Q. A uniform solid cylinder of mass $m$ and radius $r$ rolls along an inclined rough plane of inclination $45^\circ$. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be
A) $\frac1{\sqrt2}g$
B) $\frac1{3\sqrt2}g$
C) $\frac{\sqrt2}3g$
D) $\sqrt2g$
Explanation
[Solution by Dr. Abubacker] For rolling without slipping, $a=\frac{g\sin\theta}{1+k^2/r^2}=\frac{g\sin45^\circ}{1+\frac12}=\frac23\cdot\frac{g}{\sqrt2}=\frac{\sqrt2}3g$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.