Why does the pen does not move straight?

If i put a pen on a table in its horizontal position and then i try to move it horizontally by giving it a small push, so that it would fall off a table, i expect it to move horizontally but my pen ( and all other pens too! ) moves diagonally when it starts moving down the table!When i remove the notebook , the pen moves like,its shown in the picture ( if i keep it horizontally also, it gives the same result)- Why does this happen? Why does it not move horizontally ?

• Some counter questions.. Where do you apply the force? Is the chassis's structure tapered (an exaggeration would be cone)? Other reasons could be non uniform horizontal mass distribution. – physicist Jan 4 '16 at 7:34
• Do you know motion Of centre of mass which are uniform or nonuniform – Archis Welankar Jan 4 '16 at 7:37
• Your picture is not large enough or clear enough to see what the pen looks like and your statements are incomplete (and possibly wrong :-) ). You say your pen is a cylinder. Does it look like this with r constant at ALL points along its body, or at some points is R different than at others. If r is larger at and point to the left of the right hand end than it is AT the right hand end then you would expect it to move as you have shown. – Russell McMahon Jan 4 '16 at 8:06
• @RussellMcMahon My pen is cylindrical if you take out the cone part! – Aaryan Dewan Jan 4 '16 at 9:05
• Aaryan - IF it has anything other than a purely cylindrical form when it is rolled then it must roll in a "non-straight" path - for the reasons given by Pela. – Russell McMahon Jan 4 '16 at 12:26

Because your pen is not a cylinder, but a portion of a cone. Since it is also rigid, both ends have to complete one cycle of rolling simultaneously. This means that for each cycle, if the narrow and thick ends are separated by the pen's length $L$ and have radii $r_1$ and $r_2$, respectively, they roll $2\pi r_1$ and $2\pi r_2$, respectively. The only way this can be accomplished (without skidding) is for the ends to roll in a cicrcular motion around a common center.
If I haven't miscalculated, then the distance $d$ from the "small" end to this center is $$d = \frac{L}{r_2/r_1–1}$$. 