Find the final angular speed of the target

[pentazoid]

Homework Statement

A fairgorund target consists of a uniform circular disk of mass M and radius a that can turn freely about its diamerter which is fixed in a vertical position. Iniitally the target is at rest . A bullet of mass m is moving with speed u along a horizontal line at right angles to the target . The bullet embeds itself in the target at a point distance from the rotation axis . Find the final angular speed of the target.(the moment of inertica of the disk about its roation axis is Ma^2/4

Homework Equations

Equations for rotational energy

The Attempt at a Solution

T_trans., initial + T_rot., initial, + V_initial=T_rot, final . There is no translational energy now that the bullet is embedded into the block.

1/2*m*u^2++0+0=1/2(Ma^2/4
m*b^2/4)*omega^2? Is my equation correct? There is no potential energy.

 

[tiny-tim]

inelastic collision

A fairgorund target consists of a uniform circular disk of mass M and radius a that can turn freely about its diamerter which is fixed in a vertical position. Iniitally the target is at rest . A bullet of mass m is moving with speed u along a horizontal line at right angles to the target . The bullet embeds itself in the target at a point distance from the rotation axis . Find the final angular speed of the target.(the moment of inertica of the disk about its roation axis is Ma^2/4
…
Equations for rotational energy

Hi pentazoid!

Nooo … energy is not usually conserved in collisions …

and certainly not when one body embeds itself in the other!

Hint: momentum is always conserved in collisions …

and that includes angular momentum.

 

[pentazoid]

Hi pentazoid!

Nooo … energy is not usually conserved in collisions …

and certainly not when one body embeds itself in the other!

Hint: momentum is always conserved in collisions …

and that includes angular momentum.

yes I know energy isn't conserved because the collision is not elastic since the bullet becomes embedded into the block.

L_initial=L_final

m*v_bullet+0=(M+m)v_new?

and then I can plug my v_new into my energy equation?

 

[tiny-tim]

m*v_bullet+0=(M+m)v_new?

No, that's linear momentum …

you need angular momentum.

(and what energy equation? )

 

[pentazoid]

No, that's linear momentum …

you need angular momentum.

(and what energy equation? )

right , how silly of me.

m*v_bullet*a=(m+M)*v_final*(a+b)? Would I then plug v_final into my energy equation.

a being the horizontal distance and a+b being the horizontal distance plus the radius of the disk?