Question 8.15: A BALL ROLLING DOWN AN INCLINE GOAL Combine gravitational, t...
A BALL ROLLING DOWN AN INCLINE
GOAL Combine gravitational, translational, and rotational energy.
PROBLEM A uniform, solid ball of mass M and radius R starts from rest at a height of h=2.00 \mathrm{~m} and rolls down a \theta=30.0^{\circ} slope, as in Figure 8.30. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping.
STRATEGY The two points of interest are the top and bottom of the incline, with the bottom acting as the zero point of gravitational potential energy. As the ball rolls down the ramp, gravitational potential energy is converted into both translational and rotational kinetic energy without dissipation, so conservation of mechanical energy can be applied with the use of Equation 8.16.
\left(K E_{t}+K E_{r}+P E\right)_{i}=\left(K E_{t}+K E_{r}+P E\right)_{f} [8.16]
Apply conservation of energy with P E=P E_{g}, the potential energy associated with gravity:
\left(K E_{t}+K E_{r}+P E_{g}\right)_{i}=\left(K E_{t}+K E_{r}+P E_{g}\right)_{f}
Substitute the appropriate general expressions, noting that \left(K E_{t}\right)_{i}=\left(K E_{r}\right)_{i}=0 and \left(P E_{g}\right)_{f}=0 (obtain the moment of inertia of a ball from Table 8.1):
| Table 8.1 moments of Inertia for Various rigid objects of uniform Composition | |||
| Hoop or thin cylindrical shell I = MR² |  |
Solid sphere I = \frac{2}{5}MR² |  |
| Solid cylinder or disk I = \frac{1}{2}MR² |  |
Thin spherical shell I = \frac{2}{3}MR² |  |
| Long, thin rod with rotation axis through center I = \frac{1}{12}ML² |  |
Long, thin rod with rotation axis through end I = \frac{1}{3}ML² |  |
0+0+M g h=\frac{1}{2} M v^{2}+\frac{1}{2}\left(\frac{2}{5} M R^{2}\right) \omega^{2}+0
The ball rolls without slipping, so R \omega=v, the “no-slip condition,” can be applied:
M g h=\frac{1}{2} M v^{2}+\frac{1}{5} M v^{2}=\frac{7}{10} M v^{2}
Solve for v, noting that M cancels.
v=\sqrt{\frac{10 g h}{7}}=\sqrt{\frac{10\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(2.00 \mathrm{~m})}{7}}=5.29 \mathrm{~m} / \mathrm{s}
REMARKS Notice the translational speed is less than that of a block sliding down a frictionless slope, v=\sqrt{2 g h}. That’s because some of the original potential energy must go to increasing the rotational kinetic energy.