To solve this problem, let’s analyze the forces and the motion of the spherical shell. Given that it’s rolling without slipping, both translational and rotational motion are involved. Here’s the step-by-step solution:

Given Data:

• Mass of the shell, $m = 2.05 \, \text{kg}$
• Angle of incline, $\theta = 40.0^\circ$
• Moment of inertia for a hollow spherical shell: $I = \frac{2}{3} m r^2$

Goal:

Find the acceleration $a_{\text{cm}}$ of the center of mass of the shell.

Solution:

1. Determine the forces:

   • The gravitational force acting on the shell is $mg$, where $g \approx 9.81 \, \text{m/s}^2$.
   • The component of gravitational force along the slope is $mg \sin \theta$.
   • The frictional force $f$ provides the torque needed for rotational motion but does not do work on the center of mass since it rolls without slipping.

2. Apply Newton's Second Law for Translation: The net force along the slope: $F_{\text{net}} = mg \sin \theta - f = m a_{\text{cm}}$

3. Apply Newton's Second Law for Rotation: The torque $\tau$ around the center of mass due to friction $f$ is: $\tau = f r = I \alpha$ where $\alpha$ is the angular acceleration and $r$ is the radius of the shell. Since $I = \frac{2}{3} m r^2$, we have: $f r = \frac{2}{3} m r^2 \alpha$ Simplifying, we get: $f = \frac{2}{3} m r \alpha$

4. Relate Linear and Angular Acceleration: For rolling without slipping, $a_{\text{cm}} = r \alpha$. Thus, $\alpha = \frac{a_{\text{cm}}}{r}$.

5. Substitute $\alpha$ into the Friction Force: Substitute $\alpha = \frac{a_{\text{cm}}}{r}$ into the equation for $f$: $f = \frac{2}{3} m a_{\text{cm}}$

6. Combine Equations: Substitute $f = \frac{2}{3} m a_{\text{cm}}$ into the translational equation: $mg \sin \theta - \frac{2}{3} m a_{\text{cm}} = m a_{\text{cm}}$ Dividing through by $m$ and rearranging terms: $g \sin \theta = a_{\text{cm}} + \frac{2}{3} a_{\text{cm}}$ $g \sin \theta = \frac{5}{3} a_{\text{cm}}$ $a_{\text{cm}} = \frac{3}{5} g \sin \theta$

7. Calculate $a_{\text{cm}}$: Substitute $g = 9.81 \, \text{m/s}^2$ and $\theta = 40.0^\circ$: $a_{\text{cm}} = \frac{3}{5} \times 9.81 \times \sin 40^\circ$ $a_{\text{cm}} = \frac{3}{5} \times 9.81 \times 0.6428$ $a_{\text{cm}} \approx \frac{3}{5} \times 6.303 \approx 3.78 \, \text{m/s}^2$

Answer:

The magnitude of the acceleration $a_{\text{cm}}$ of the center of mass of the spherical shell is approximately: $\boxed{3.78 \, \text{m/s}^2}$

Would you like further details on any step or additional questions?

1. How does the moment of inertia affect the rolling motion?
2. Why does rolling without slipping mean $a_{\text{cm}} = r \alpha$?
3. What would change if it were a solid sphere instead of a hollow one?
4. How does the incline angle influence the acceleration?
5. How is energy distributed in rolling motion?

Tip: Always check if a problem involves "rolling without slipping," as this connects linear and rotational motion through $a = r \alpha$.