Sure, I'd be happy to help explain this step-by-step.
When a sphere rolls down an inclined plane without slipping, its kinetic energy is a combination of translational kinetic energy and rotational kinetic energy. Let's break it down:
1. Translational Kinetic Energy:
This is the energy due to the motion of the sphere's center of mass. It is given by the formula:
[tex]\[
KE_{\text{trans}} = \frac{1}{2} m v^2
\][/tex]
where [tex]\( m \)[/tex] is the mass of the sphere and [tex]\( v \)[/tex] is the velocity of the center of mass.
2. Rotational Kinetic Energy:
This is the energy due to the rotation of the sphere about its own axis. The formula for rotational kinetic energy is:
[tex]\[
KE_{\text{rot}} = \frac{1}{2} I \omega^2
\][/tex]
where [tex]\( I \)[/tex] is the moment of inertia of the sphere about its center of mass and [tex]\( \omega \)[/tex] is the angular velocity.
Since the sphere rolls without slipping, there is a relationship between the translational velocity [tex]\( v \)[/tex] and the angular velocity [tex]\( \omega \)[/tex]. Specifically, this relationship is given by:
[tex]\[
v = \omega R
\][/tex]
where [tex]\( R \)[/tex] is the radius of the sphere.
Combining both types of kinetic energy, the total kinetic energy ([tex]\( KE_{\text{total}} \)[/tex]) of the rolling sphere is:
[tex]\[
KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}}
\][/tex]
[tex]\[
KE_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2
\][/tex]
Thus, the kinetic energy when a sphere rolls down an inclined plane without slipping is:
[tex]\[
\frac{1}{2} I \omega^2 + \frac{1}{2} m v^2
\][/tex]
So the correct option is:
D. [tex]\(\frac{1}{2} I \omega^2 + \frac{1}{2} m v^2\)[/tex]
