To solve for the velocity of the center of mass of the solid sphere at the bottom of the incline, we can apply the principle of conservation of energy, considering both translational and rotational energy as the sphere rolls without slipping.

Step 1: Setting up the energy conservation equation

The sphere starts from rest, so initially, it only has potential energy due to its height on the incline.

1. Potential energy at the top of the incline (PE):
   $PE = mgh$ where $m = 4.00 \ \text{kg}$, $g = 9.8 \ \text{m/s}^2$, and $h$ is the height of the incline.

2. Finding $h$:
   Since the incline makes an angle of $30.0^\circ$ and has a length of $50.0 \ \text{cm} = 0.5 \ \text{m}$: $h = L \sin \theta = 0.5 \times \sin(30^\circ) = 0.5 \times 0.5 = 0.25 \ \text{m}$

   So, $PE = mgh = 4.00 \times 9.8 \times 0.25 = 9.8 \ \text{J}$

3. Total kinetic energy at the bottom (KE):
   At the bottom, the sphere has both translational kinetic energy ($KE_{\text{trans}}$) and rotational kinetic energy ($KE_{\text{rot}}$): $KE = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ where $v$ is the linear velocity of the center of mass and $\omega$ is the angular velocity.

4. Moment of inertia for a solid sphere (I):
   For a solid sphere, $I = \frac{2}{5}mr^2$.

5. Relating $\omega$ and $v$:
   Since the sphere rolls without slipping, $v = r\omega$ or $\omega = \frac{v}{r}$.

Step 2: Substitute $I$ and $\omega$ into the energy equation

Substituting $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$ into the equation for total kinetic energy: $KE = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left( \frac{v}{r} \right)^2$

Simplifying: $KE = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

Step 3: Set up the conservation of energy equation

Using conservation of energy $PE = KE$, we have: $9.8 = \frac{7}{10} \cdot 4.00 \cdot v^2$

Solving for $v$: $v^2 = \frac{9.8}{\frac{7}{10} \cdot 4.00} = \frac{9.8}{2.8} = 3.5$ $v = \sqrt{3.5} \approx 1.87 \ \text{m/s}$

Answer

The correct answer is:

1.87 m/s

Would you like further explanation, or do you have any questions?

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Here are 5 related questions to deepen your understanding:

1. How does the angle of incline affect the velocity of the sphere at the bottom?
2. What would happen to the final velocity if the sphere rolled on a frictionless incline?
3. How would the result change if we used a hollow sphere instead of a solid one?
4. Why is the rotational inertia different for solid and hollow spheres?
5. How does the radius of the sphere impact its final velocity on an incline?

Tip: When dealing with rolling objects, always account for both translational and rotational kinetic energy in energy conservation calculations.