Question

A cylinder of mass $10kg$ and radius $15cm$ is rolling perfectly on a plane of inclinato of ${30}^o$. The coefficient of static friction ${\mu }_k=0.25$. If the inclination $\theta$ of the plane is increased, at what value of $\theta$ does the cylinder begin to skid, and not roll perfectly?

• A. ${37}^o$
• B. ${53}^o$
• C. ${90}^o$
• D. $0^o$

Answer 1

The correct option is A ${37}^o$

Given: a cylinder of mass 10kg and radius 15cm is rolling perfectly on a plane of inclination of ${30}^o$. The coefficient of static friction ${\mu }_k=0.25$

To find: if the inclination $\theta$ of the plane is increased, at what value of $\theta$ does the cylinder begin to skid, and not roll perfectly

Solution:

As per the given condition,

$m=10kg$

$r=0.15m$

${\mu }_k=0.25$

$\theta ={30}^{\circ }$

$I=\frac{1}{2}m{r}^2$

Acceleration of the cylinder is given as:

$a=\frac{mg\sin \theta }{m+\frac{I}{r^2}}⟹a=\frac{mg\sin \theta }{m+\frac{\frac{1}{2}m{r}^2}{r^2}}⟹a=\frac{mg\sin \theta }{m+\frac{1}{2}m}⟹a=\frac{mg\sin \theta }{\frac{3}{2}m}⟹a=\frac{2}{3}g\sin {30}^{\circ }⟹a=\frac{2}{3}\times 9.8\times \frac{1}{2}⟹a=3.27m/s^2$

For rolling without skid, we have the relation

$\mu =\frac{1}{3}\tan \theta ⟹\tan \theta =3\mu =3\times 0.5⟹\theta ={\tan }^{-1}\left(0.75\right)={36.87}^{\circ }\approx {37}^{\circ }$

is the value of $\theta$ for which the cylinder begin to skid, and not roll perfectly