Acceleration of Rolling Body on Incline

The formula

A body of mass $m$, radius $R$, radius of gyration $k$ rolls without slipping down an incline of angle $\theta$:

$a = \frac{g\sin\theta}{1 + k^2/R^2}$

Derivation

Forces on the rolling body:

• Weight component along incline: $mg\sin\theta$ (down)
• Friction: $f$ (up the incline — static, enforces rolling)
• Normal force: $N = mg\cos\theta$

Translational equation (along incline, down positive):

$mg\sin\theta - f = ma \quad \text{...(1)}$

Rotational equation (about CM, torque from friction):

$fR = I\alpha = mk^2\alpha \quad \text{...(2)}$

Rolling condition:

$a = R\alpha \implies \alpha = \frac{a}{R} \quad \text{...(3)}$

From (2) and (3):

$f = \frac{mk^2 a}{R^2}$

Substitute into (1):

$mg\sin\theta - \frac{mk^2 a}{R^2} = ma$

$g\sin\theta = a\left(1 + \frac{k^2}{R^2}\right)$

$\boxed{a = \frac{g\sin\theta}{1 + k^2/R^2}}$

Comparison with sliding (no friction)

Sliding body: $a_{slide} = g\sin\theta$

Rolling body: $a_{roll} = \frac{g\sin\theta}{1+k^2/R^2} < g\sin\theta$

Rolling is always slower than sliding on the same incline. The rotational inertia "uses up" some of the gravitational force — the friction force that creates the torque is not available for translation.

Friction force required

$f = \frac{mk^2 a}{R^2} = \frac{mk^2 g\sin\theta}{R^2(1+k^2/R^2)} = \frac{mg\sin\theta \cdot k^2/R^2}{1+k^2/R^2}$

For rolling to occur without slipping: $f \leq \mu_s N = \mu_s mg\cos\theta$

$\frac{k^2/R^2}{1+k^2/R^2}\tan\theta \leq \mu_s$

Accelerations for standard bodies (on incline at angle $\theta$)