Mouse falls on rotating disc, find the work it needs to go to the center of it

[tonit]

Homework Statement

mass of the mouse = 0.05 kg
disc's radius = 0.2m
disc's angular speed = 33 rev / min
assume that the angular speed ω doesn't change

Homework Equations

tangential speed = ω * r

The Attempt at a Solution

well, what i did was: drew the vectors, one was the tangential speed of the circle, another one was the radius which is the displacement of the mouse.
the vector connecting the ends of both these vectors is supposed to be the speed of the mouse

since tangential speed and radius are perpendicular, then with the Pythagoras's theorem the speed of the mouse is v^2 = vt^2 + r^2

then I calculated the work done by the change of energy
ΔE = [((1/2)mv^2 ) + Iω^2] - Iω^2 (since the omega doesn't change)

am I right, if not any suggestions. thanks

 

[Simon Bridge]

How would you figure out the work for the mouse to climb to a height h? Would you factor in the speed that the mouse climbs the hill?
In the derived equation "v^2=vt^2+r^2", the LHS has units of speed while the RHS has units of distance (sort of) so there is something wrong with your reasoning here.
What usually helps is a clear description of the important part of the physics - eg:

The mouse starts out going in a big circle and finishes rotating on the spot.
What is the difference in energy of these two states?

 

[tonit]

conservation of angular momentum:

[itex]I[/itex]_i[itex]\omega[/itex]_i [itex]=[/itex] [itex]I[/itex]_f[itex]\omega[/itex]_f [itex](1)[/itex]

rotational intertia of mouse is

[itex]I[/itex]_m [itex]=[/itex] [itex]mr^2[/itex] [itex](2)[/itex]

since at the end the mouse rotates at place, it shouldn't contribute anything so

[itex](I[/itex]_disc+[itex]I[/itex]_mouse[itex])\omega[/itex] = [itex]I[/itex]_disc[itex]\omega[/itex]

can anyone help me I'm stuck here