Impulse applied to a disk at the end of a pole

[srecko97]

Homework Statement

May I ask you something about a task from the last years test at faculty of mathematics and physics, University of Ljubljana, Slovenia ...
There is a disk at the end of a pole. We cause some F dt tangentally on the disk, which causes the change of momentum. I need to calculate ω for pole and for disk too. I have given mass of a pole and mass of a disk and also length and radius.
http://Fdt causes the change of angular inertia ΔG ⋅r = J ω Is it Ok if i first calculate the ω of disk and than put into equation that Jdiskωdisk = J poleωpole or is it Jdiskωdisk+J poleωpole=ΔG ⋅r and the second equation Jdiskωdisk2 +J poleωpole2=mv2 but i do not know v ... I do not have answers so i am so sceptical about my calculation "os A" means axis in point A I would be really greatful if you answer me. If you can't, it is ok too.[/PLAIN]

Homework Equations

ΔG ⋅r = J ω

The Attempt at a Solution

Fdt causes the change of angular inertia
ΔG ⋅r = J ω

Is it Ok if i first calculate the ω of disk and than put into equation that Jdiskωdisk = J _poleω_pole

or is it J_diskω_disk+J _poleω_pole=ΔG ⋅(r+L)
and the second equation J_diskω_disk² +J _poleω_pole²=mv²

but i do not know v ...
I do not have answers so i am so sceptical about my calculation
"os A" means axis in point A

I would be really greatful if you answer me. If you can't, it is ok too.

 

[mfb]

Is the disk fixed to the pole?
Is the end of the pole fixed?

 

[srecko97]

the end of the pole can rotate around axis "os A" in the picture. The disk can rotate around the end of the pole

 

[haruspex]

Jdiskωdisk = J poleωpole

On what basis?

Jdiskωdisk+J poleωpole=ΔG ⋅(r+L)

Yes, but be careful. Angular momentum is always in reference to some axis. Make sure you are using the same axis throughout the equation.

Jdiskωdisk2 +J poleωpole2=mv2

Again, on what basis? Why would the imparted energy be shared in exactly this way?

 

[srecko97]

there is no other energy after except 2 rotational energies. At the biginning it is only mv²/2

 

[haruspex]

there is no other energy after except 2 rotational energies. At the biginning it is only mv²/2

At the beginning there is no energy. Then there is an impulse, and suddenly there are energies. You can count these in various different ways, e.g. the disc and the rod each having rotational KE about their centres, and also having the linear KEs of their mass centres; or you could take the rod as just having rotational KE about one end, etc. But I see no basis for writing any kind of energy balance equation here.