# Find the final angular speed of the target

• pentazoid
In summary, a bullet of mass m moves with speed u along a horizontal line at right angles to a rotating fairgorund target. The target has rotational energy, and when the bullet enters the target it embeds itself at a point distance from the rotation axis. The final angular speed of the target is found by multiplying the moment of inertia of the target about its rotation axis by 1/2.
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.

inelastic collision

pentazoid said:
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.

tiny-tim said:
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?

pentazoid said:
m*v_bullet+0=(M+m)v_new?

No, that's linear momentum …

you need angular momentum.

(and what energy equation? )

tiny-tim said:
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?

## 1. What is the formula for finding the final angular speed of the target?

The formula for finding the final angular speed of the target is ωf = ωi + αt, where ωf is the final angular speed, ωi is the initial angular speed, α is the angular acceleration, and t is the time.

## 2. How is angular acceleration calculated?

Angular acceleration is calculated by dividing the change in angular velocity by the change in time. The formula for angular acceleration is α = (ωf - ωi) / t.

## 3. What is the unit of measurement for angular speed and acceleration?

The unit of measurement for angular speed is radians per second (rad/s), while the unit for angular acceleration is radians per second squared (rad/s^2).

## 4. Can you use the same formula to find the final angular speed of any rotating object?

Yes, the formula for finding the final angular speed of the target can be used for any rotating object, as long as the initial angular speed and acceleration are known.

## 5. How does the final angular speed of the target affect its overall rotation?

The final angular speed of the target determines how fast it will rotate. A higher final angular speed means the target will rotate faster, while a lower final angular speed means it will rotate slower. However, the direction of rotation is also determined by the sign of the final angular speed.

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