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  • A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30^ o(a).

Q7.31     A cylinder of mass 10kg  and radius 15cm  is rolling perfectly on a plane of inclination 30^{0}. The coefficient
              of static friction \mu _{s}=0.25. How much force of friction is acting on the cylinder?

Answers (1)

best_answer

Consider the following figure :

                                      Rotational motion, 20214

The moment of inertia of the cylinder is :

                                                              I\ =\ \frac{1}{2}mr^2

Thus acceleration is given by :

                                                              a\ =\ \frac{mg \sin \Theta}{m\ +\ \frac{1}{2}\times \frac{ mr^2}{r^2}}

or                                                                 =\ \frac{2}{3}\ g \sin 30^{\circ}\ =\ 3.27\ m/s^2

Now using Newton's law of motion :

                                                        F\ =\ ma

or                                                 mg \sin 30^{\circ}\ -\ f\ =\ ma

or                                                        f\ =\ mg \sin 30^{\circ}\ -\ ma

or                                                              =\ 49\ -\ 32.7\ =\ 16.3\ N

Hence frictional force is 16.3 N.

Posted by

Devendra Khairwa

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