ball 2 . $K_{\text{trans,2 }}=\frac{1}{2}m?\vec{v}{?}^2=$ (g) By adding the translational kinetic energy of ball 1 and the translational kinetic energy of ball 2, calculate the total kinetic energy of the barbell. $K_{\text{total }}=\mathrm{J}$ II: Treat the object as one barbell (h) Calculate the moment of inertia $I$ of the barbell. $I=\mathrm{kg}\hat{A}?{\mathrm{m}}^2$ (i) What is the direction of the angular velocity vector $\overrightarrow{?}$ ? out of page $?$ into page zero magnitude; no direction (j) Use the moment of inertia $I$ and the angular speed $?\overrightarrow{?}?=120\mathrm{rad}/\mathrm{s}$ to calculate the rotational angular momentum of the barbell: $?{\vec{I}}_{\text{rot }}?=I?\overrightarrow{?}?=\mathrm{kg}\mathrm{A}?{\mathrm{m}}^2/\mathrm{s}$ $O_{\text{out }}$ of page zero magnitude; no direction $?$ into page (k) How does this value, $?{\vec{L}}_{\text{rot }}?$, compare to the angular momentum $?{\vec{L}}_{\text{tot, }}$ A $?$ calculated earlier by adding the translational angular momenta of the two balls? $\begin{array}{l}?{\vec{L}}_{\text{rot }}?<?{\vec{L}}_{\text{tot, }A}?\\ ?{\vec{L}}_{\text{rot }}?=?{\vec{L}}_{\text{tot, }A}?\\ ?{\vec{L}}_{\text{rot }}?>?{\vec{L}}_{\text{tot },A}?\end{array}$ (I) Use the moment of inertia $I$ and the angular speed $?\overrightarrow{?}?=120\mathrm{rad}/\mathrm{s}$ to calculate the rotational kinetic energy of the barbell: $K_{\mathrm{rot}}=\frac{1}{2}I{?}^2=$ (m) How does this value, $K_{\text{rot }}$, compare to the kinetic energy $K_{\text{total }}$ calculated earlier by adding the translational kinetic energies of the two balls? $\begin{array}{l}{K}_{\text{rot }}>K_{\text{total }}\\ K_{\text{rot }}=K_{\text{total }}\\ K_{\text{rot }}<K_{\text{total }}\end{array}$