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Step-by-Step Solution
Step 1: Identify the known quantities
• Mass of the sphere, $m = 500\,\text{g} = 0.5\,\text{kg}$.
• Speed of the center of the sphere, $V = 5.00\,\text{cm/s} = 0.05\,\text{m/s}$.
• The sphere is rolling without slipping.
Step 2: Write the expression for total kinetic energy of a rolling sphere
When a sphere rolls without slipping, its total kinetic energy ($K.E_{\text{total}}$) is the sum of its translational kinetic energy and rotational kinetic energy about its center of mass. Mathematically,
$$
K.E_{\text{total}} = \frac{1}{2} m V^2 + \frac{1}{2} I_{\text{cm}} \omega^2.
$$
Step 3: Use the moment of inertia for a solid sphere
For a solid sphere of mass $m$ and radius $R$, the moment of inertia about its center of mass is
$$
I_{\text{cm}} = \frac{2}{5} m R^2.
$$
Step 4: Relate angular speed $\omega$ to linear speed $V$
Because the sphere rolls without slipping, the linear speed $V$ of the center of mass is related to the angular speed $\omega$ by
$$
V = \omega R \quad \Longrightarrow \quad \omega = \frac{V}{R}.
$$
Step 5: Substitute $I_{\text{cm}}$ and $\omega$ into the kinetic energy expression
Substituting $I_{\text{cm}} = \frac{2}{5} m R^2$ and $\omega = \frac{V}{R}$ into the expression for $K.E_{\text{total}}$:
$$
K.E_{\text{total}}
= \frac{1}{2} m V^2 + \frac{1}{2} \left( \frac{2}{5} m R^2 \right) \left( \frac{V}{R} \right)^2
= \frac{1}{2} m V^2 + \frac{1}{2} \cdot \frac{2}{5} m V^2
= \frac{1}{2} m V^2 + \frac{1}{5} m V^2
= \frac{7}{10} m V^2.
$$
Step 6: Perform the numerical substitution
• $m = 0.5\,\text{kg}$
• $V = 0.05\,\text{m/s}$
Therefore,
$$
K.E_{\text{total}} = \frac{7}{10} \times 0.5 \times \left(0.05\right)^2.
$$
Step 7: Calculate the result
$$
K.E_{\text{total}} = \frac{7}{10} \times 0.5 \times 0.0025
= 0.35 \times 0.0025
= 8.75 \times 10^{-4}\,\text{J}.
$$
Final Answer
The kinetic energy of the rolling sphere is $8.75 \times 10^{-4}\,\text{J}$.
