pure rolling acceleration

Question

A uniform solid cylinder (I = (1/2) m r^2) rolls down incline of angle θ without slipping. Linear acceleration is (g=gravity):

RankGuru · Accepted Answer

The linear acceleration $a$ for an object rolling down an incline without slipping is:
$a = \frac{g \sin 	heta}{1 + \beta}$ where $\beta = \frac{I}{m r^2}$
For a solid cylinder, $I = \frac{1}{2} m r^2$, so $\beta = \frac{1}{2}$.
$a = \frac{g \sin 	heta}{1 + 1/2} = \frac{g \sin 	heta}{3/2} = \frac{2}{3} g \sin 	heta$.
Note: This acceleration is less than $g \sin 	heta$ (acceleration of sliding) because some of the potential energy is converted into rotational kinetic energy.

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pure rolling acceleration

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