Stick thrown in air physics problem

[ritwik06]

Homework Statement

A student throws a stick of length L up in the air. At the moment the stick leaves his hand, the speed of the stick's end is zero. The stick completes N turns as its caught by the student at the initial release point. Show that the height to which the centre of mass of the stick rose is $$\pi NL/4$$

The Attempt at a Solution

All that I can make out of the problem is that t if the velocity of the centre of mass of the stick was v then by energy conservation the centre of mass of the stick would have risen by $$v^{2}/2g$$

I cannot make out what can I do with that N and all. Help me please.

 

[tiny-tim]

A student throws a stick of length L up in the air. At the moment the stick leaves his hand, the speed of the stick's end is zero. The stick completes N turns as its caught by the student at the initial release point. Show that the height to which the centre of mass of the stick rose is $$\pi NL/4$$

Hi ritwik06!

The way to approach a question like this is to start:

"Assume the velocity of the two ends are v and 0, vertically …",

and then calculate both the maximum height and the number of turns.

 

[ritwik06]

Hi ritwik06!

The way to approach a question like this is to start:

"Assume the velocity of the two ends are v and 0, vertically …",

and then calculate both the maximum height and the number of turns.

I tried this.
Assuming the velocity of the two ends are v and 0
Using conservation of angular momentum:
mvL/2=I $$\omega$$

then 0.5 I $$\omega^{2}$$=mgh
?

 

[Redbelly98]

I think you'll also need the time taken to reach the maximum height. And from that figure out the time taken to come back down as well.

 

[tiny-tim]

exactly what it says on the box! …

I tried this.
Assuming the velocity of the two ends are v and 0
Using conservation of angular momentum:
mvL/2=I $$\omega$$

then 0.5 I $$\omega^{2}$$=mgh
?

Hi ritwik06!

(have an omega: ω and a pi: π and a squared: ² )

uh-uh … "conservation of angular momentum" means exactly what it says on the box! …

the angular momentum will carry on the same, no matter where the stick goes (unless there's a torque, which there isn't … the weight acts through the c.o.m) …

so this is two separate problems …

for the height you can ignore the angular momentum, and for the angular momentum you can ignore the height.