Answer:
Approximately [tex]3.5 \times 10^{-4}\; {\rm J}[/tex], assuming that the density of the apple is uniform.
Explanation:
The rotational kinetic energy of an object is [tex](1/2)\, I\, \omega^{2}[/tex], where [tex]I[/tex] is the moment of inertia and [tex]\omega[/tex] is the angular velocity of the motion.
Let [tex]m[/tex] denote the mass of the sphere, and let [tex]r[/tex] denote the radius of the sphere.
If the apple rotates along an axis that goes through the center of mass, the moment of inertia would be [tex](2/5)\, m\, r^{2}[/tex]. However, in this question, the actual axis of rotation is parallel to the axis that goes through the center of mass, but at a distance of [tex]d = (r/2)[/tex] ("halfway out") from that axis. The actual moment of inertia would be greater than [tex](2/5)\, m\, r^{2}[/tex].
Apply the parallel-axis theorem to find the exact value for the moment of inertia in this scenario. By the parallel-axis theorem, the moment of inertia would be the sum of:
- [tex](2/5)\, m\, r^{2}[/tex], representing the moment of inertia along the axis of rotation that goes through the center of mass, and
- [tex]m\, d^{2}[/tex] where [tex]d = (r/2)[/tex] is the distance between the actual axis of rotation and the parallel axis that goes through the center of mass. This term accounts for the additional moment of inertia from rotation along an axis that is off-center.
Hence, the moment of inertia along the off-center axis would be:
[tex]\begin{aligned} I &= \frac{2}{5}\, m\, r^{2} + m\, \left(\frac{1}{2}\, r\right)^{2} = \frac{13}{20}\, m\, r^{2}\end{aligned}[/tex].
Make sure that all quantities are measured in standard units:
- [tex]m = 270\; {\rm g} = 0.270\; {\rm kg}[/tex],
- [tex]r = 3\; {\rm cm} = 0.03\; {\rm m}[/tex], and
- Angular velocity should be measured in radians per second:
[tex]\begin{aligned}\omega &= 20\; \text{revolution} \cdot \text{minute}^{-1} \times \frac{2\, \pi}{1\; \text{revolution}} \times \frac{1\; \text{minute}}{60\; {\text{second}}} \\ &= \frac{2}{3}\, \pi\; {\rm s^{-1}}\end{aligned}[/tex].
Substitute the values into the expression to find the rotational kinetic energy:
[tex]\begin{aligned}\frac{1}{2}\, I\, \omega^{2} &= \frac{1}{2}\, \left(\frac{13}{20}\, m\, r^{2}\right)\, \omega^{2} \\ &= \frac{1}{2}\, \left(\frac{13}{20}\, (0.270\; {\rm kg})\, (0.03\; {\rm m})^{2}\right)\, \left(\frac{2\, \pi}{3}\; {\rm s^{-1}}\right)^{2} \\ &\approx 3.5\times 10^{-4}\; {\rm J}\end{aligned}[/tex].
