# The height traveled by the mass center of a thrown rod

I'm learning Newtonian mechanics and want to understand the following example and the methods I can use for it.

Suppose I throw a rod in the air with length $L$. The time I'm letting go of the rod the closest edge of the rod to my hand has zero velocity. If the rod does $N$ rotation until I catch it again at the same point I threw it, show that the height its mass centre traveled in the air is $$h=\frac{\pi NL}{4}.$$

Maybe I can try Energy conservation. But not sure how to. Or just use Newton's laws but I don't know how because how it is possible for the starting velocity to be zero since I throw something in the air like a projectile?

• Interesting question. You have some ideas how you might solve it - so why don't you try and show us your calculation? I think you will have to use the eqns for projectile motion to relate time of flight and max. height to initial velocity u of the CM. You can also relate u to the constant rate of rotation - ie the number of rotations during the flight time. (I haven't worked through the problem myself yet, but I think this will work.) – sammy gerbil Sep 13 '16 at 23:34
• The second sentence in your question is not much clear. Can you elaborate more? – Kosala Sep 15 '16 at 16:10
• @Kosala i mean my had is moving i dont drop the object i throw it how can it have zero velocity? – Jam Sep 15 '16 at 17:04
• The end of the rod which is in your hand has zero velocity, but the centre of mass of the rod does not have zero velocity. – sammy gerbil Oct 1 '16 at 14:46

## 1 Answer

It sounds like you start with a horizontal rod with one end in your hand at zero speed, and the center of mass moving upward at some speed u. You can just call it u, it will be fine. Now find the time of flight, since you will find the number of rotations from the angular velocity and the time of flight. The easiest way to get time of flight is to notice it will be twice the time to decelerate from speed u to speed 0, so t = 2u/g. The angular velocity is u/L. Put that together and get the number of rotations. then see the u^2 in there, and use a kinematics formula to replace u^2 with 2gh. Interestingly, g cancels out, so the answer is the same on the Moon.