Question 17.SP.5: A large box with a mass m and a flat bottom rests on two ide...

ANALYSIS:
Conservation of Energy.

$T_1  +  V_{\mathrm{g}_1}  +  V_{\mathrm{e}_1}=T_2  +  V_{\mathrm{g}_2}  +  V_{\mathrm{e}_2}$              (1)

You need to calculate the energy at position 1 and position 2.

Potential Energy. Because there is no spring in the system,  $V_{\mathrm{e}_1}=V_{\mathrm{e}_2}=0$.  If you place your datum at the center of mass of each object when the system is at position 2, you have  $V_{\mathrm{g}_2}=0$.  The vertical distance a roller moves is  $h=r \theta \sin \phi$,  so

$V_{g_1}=m g(2 h)  +  2\left(\frac{m}{2}\right) g(h)=3 m g h=3 m g r \theta \sin \phi$

Kinetic Energy. The velocity in position 1 is zero, so  $T_1=0$.  At position 2,

$T_2=2\left(\frac{1}{2} m_R \nu_R^2  +  \frac{1}{2} \bar{I} \omega^2\right)  +  \frac{1}{2} m \nu_B^2$

where  $m_R$  is the mass of the roller and  $\bar{I}$  is the mass moment of inertia of the roller about its center of gravity. Substituting  $\nu_B=2 \omega r, \nu_R=\omega r  ,$,  and  $\bar{I}=\frac{1}{2} m_R r^2=\frac{1}{2}\left(\frac{m}{2}\right) r^2=\frac{1}{4} m r^2 \text { into } T_2$  gives

$T_2=\frac{11}{4} m r^2 \omega^2$

Substituting these expressions into Eq. (1), you find

$0  +  3 m g r \theta \sin \phi  +  0=\frac{11}{4} m r^2 \omega^2  +  0  +  0$

Solving for the angular velocity,  $\omega=\sqrt{\frac{12 g \theta \sin \phi}{11 r}}$,  so the velocity of the box at  $\nu_B=2 \omega r$  is

$\nu_B=4 \sqrt{\frac{3}{11} g r \theta \sin \phi}$

REFLECT and THINK: If the rollers had been attached to the box by brackets, they would have traveled the same vertical distance as the box and the change in height of the centers of gravity of rollers and of the box would have been equal.