# 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?

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.

## Related Chapters

### Preparation Products

##### Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
##### Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Test Series NEET Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-