Question 17.SP.5: A large box with a mass m and a flat bottom rests on two ide...
A large box with a mass m and a flat bottom rests on two identical homogeneous cylindrical rollers, where each has radius r and a mass half that of the crate. The system is released from rest on a plane that is inclined at angle f to the horizontal. Determine the speed of the box at the instant when the rollers have turned through an angle θ. Neglect rolling resistance and assume that the rollers do not slide.
STRATEGY: You are interested in the velocity after the rollers have moved a specified distance, rθ, and the friction force in rolling motion does no work, so use the conservation of energy.
MODELING: Choose the box and the two cylindrical rollers as your system and model them as rigid bodies. In order for you to draw the system in its initial and final positions, you need to know how far each mass travels. You can determine this by using the instantaneous center of rotation. The rollers do not slide, so the instantaneous center of rotation for each roller is located at the point of contact, C, between the roller and the ground (Fig. 1). Using this instantaneous center of velocity, you know \nu_B=2 \omega r and \nu_R=\omega r. Therefore, the box moves down a distance 2h when the rollers move a distance h (Fig. 2). Since you have three masses in the system (the two rollers and the box), you may define an individual datum for each mass to simplify the calculation of the gravitational potential energy.
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.