To solve this problem, we need to apply the principles of energy conservation, considering both translational and rotational kinetic energy. The total mechanical energy of the sphere is conserved because no external work is done (i.e., there is no slipping, and friction is static).

Step 1: Initial energy at the top of the ramp

At the top of the ramp, the sphere has both translational and rotational kinetic energy due to its motion across the horizontal surface. The total mechanical energy is the sum of the translational kinetic energy and the rotational kinetic energy:

$E_{\text{initial}} = K_{\text{trans}} + K_{\text{rot}}$

The translational kinetic energy is given by:

$K_{\text{trans}} = \frac{1}{2} m v^2$

The rotational kinetic energy for a solid sphere is:

$K_{\text{rot}} = \frac{1}{2} I \omega^2$

For a solid sphere, the moment of inertia $I$ is:

$I = \frac{2}{5} m r^2$

The angular velocity $\omega$ is related to the translational velocity $v$ by the no-slip condition:

$v = \omega r \quad \text{or} \quad \omega = \frac{v}{r}$

Thus, the rotational kinetic energy becomes:

$K_{\text{rot}} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2$

So, the total initial energy is:

$E_{\text{initial}} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2$

Step 2: Final energy at the bottom of the ramp

When the sphere reaches the bottom of the ramp, it will have both translational and rotational kinetic energy again, but we also need to account for the gravitational potential energy it loses as it rolls down the ramp. The gravitational potential energy lost is converted into kinetic energy.

The height $h$ of the ramp is related to its length $x$ and angle $\theta$ by:

$h = x \sin \theta = 3.00 \, \text{m} \times \sin 30.0^\circ = 1.50 \, \text{m}$

The potential energy lost is:

$U = mgh$

At the bottom, all of the potential energy is converted into kinetic energy. So, the total energy at the bottom is:

$E_{\text{final}} = \frac{7}{10} m v_f^2$

By conservation of energy, the initial energy plus the potential energy lost equals the final kinetic energy:

$E_{\text{initial}} + U = E_{\text{final}}$

Substituting:

$\frac{7}{10} m v_i^2 + mgh = \frac{7}{10} m v_f^2$

Step 3: Solving for the final velocity

Canceling $m$ from the equation and solving for $v_f$:

$\frac{7}{10} v_i^2 + gh = \frac{7}{10} v_f^2$

$v_f^2 = v_i^2 + \frac{10}{7} gh$

Now, substitute the known values:

• $v_i = 3.00 \, \text{m/s}$
• $g = 9.81 \, \text{m/s}^2$
• $h = 1.50 \, \text{m}$

$v_f^2 = (3.00)^2 + \frac{10}{7} \times 9.81 \times 1.50$

$v_f^2 = 9.00 + \frac{10}{7} \times 14.715$

$v_f^2 = 9.00 + 21.021$

$v_f^2 = 30.021$

$v_f = \sqrt{30.021} \approx 5.48 \, \text{m/s}$

Final Answer:

The translational speed of the sphere when it reaches the bottom of the ramp is approximately:

$v_f = 5.48 \, \text{m/s}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the moment of inertia affect the rotational kinetic energy?
2. What happens if the sphere slips instead of rolling without slipping?
3. How would the result change if the ramp were longer or steeper?
4. Can you explain the energy distribution between rotational and translational kinetic energy?
5. How do we derive the moment of inertia for a solid sphere?

Tip: Always check whether objects are rolling or sliding, as this influences the energy calculations.