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  • A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s(a).

Q7.21     A solid cylinder rolls up an inclined plane of angle of inclination 30^{0}. At the bottom
              of the inclined plane the centre of mass of the cylinder has a speed of 5m/s .

            (a) How far will the cylinder go up the plane?

Answers (1)

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The rotational energy is converted into the translational energy.(Law of conservation of energy)

                                           \frac{1}{2}I \omega ^2\ =\ \frac{1}{2}mv^2\ =\ mgh

Since the moment of inertia for the cylinder is  :

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

Putting the value of MI and v = wr in the above equation, we get : 

                                                  \frac{1}{4}v^2\ +\ \frac{1}{2}v^2\ =\ gh

or                                                h\ =\ \frac{3}{4}\times \frac{v^2}{g}

or                                                       =\ \frac{3}{4}\times \frac{5\times 5}{g}\ =\ 1.91\ m

Now using the geometry of the cylinder we can write :

                                                    \sin \Theta \ =\ \frac{h}{d}

or                                                 \sin 30^{\circ}\ =\ \frac{h}{d}

or                                                  d\ =\ \frac{1.91}{0.5}

                                                           =\ 3.82\ m

Thus cylinder will travel up to 3.82 m up the incline.

Posted by

Devendra Khairwa

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