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## Homework Statement

A disk [m=0.1 Kg; R=0.1 m] rotates about its center of mass [w=40 rad/s], on a smooth floor. A bar [m=0.1 Kg; lenght=R=0.1 m] moves on the floor with a speed Vb=4 m/s. At one point, the bar hits disk's edge in an inelastic collision, and they start rotating together.

A)[Fixed axis passing through disk's center of mass and perperdicular to the floor]

-Find the new angular speed [w'] of the system disk+bar]

B)[System can translate after the collsion]

-Find the new angular speed [w''] of the system disk+bar];

-Find the CM's speed

## Homework Equations

## The Attempt at a Solution

A)

When the bar hits disk's edge, same forces on the axis through the disk's CM prevent the system start rotating about the new CM. The momentum of these forces is zero because they have the same direction of the vector

**r**, so the angular momentum (P) is costant:

[I choose disk's CM as pole for the momentum, so I'll calculate the moment of inertia from that pole]

-P1=I1*w1=1/2*m*R^2*w1

-P2=I2*w2=(I(disk)+I(bar))*w2=(1/2*m*R^2+1/12*m*R^2+m*(R+1/2R)^2)w2=17/6*m*R^2*w2

-P1=P2

- 1/2*m*R^2*w1=17/6*m*R^2*w2--------->w2=0.5*w1*6/17--->w2=0.5*40*6/17=7.06 rad/s

[Right solution]

B) In this case the rotation axis isn't fixed so, after the collision, the system starts translating and rotating about the new CM. Also, since the axis is not fixed, there are no external forces, so the traslational momentum [Q] is costant.

**New CM's coordinates:**

-DISK:

SUM(m(i)*x(i)^2)/m=0------>SUM(m(i)*x(i)^2)=0

-BAR:

SUM(m(i)*x(i)^2)/m=3/2*R------>SUM(m(i)*x(i)^2)=3/2*R*m

-DISK+BAR

Xcm=SUM(m(i)*x(i)^2)/2m=(0+3/2*R*m)/2m=3/4*R

[Ycm=0]

CM=(3/4*R;0);

**CM's speed**

Q=costant

Qi=m*Vb

Qf=Mcm*Vcm=2*m*Vcm

Qi=Qf

m*Vb=2*m*Vcm---->Vcm=Vb/2---->Vcm=2m/s [Right solution]

**Angular speed of the system (w'')**

Initially I thought that

**w''**was the same of

**w'**, which I found in the previous point.

Also in this case the momentum of external forces is zero [there are no external forces bisides weight, actually] so P is costant. If I choose the same pole for the angular momentum I find the previous equation. However book's solution is different, since it calculate the angular momentum about the CM.

I tryed to motivate this solution in this way: in general, the angular momentum about a generic pole

**O**(in this case disk's CM) can be written and the sum of two terms: the angular momentum about the CM's of the system and the term

**r**^(M*

**V**cm) wher

**V**cm is the CM's speed aboout the pole

**O.**So:

P

**o**=P

**cm**+

**r**^(M*

**V**cm)

In this case

**r**parallel to

**V**cm so that

**r**^(M*

**V**cm)=0 and P

**o**=P

**cm**

**My finally question:**

In point A, I tried to calcute P using P

**o**=P

**cm**+

**r**^(M*

**V**cm), and I found the same result of calculating the moment of inertia about the disk's CM and using P=I*w both before and after the collision.

In point B P=I*w can be used only before the collision.

So, I don't understand why it can't be used also after the collision and why in the case A

P

**o**=P

**cm**+

**r**^(M*

**V**cm) and P

**o=**I

**o***w provides the same result, so when I can use both of theme