Question 773: Centre of Mass and Rotational Motion

Question

A solid cylinder of mass m and radius $r$ is rolling on rough inclined plane of inclination $	heta$. The coefficient of friction between the cylinder and incline is $\mu$. then

Accepted Answer

$$ 	ext { From the diagram, } $$
We observe $f$ opposes translational motion and (option d) supports rotational motion.
$$ \begin{aligned} m g \sin 	heta-f & =m a \ f \mathrm{R} & =\mathrm{I} \alpha=\frac{1}{2} \mathrm{MR}^2\left(\frac{a}{\mathrm{R}}ight)=\frac{1}{2} m a \mathrm{R} \end{aligned} $$
$$ \begin{array}{rlrl} m g \sin 	heta-f & =2 f & (m a & =2 f) \ m g \sin 	heta & =3 f & {[a} & =r \alpha \ \alpha & \left.=\frac{a}{r}ight] & \end{array} $$
$$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, f=\frac{1}{3} m g \sin 	heta \ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  f \propto \sin 	heta \end{aligned} $$
Thus, if we reduce $	heta, f$ will be decreased Option (c) is correct.

Answer

frictional force is always $\mu \mathrm{mg} \cos 	heta$.

Answer

friction is a dissipative force.

Answer

by decreasing $	heta$, frictional force decreases.

Answer

friction opposes translation and supports rotation.

Centre of Mass and Rotational Motion

A solid cylinder of mass m and radius r is rolling on rough inclined plane of inclination \theta. The coefficient of friction between the cylinder and incline is \mu. then

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