Question 11.8: Disk and Stick Collision A 2.0-kg disk traveling at 3.0 m/s ...
Disk and Stick Collision
A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick of length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of Figure 11.11a. The disk strikes at the endpoint of the stick, at a distance r = 2.0 m from the stick’s center. Assume the collision is elastic and the disk does not deviate from its original line of motion. Find the translational speed of the disk, the translational speed of the stick, and the angular speed of the stick after the collision. The moment of inertia of the stick about its center of mass is 1.33 kg . m².
Conceptualize Examine Figure 11.11a and imagine what happens after the disk hits the stick. Figure 11.11b shows what you might expect: the disk continues to move at a slower speed, and the stick is in both translational and rotational motion. We assume the disk does not deviate from its original line of motion because the force exerted by the stick on the disk is parallel to the original path of the disk.
Categorize Because the ice is frictionless, the disk and stick form an isolated system in terms of momentum and angular momentum. Ignoring the sound made in the collision, we also model the system as an isolated system in terms of energy. In addition, because the collision is assumed to be elastic, the kinetic energy of the system is constant.
Analyze First notice that we have three unknowns, so we need three equations to solve simultaneously.
Apply the isolated system model for momentum to the system and then rearrange the result:
\Delta \overrightarrow{p}_{\text {tot }}=0 \rightarrow\left(m_d v_{d f}+m_s v_s\right)-m_d v_{d i}=0(1) m_d\left(v_{d i}-v_{d f}\right)=m_s v_s
Apply the isolated system model for angular momentum to the system and rearrange the result. Use an axis passing through the center of the stick as the rotation axis so that the path of the disk is a distance r from the rotation axis:
\Delta \overrightarrow{L}_{\text {tot }}=0 \rightarrow\left(-r m_d v_{d f}+I \omega\right)-\left(-r m_d v_{d i}\right)=0(2) -r m_d\left(v_{d i}-v_{d f}\right)=I \omega
Apply the isolated system model for energy to the system, rearrange the equation, and factor the combination of terms related to the disk:
\Delta K=0 \rightarrow\left(\frac{1}{2} m_d v_{d f}{}^2+\frac{1}{2} m_s v_s{}^2+\frac{1}{2} I \omega^2\right)-\frac{1}{2} m_d v_{d i}{}^2=0(3) m_d\left(v_{d i}-v_{d f}\right)\left(v_{d i}+v_{d f}\right)=m_s v_s{}^2+I \omega^2
Multiply Equation (1) by r and add to Equation (2):
\begin{aligned}r m_d\left(v_{d i}-v_{d f}\right) & =r m_s v_s \\-r m_d\left(v_{d i}-v_{d f}\right) & =I \omega \\0 & =r m_s v_s+I \omega\end{aligned}Solve for ω:
(4) \omega=-\frac{r m_s v_s}{I}
Divide Equation (3) by Equation (1):
\frac{m_d\left(v_{d i}-v_{d f}\right)\left(v_{d i}+v_{d f}\right)}{m_d\left(v_{d i}-v_{d f}\right)}=\frac{m_s v_s{}^2+I \omega^2}{m_s v_s}(5) v_{d i}+v_{d f}=v_s+\frac{I \omega^2}{m_s v_s}
Substitute Equation (4) into Equation (5):
(6) v_{d i}+v_{d f}=v_s\left(1+\frac{r^2 m_s}{I}\right)
Substitute v_{d f} from Equation (1) into Equation (6):
v_{d i}+\left(v_{d i}-\frac{m_s}{m_d} v_s\right)=v_s\left(1+\frac{r^2 m_s}{I}\right)Solve for v_s and substitute numerical values:
\begin{aligned} v_s&=\frac{2 v_{d i}}{1+\left(m_s / m_d\right)+\left(r^2 m_s / I\right)} \\ &=\frac{2(3.0 m/s)}{1+(1.0 kg / 2.0 kg)+\left[(2.0 m)^2(1.0 kg) / 1.33 kg \cdot m^2\right]}=1.3 m/s\end{aligned}Substitute numerical values into Equation (4):
\omega=-\frac{(2.0 m)(1.0 kg)(1.3 m/s)}{1.33 kg \cdot m^2}=-2.0 \text{ rad} / sSolve Equation (1) for v_{d f} and substitute numerical values:
v_{d f}=v_{d i}-\frac{m_s}{m_d} v_s=3.0 m/s-\frac{1.0 kg}{2.0 kg}(1.3 m/s)=2.3 m/sFinalize These values seem reasonable. The disk is moving more slowly after the collision than it was before the collision. The stick has a small translational speed and is rotating clockwise. Table 11.1 summarizes the initial and final values of variables for the disk and the stick, and it verifies the conservation of linear momentum, angular momentum, and kinetic energy for the isolated system.
| Table 11.1 Comparison of Values in Example 11.8 Before and After the Collision | ||||||
| v (m/s) | ω (rad/s) | p (kg . m/s) | L (kg . m²/s) | K_{\text{trans}} (J) | K_{\text{rot}} (J) | |
| Before | ||||||
| Disk | 3.0 | — | 6.0 | -12 | 9.0 | — |
| Stick | 0 | 0 | 0 | 0 | 0 | 0 |
| Total for system | — | — | 6.0 | -12 | 9.0 | 0 |
| After | ||||||
| Disk | 2.3 | — | 4.7 | -9.3 | 5.4 | — |
| Stick | 1.3 | -2.0 | 1.3 | -2.7 | 0.9 | 2.7 |
| Total for system | — | — | 6.0 | -12 | 6.3 | 2.7 |
| Note: Linear momentum, angular momentum, and total kinetic energy of the system are all conserved. | ||||||