RBD BYU 29

1. Force Equilibrium

Resolving forces horizontally and vertically. Let $\theta$ be the angle with the vertical.

Vertical (Balancing weight):

$$ f \sin\theta + N \cos\theta = mg \quad \dots(1) $$

Horizontal (Balancing normal component):

$$ f \cos\theta = N \sin\theta $$

From the horizontal equation, we find the relation between friction ($f$) and Normal force ($N$):

$$ N = \frac{f \cos\theta}{\sin\theta} = f \cot\theta $$

2. Solving for Friction

Substitute $N = f \cot\theta$ into equation (1):

$$ f \sin\theta + (f \cot\theta)\cos\theta = mg $$ $$ f \sin\theta + f \frac{\cos^2\theta}{\sin\theta} = mg $$ $$ f \left( \frac{\sin^2\theta + \cos^2\theta}{\sin\theta} \right) = mg $$ $$ f \left( \frac{1}{\sin\theta} \right) = mg \implies f = mg \sin\theta $$

3. Rotational Dynamics

For the small cylinder of radius $r$ and mass $m$, the torque $\tau$ about its center is provided by friction:

$$ \tau = I \alpha’ $$ $$ f \cdot r = \left(\frac{mr^2}{2}\right) \alpha’ $$

Where $\alpha’$ is the angular acceleration of the small cylinder.

4. Constraint Equation

For rolling without slipping (or matching the acceleration at the contact point driven by the outer shell), the tangential acceleration must match:

$$ \alpha R = \alpha’ r \implies \alpha’ = \frac{\alpha R}{r} $$

5. Final Calculation

Substitute the value of $f$ (from step 2) and $\alpha’$ (from step 4) into the torque equation:

$$ (mg \sin\theta) r = \frac{mr^2}{2} \left( \frac{\alpha R}{r} \right) $$

Simplify terms:

$$ mg \sin\theta \cdot r = \frac{1}{2} m r (\alpha R) $$

Cancel $m$ and $r$ from both sides:

$$ g \sin\theta = \frac{\alpha R}{2} $$